{"id":14449,"date":"2018-09-27T18:08:55","date_gmt":"2018-09-27T18:08:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/polar-form-of-complex-numbers\/"},"modified":"2025-04-16T13:08:21","modified_gmt":"2025-04-16T13:08:21","slug":"polar-form-of-complex-numbers","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/polar-form-of-complex-numbers\/","title":{"raw":"Polar Form of Complex Numbers","rendered":"Polar Form of Complex Numbers"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Plot complex numbers in the complex plane.<\/li>\r\n \t<li>Find the absolute value of a complex number.<\/li>\r\n \t<li>Write complex numbers in polar form.<\/li>\r\n \t<li>Convert a complex number from polar to rectangular form.<\/li>\r\n \t<li>Find products of complex numbers in polar form.<\/li>\r\n \t<li>Find quotients of complex numbers in polar form.<\/li>\r\n \t<li>Find powers and roots of complex numbers in polar form.<\/li>\r\n<\/ul>\r\n<\/div>\r\n\"God made the integers; all else is the work of man.\" This rather famous quote by nineteenth-century German mathematician Leopold <strong>Kronecker<\/strong> sets the stage for this section on the polar form of a complex number. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as <strong>Pythagoras<\/strong>, <strong>Descartes<\/strong>, De Moivre, <strong>Euler<\/strong>, <strong>Gauss<\/strong>, and others. Complex numbers answered questions that for centuries had puzzled the greatest minds in science.\r\n\r\nWe first encountered complex numbers in Precalculus I. In this section, we will focus on the mechanics of working with complex numbers: translation of complex numbers from polar form to rectangular form and vice versa, interpretation of complex numbers in the scheme of applications, and application of De Moivre\u2019s Theorem.\r\n<h2>Plotting Complex Numbers in the Complex Plane<\/h2>\r\nPlotting a <strong>complex number<\/strong> [latex]a+bi[\/latex] is similar to plotting a real number, except that the horizontal axis represents the real part of the number, [latex]a[\/latex], and the vertical axis represents the imaginary part of the number, [latex]bi[\/latex].\r\n<div class=\"textbox\">\r\n<h3>How To: Given a complex number [latex]a+bi[\/latex], plot it in the complex plane.<\/h3>\r\n<ol>\r\n \t<li>Label the horizontal axis as the <em>real<\/em> axis and the vertical axis as the <em>imaginary axis.<\/em><\/li>\r\n \t<li>Plot the point in the complex plane by moving [latex]a[\/latex] units in the horizontal direction and [latex]b[\/latex] units in the vertical direction.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Plotting a Complex Number in the Complex Plane<\/h3>\r\nPlot the complex number [latex]2 - 3i[\/latex] in the <strong>complex plane<\/strong>.\r\n\r\n[reveal-answer q=\"821515\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"821515\"]\r\n\r\nFrom the origin, move two units in the positive horizontal direction and three units in the negative vertical direction.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180842\/CNX_Precalc_Figure_08_05_0012.jpg\" alt=\"Plot of 2-3i in the complex plane (2 along the real axis, -3 along the imaginary axis).\" width=\"487\" height=\"331\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nPlot the point [latex]1+5i[\/latex] in the complex plane.\r\n\r\n[reveal-answer q=\"522113\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"522113\"]\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180857\/CNX_Precalc_Figure_08_05_0022.jpg\" alt=\"Plot of 1+5i in the complex plane (1 along the real axis, 5 along the imaginary axis).\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Finding the Absolute Value of a Complex Number<\/h2>\r\nThe first step toward working with a complex number in <strong>polar form<\/strong> is to find the absolute value. The absolute value of a complex number is the same as its <strong>magnitude<\/strong>, or [latex]|z|[\/latex]. It measures the distance from the origin to a point in the plane. For example, the graph of [latex]z=2+4i[\/latex], in Figure 2, shows [latex]|z|[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180845\/CNX_Precalc_Figure_08_05_0032.jpg\" alt=\"Plot of 2+4i in the complex plane and its magnitude, |z| = rad 20.\" width=\"487\" height=\"368\" \/> <b>Figure 2<\/b>[\/caption]\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Absolute Value of a Complex Number<\/h3>\r\nGiven [latex]z=x+yi[\/latex], a complex number, the absolute value of [latex]z[\/latex] is defined as\r\n<p style=\"text-align: center;\">[latex]|z|=\\sqrt{{x}^{2}+{y}^{2}}[\/latex]<\/p>\r\nIt is the distance from the origin to the point [latex]\\left(x,y\\right)[\/latex].\r\n\r\nNotice that the absolute value of a real number gives the distance of the number from 0, while the absolute value of a complex number gives the distance of the number from the origin, [latex]\\left(0,\\text{ }0\\right)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Finding the Absolute Value of a Complex Number with a Radical<\/h3>\r\nFind the absolute value of [latex]z=\\sqrt{5}-i[\/latex].\r\n\r\n[reveal-answer q=\"219333\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"219333\"]\r\n\r\nUsing the formula, we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;|z|=\\sqrt{{x}^{2}+{y}^{2}}\\\\ &amp;|z|=\\sqrt{{\\left(\\sqrt{5}\\right)}^{2}+{\\left(-1\\right)}^{2}} \\\\ &amp;|z|=\\sqrt{5+1} \\\\ &amp;|z|=\\sqrt{6} \\end{align}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180847\/CNX_Precalc_Figure_08_05_0042.jpg\" alt=\"Plot of z=(rad5 - i) in the complex plane and its magnitude rad6.\" width=\"487\" height=\"331\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the absolute value of the complex number [latex]z=12 - 5i[\/latex].\r\n\r\n[reveal-answer q=\"514025\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"514025\"]\r\n\r\n13\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Finding the Absolute Value of a Complex Number<\/h3>\r\nGiven [latex]z=3 - 4i[\/latex], find [latex]|z|[\/latex].\r\n\r\n[reveal-answer q=\"748063\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"748063\"]\r\n\r\nUsing the formula, we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;|z|=\\sqrt{{x}^{2}+{y}^{2}} \\\\ &amp;|z|=\\sqrt{{\\left(3\\right)}^{2}+{\\left(-4\\right)}^{2}} \\\\ &amp;|z|=\\sqrt{9+16} \\\\ &amp;|z|=\\sqrt{25}\\\\ &amp;|z|=5 \\end{align}[\/latex]<\/p>\r\nThe absolute value [latex]z[\/latex] is 5.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180849\/CNX_Precalc_Figure_08_05_0052-1.jpg\" alt=\"Plot of (3-4i) in the complex plane and its magnitude |z| =5.\" width=\"487\" height=\"331\" \/> <b>Figure 4<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]z=1 - 7i[\/latex], find [latex]|z|[\/latex].\r\n\r\n[reveal-answer q=\"473571\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"473571\"]\r\n\r\n[latex]|z|=\\sqrt{50}=5\\sqrt{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173801[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>\u00a0Writing Complex Numbers in Polar Form<\/h2>\r\nThe <strong>polar form of a complex number<\/strong> expresses a number in terms of an angle [latex]\\theta [\/latex] and its distance from the origin [latex]r[\/latex]. Given a complex number in <strong>rectangular form<\/strong> expressed as [latex]z=x+yi[\/latex], we use the same conversion formulas as we do to write the number in trigonometric form:\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}x=r\\cos \\theta \\\\ y=r\\sin \\theta \\\\ r=\\sqrt{{x}^{2}+{y}^{2}} \\end{gathered}[\/latex]<\/div>\r\nWe review these relationships in Figure 5.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180851\/CNX_Precalc_Figure_08_05_0062.jpg\" alt=\"Triangle plotted in the complex plane (x axis is real, y axis is imaginary). Base is along the x\/real axis, height is some y\/imaginary value in Q 1, and hypotenuse r extends from origin to that point (x+yi) in Q 1. The angle at the origin is theta. There is an arc going through (x+yi).\" width=\"487\" height=\"331\" \/> <b>Figure 5<\/b>[\/caption]\r\n\r\nWe use the term <strong>modulus<\/strong> to represent the absolute value of a complex number, or the distance from the origin to the point [latex]\\left(x,y\\right)[\/latex]. The modulus, then, is the same as [latex]r[\/latex], the radius in polar form. We use [latex]\\theta [\/latex] to indicate the angle of direction (just as with polar coordinates). Substituting, we have\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;z=x+yi \\\\ &amp;z=r\\cos \\theta +\\left(r\\sin \\theta \\right)i \\\\ &amp;z=r\\left(\\cos \\theta +i\\sin \\theta \\right) \\end{align}[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Polar Form of a Complex Number<\/h3>\r\nWriting a complex number in polar form involves the following conversion formulas:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} x=r\\cos \\theta \\\\ y=r\\sin \\theta \\\\ r=\\sqrt{{x}^{2}+{y}^{2}} \\end{gathered}[\/latex]<\/p>\r\nMaking a direct substitution, we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;z=x+yi \\\\ &amp;z=\\left(r\\cos \\theta \\right)+i\\left(r\\sin \\theta \\right) \\\\ &amp;z=r\\left(\\cos \\theta +i\\sin \\theta \\right) \\end{align}[\/latex]<\/p>\r\nwhere [latex]r[\/latex] is the <strong>modulus<\/strong> and [latex]\\theta [\/latex] is the <strong>argument<\/strong>. We often use the abbreviation [latex]r\\text{cis}\\theta [\/latex] to represent [latex]r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Expressing a Complex Number Using Polar Coordinates<\/h3>\r\nExpress the complex number [latex]4i[\/latex] using polar coordinates.\r\n\r\n[reveal-answer q=\"3488\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"3488\"]\r\n\r\nOn the complex plane, the number [latex]z=4i[\/latex] is the same as [latex]z=0+4i[\/latex]. Writing it in polar form, we have to calculate [latex]r[\/latex] first.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;r=\\sqrt{{x}^{2}+{y}^{2}} \\\\ &amp;r=\\sqrt{{0}^{2}+{4}^{2}} \\\\ &amp;r=\\sqrt{16} \\\\ &amp;r=4 \\end{align}[\/latex]<\/p>\r\nNext, we look at [latex]x[\/latex]. If [latex]x=r\\cos \\theta [\/latex], and [latex]x=0[\/latex], then [latex]\\theta =\\frac{\\pi }{2}[\/latex]. In polar coordinates, the complex number [latex]z=0+4i[\/latex] can be written as [latex]z=4\\left(\\cos \\left(\\frac{\\pi }{2}\\right)+i\\sin \\left(\\frac{\\pi }{2}\\right)\\right)[\/latex] or [latex]4\\text{cis}\\left(\\frac{\\pi }{2}\\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180854\/CNX_Precalc_Figure_08_05_0072.jpg\" alt=\"Plot of z=4i in the complex plane, also shows that the in polar coordinate it would be (4,pi\/2).\" width=\"487\" height=\"294\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nExpress [latex]z=3i[\/latex] as [latex]r\\text{cis}\\theta [\/latex] in polar form.\r\n\r\n[reveal-answer q=\"364839\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"364839\"]\r\n\r\n[latex]z=3\\left(\\cos \\left(\\frac{\\pi }{2}\\right)+i\\sin \\left(\\frac{\\pi }{2}\\right)\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Finding the Polar Form of a Complex Number<\/h3>\r\nFind the polar form of [latex]-4+4i[\/latex].\r\n\r\n[reveal-answer q=\"871564\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"871564\"]\r\n\r\nFirst, find the value of [latex]r[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;r=\\sqrt{{x}^{2}+{y}^{2}} \\\\ &amp;r=\\sqrt{{\\left(-4\\right)}^{2}+\\left({4}^{2}\\right)} \\\\ &amp;r=\\sqrt{32} \\\\ &amp;r=4\\sqrt{2} \\end{align}[\/latex]<\/p>\r\nFind the angle [latex]\\theta [\/latex] using the formula:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\cos \\theta =\\frac{x}{r} \\\\ &amp;\\cos \\theta =\\frac{-4}{4\\sqrt{2}} \\\\ &amp;\\cos \\theta =-\\frac{1}{\\sqrt{2}} \\\\ &amp;\\theta ={\\cos }^{-1}\\left(-\\frac{1}{\\sqrt{2}}\\right)=\\frac{3\\pi }{4} \\end{align}[\/latex]<\/p>\r\nThus, the solution is [latex]4\\sqrt{2}\\text{cis}\\left(\\frac{3\\pi }{4}\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWrite [latex]z=\\sqrt{3}+i[\/latex] in polar form.\r\n\r\n[reveal-answer q=\"164997\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"164997\"]\r\n\r\n[latex]z=2\\left(\\cos \\left(\\frac{\\pi }{6}\\right)+i\\sin \\left(\\frac{\\pi }{6}\\right)\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173845[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>\u00a0Converting a Complex Number from Polar to Rectangular Form<\/h2>\r\nConverting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. In other words, given [latex]z=r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex], first evaluate the trigonometric functions [latex]\\cos \\theta [\/latex] and [latex]\\sin \\theta [\/latex]. Then, multiply through by [latex]r[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Converting from Polar to Rectangular Form<\/h3>\r\nConvert the polar form of the given complex number to rectangular form:\r\n<p style=\"text-align: center;\">[latex]z=12\\left(\\cos \\left(\\frac{\\pi }{6}\\right)+i\\sin \\left(\\frac{\\pi }{6}\\right)\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"145161\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"145161\"]\r\n\r\nWe begin by evaluating the trigonometric expressions.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(\\frac{\\pi }{6}\\right)=\\frac{\\sqrt{3}}{2}\\\\\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}\\end{gathered}[\/latex]<\/p>\r\nAfter substitution, the complex number is\r\n<p style=\"text-align: center;\">[latex]z=12\\left(\\frac{\\sqrt{3}}{2}+\\frac{1}{2}i\\right)[\/latex]<\/p>\r\nWe apply the distributive property:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}z&amp;=12\\left(\\frac{\\sqrt{3}}{2}+\\frac{1}{2}i\\right) \\\\ &amp;=\\left(12\\right)\\frac{\\sqrt{3}}{2}+\\left(12\\right)\\frac{1}{2}i \\\\ &amp;=6\\sqrt{3}+6i \\end{align}[\/latex]<\/p>\r\nThe rectangular form of the given point in complex form is [latex]6\\sqrt{3}+6i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Finding the Rectangular Form of a Complex Number<\/h3>\r\nFind the rectangular form of the complex number given [latex]r=13[\/latex] and [latex]\\tan \\theta =\\frac{5}{12}[\/latex].\r\n\r\n[reveal-answer q=\"11326\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"11326\"]\r\n\r\nIf [latex]\\tan \\theta =\\frac{5}{12}[\/latex], and [latex]\\tan \\theta =\\frac{y}{x}[\/latex], we first determine [latex]r=\\sqrt{{x}^{2}+{y}^{2}}=\\sqrt{{12}^{2}+{5}^{2}}=13\\text{.}[\/latex] We then find [latex]\\cos \\theta =\\frac{x}{r}[\/latex] and [latex]\\sin \\theta =\\frac{y}{r}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}z&amp;=13\\left(\\cos \\theta +i\\sin \\theta \\right) \\\\ &amp;=13\\left(\\frac{12}{13}+\\frac{5}{13}i\\right) \\\\ &amp;=12+5i \\end{align}[\/latex]<\/p>\r\nThe rectangular form of the given number in complex form is [latex]12+5i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nConvert the complex number to rectangular form:\r\n<p style=\"text-align: center;\">[latex]z=4\\left(\\cos \\frac{11\\pi }{6}+i\\sin \\frac{11\\pi }{6}\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[reveal-answer q=\"9609\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"9609\"]<\/p>\r\n<p style=\"text-align: left;\">[latex]z=2\\sqrt{3}-2i[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173840[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>\u00a0Finding Products and Quotients of Complex Numbers in Polar Form<\/h2>\r\nNow that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham <strong>de Moivre<\/strong> (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Products of Complex Numbers in Polar Form<\/h3>\r\nIf [latex]{z}_{1}={r}_{1}\\left(\\cos {\\theta }_{1}+i\\sin {\\theta }_{1}\\right)[\/latex] and [latex]{z}_{2}={r}_{2}\\left(\\cos {\\theta }_{2}+i\\sin {\\theta }_{2}\\right)[\/latex], then the product of these numbers is given as:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{z}_{1}{z}_{2}&amp;={r}_{1}{r}_{2}\\left[\\cos \\left({\\theta }_{1}+{\\theta }_{2}\\right)+i\\sin \\left({\\theta }_{1}+{\\theta }_{2}\\right)\\right] \\\\ {z}_{1}{z}_{2}&amp;={r}_{1}{r}_{2}\\text{cis}\\left({\\theta }_{1}+{\\theta }_{2}\\right) \\end{align}[\/latex]<\/p>\r\nNotice that the product calls for multiplying the moduli and adding the angles.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Finding the Product of Two Complex Numbers in Polar Form<\/h3>\r\nFind the product of [latex]{z}_{1}{z}_{2}[\/latex], given [latex]{z}_{1}=4\\left(\\cos \\left(80^\\circ \\right)+i\\sin \\left(80^\\circ \\right)\\right)[\/latex] and [latex]{z}_{2}=2\\left(\\cos \\left(145^\\circ \\right)+i\\sin \\left(145^\\circ \\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"385516\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"385516\"]\r\n\r\nFollow the formula\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{z}_{1}{z}_{2}=4\\cdot 2\\left[\\cos \\left(80^\\circ +145^\\circ \\right)+i\\sin \\left(80^\\circ +145^\\circ \\right)\\right] \\\\ &amp;{z}_{1}{z}_{2}=8\\left[\\cos \\left(225^\\circ \\right)+i\\sin \\left(225^\\circ \\right)\\right] \\\\ &amp;{z}_{1}{z}_{2}=8\\left[\\cos \\left(\\frac{5\\pi }{4}\\right)+i\\sin \\left(\\frac{5\\pi }{4}\\right)\\right] \\\\ {z}_{1}{z}_{2}=8\\left[-\\frac{\\sqrt{2}}{2}+i\\left(-\\frac{\\sqrt{2}}{2}\\right)\\right] \\\\ &amp;{z}_{1}{z}_{2}=-4\\sqrt{2}-4i\\sqrt{2} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding Quotients of Complex Numbers in Polar Form<\/h2>\r\nThe quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Quotients of Complex Numbers in Polar Form<\/h3>\r\nIf [latex]{z}_{1}={r}_{1}\\left(\\cos {\\theta }_{1}+i\\sin {\\theta }_{1}\\right)[\/latex] and [latex]{z}_{2}={r}_{2}\\left(\\cos {\\theta }_{2}+i\\sin {\\theta }_{2}\\right)[\/latex], then the quotient of these numbers is\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\frac{{z}_{1}}{{z}_{2}}=\\frac{{r}_{1}}{{r}_{2}}\\left[\\cos \\left({\\theta }_{1}-{\\theta }_{2}\\right)+i\\sin \\left({\\theta }_{1}-{\\theta }_{2}\\right)\\right],{z}_{2}\\ne 0\\\\ &amp;\\frac{{z}_{1}}{{z}_{2}}=\\frac{{r}_{1}}{{r}_{2}}\\text{cis}\\left({\\theta }_{1}-{\\theta }_{2}\\right),{z}_{2}\\ne 0\\end{align}[\/latex]<\/p>\r\nNotice that the moduli are divided, and the angles are subtracted.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given two complex numbers in polar form, find the quotient.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Divide [latex]\\frac{{r}_{1}}{{r}_{2}}[\/latex].<\/li>\r\n \t<li>Find [latex]{\\theta }_{1}-{\\theta }_{2}[\/latex].<\/li>\r\n \t<li>Substitute the results into the formula: [latex]z=r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex]. Replace [latex]r[\/latex] with [latex]\\frac{{r}_{1}}{{r}_{2}}[\/latex], and replace [latex]\\theta [\/latex] with [latex]{\\theta }_{1}-{\\theta }_{2}[\/latex].<\/li>\r\n \t<li>Calculate the new trigonometric expressions and multiply through by [latex]r[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Finding the Quotient of Two Complex Numbers<\/h3>\r\nFind the quotient of [latex]{z}_{1}=2\\left(\\cos \\left(213^\\circ \\right)+i\\sin \\left(213^\\circ \\right)\\right)[\/latex] and [latex]{z}_{2}=4\\left(\\cos \\left(33^\\circ \\right)+i\\sin \\left(33^\\circ \\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"244676\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"244676\"]\r\n\r\nUsing the formula, we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\frac{{z}_{1}}{{z}_{2}}=\\frac{2}{4}\\left[\\cos \\left(213^\\circ -33^\\circ \\right)+i\\sin \\left(213^\\circ -33^\\circ \\right)\\right] \\\\ &amp;\\frac{{z}_{1}}{{z}_{2}}=\\frac{1}{2}\\left[\\cos \\left(180^\\circ \\right)+i\\sin \\left(180^\\circ \\right)\\right] \\\\ &amp;\\frac{{z}_{1}}{{z}_{2}}=\\frac{1}{2}\\left[-1+0i\\right] \\\\ &amp;\\frac{{z}_{1}}{{z}_{2}}=-\\frac{1}{2}+0i \\\\ &amp;\\frac{{z}_{1}}{{z}_{2}}=-\\frac{1}{2} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the product and the quotient of [latex]{z}_{1}=2\\sqrt{3}\\left(\\cos \\left(150^\\circ \\right)+i\\sin \\left(150^\\circ \\right)\\right)[\/latex] and [latex]{z}_{2}=2\\left(\\cos \\left(30^\\circ \\right)+i\\sin \\left(30^\\circ \\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"381268\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"381268\"]\r\n\r\n[latex]{z}_{1}{z}_{2}=-4\\sqrt{3};\\frac{{z}_{1}}{{z}_{2}}=-\\frac{\\sqrt{3}}{2}+\\frac{3}{2}i[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173851[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>\u00a0Finding Powers and Roots of Complex Numbers in Polar Form<\/h2>\r\nFinding powers of complex numbers is greatly simplified using <strong>De Moivre\u2019s Theorem<\/strong>. It states that, for a positive integer [latex]n,{z}^{n}[\/latex] is found by raising the modulus to the [latex]n\\text{th}[\/latex] power and multiplying the argument by [latex]n[\/latex]. It is the standard method used in modern mathematics.\r\n<div class=\"textbox\">\r\n<h3>A General Note: De Moivre\u2019s Theorem<\/h3>\r\nIf [latex]z=r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex] is a complex number, then\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{z}^{n}={r}^{n}\\left[\\cos \\left(n\\theta \\right)+i\\sin \\left(n\\theta \\right)\\right]\\\\ &amp;{z}^{n}={r}^{n}\\text{cis}\\left(n\\theta \\right)\\end{align}[\/latex]<\/p>\r\nwhere [latex]n[\/latex]\u00a0is a positive integer.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 10: Evaluating an Expression Using De Moivre\u2019s Theorem<\/h3>\r\nEvaluate the expression [latex]{\\left(1+i\\right)}^{5}[\/latex] using De Moivre\u2019s Theorem.\r\n\r\n[reveal-answer q=\"728625\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"728625\"]\r\n\r\nSince De Moivre\u2019s Theorem applies to complex numbers written in polar form, we must first write [latex]\\left(1+i\\right)[\/latex] in polar form. Let us find [latex]r[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;r=\\sqrt{{x}^{2}+{y}^{2}} \\\\ &amp;r=\\sqrt{{\\left(1\\right)}^{2}+{\\left(1\\right)}^{2}} \\\\ &amp;r=\\sqrt{2} \\end{align}[\/latex]<\/p>\r\nThen we find [latex]\\theta [\/latex]. Using the formula [latex]\\tan \\theta =\\frac{y}{x}[\/latex] gives\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\tan \\theta =\\frac{1}{1} \\\\ &amp;\\tan \\theta =1 \\\\ &amp;\\theta =\\frac{\\pi }{4} \\end{align}[\/latex]<\/p>\r\nUse De Moivre\u2019s Theorem to evaluate the expression.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{\\left(a+bi\\right)}^{n}={r}^{n}\\left[\\cos \\left(n\\theta \\right)+i\\sin \\left(n\\theta \\right)\\right]\\\\ &amp;{\\left(1+i\\right)}^{5}={\\left(\\sqrt{2}\\right)}^{5}\\left[\\cos \\left(5\\cdot \\frac{\\pi }{4}\\right)+i\\sin \\left(5\\cdot \\frac{\\pi }{4}\\right)\\right] \\\\ &amp;{\\left(1+i\\right)}^{5}=4\\sqrt{2}\\left[\\cos \\left(\\frac{5\\pi }{4}\\right)+i\\sin \\left(\\frac{5\\pi }{4}\\right)\\right] \\\\ &amp;{\\left(1+i\\right)}^{5}=4\\sqrt{2}\\left[-\\frac{\\sqrt{2}}{2}+i\\left(-\\frac{\\sqrt{2}}{2}\\right)\\right] \\\\ &amp;{\\left(1+i\\right)}^{5}=-4 - 4i \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding Roots of Complex Numbers in Polar Form<\/h2>\r\nTo find the <strong><em>n<\/em>th root of a complex number<\/strong> in polar form, we use the [latex]n\\text{th}[\/latex] Root Theorem or <strong>De Moivre\u2019s Theorem<\/strong> and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding [latex]n\\text{th}[\/latex] roots of complex numbers in polar form.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The <em>n<\/em>th Root Theorem<\/h3>\r\nTo find the [latex]n\\text{th}[\/latex] root of a complex number in polar form, use the formula given as\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{z}^{\\frac{1}{n}}={r}^{\\frac{1}{n}}\\left[\\cos \\left(\\frac{\\theta }{n}+\\frac{2k\\pi }{n}\\right)+i\\sin \\left(\\frac{\\theta }{n}+\\frac{2k\\pi }{n}\\right)\\right]\\end{align}[\/latex]<\/p>\r\nwhere [latex]k=0,1,2,3,...,n - 1[\/latex]. We add [latex]\\frac{2k\\pi }{n}[\/latex] to [latex]\\frac{\\theta }{n}[\/latex] in order to obtain the periodic roots.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 11: Finding the <em>n<\/em>th Root of a Complex Number<\/h3>\r\nEvaluate the cube roots of [latex]z=8\\left(\\cos \\left(\\frac{2\\pi }{3}\\right)+i\\sin \\left(\\frac{2\\pi }{3}\\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"791675\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"791675\"]\r\n\r\nWe have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{z}^{\\frac{1}{3}}={8}^{\\frac{1}{3}}\\left[\\cos \\left(\\frac{\\frac{2\\pi }{3}}{3}+\\frac{2k\\pi }{3}\\right)+i\\sin \\left(\\frac{\\frac{2\\pi }{3}}{3}+\\frac{2k\\pi }{3}\\right)\\right] \\\\ &amp;{z}^{\\frac{1}{3}}=2\\left[\\cos \\left(\\frac{2\\pi }{9}+\\frac{2k\\pi }{3}\\right)+i\\sin \\left(\\frac{2\\pi }{9}+\\frac{2k\\pi }{3}\\right)\\right] \\end{align}[\/latex]<\/p>\r\nThere will be three roots: [latex]k=0,1,2[\/latex]. When [latex]k=0[\/latex], we have\r\n<p style=\"text-align: center;\">[latex]{z}^{\\frac{1}{3}}=2\\left(\\cos \\left(\\frac{2\\pi }{9}\\right)+i\\sin \\left(\\frac{2\\pi }{9}\\right)\\right)[\/latex]<\/p>\r\nWhen [latex]k=1[\/latex], we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{z}^{\\frac{1}{3}}=2\\left[\\cos \\left(\\frac{2\\pi }{9}+\\frac{6\\pi }{9}\\right)+i\\sin \\left(\\frac{2\\pi }{9}+\\frac{6\\pi }{9}\\right)\\right] &amp;&amp; \\text{ Add }\\frac{2\\left(1\\right)\\pi }{3}\\text{ to each angle.} \\\\ &amp;{z}^{\\frac{1}{3}}=2\\left(\\cos \\left(\\frac{8\\pi }{9}\\right)+i\\sin \\left(\\frac{8\\pi }{9}\\right)\\right) \\end{align}[\/latex]<\/p>\r\nWhen [latex]k=2[\/latex], we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{z}^{\\frac{1}{3}}=2\\left[\\cos \\left(\\frac{2\\pi }{9}+\\frac{12\\pi }{9}\\right)+i\\sin \\left(\\frac{2\\pi }{9}+\\frac{12\\pi }{9}\\right)\\right]&amp;&amp; \\text{Add }\\frac{2\\left(2\\right)\\pi }{3}\\text{ to each angle.} \\\\ &amp;{z}^{\\frac{1}{3}}=2\\left(\\cos \\left(\\frac{14\\pi }{9}\\right)+i\\sin \\left(\\frac{14\\pi }{9}\\right)\\right)\\end{align}[\/latex]<\/p>\r\nRemember to find the common denominator to simplify fractions in situations like this one. For [latex]k=1[\/latex], the angle simplification is\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{\\frac{2\\pi }{3}}{3}+\\frac{2\\left(1\\right)\\pi }{3}&amp;=\\frac{2\\pi }{3}\\left(\\frac{1}{3}\\right)+\\frac{2\\left(1\\right)\\pi }{3}\\left(\\frac{3}{3}\\right)\\\\ &amp;=\\frac{2\\pi }{9}+\\frac{6\\pi }{9} \\\\ &amp;=\\frac{8\\pi }{9} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the four fourth roots of [latex]16\\left(\\cos \\left(120^\\circ \\right)+i\\sin \\left(120^\\circ \\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"929013\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"929013\"]\r\n\r\n[latex]{z}_{0}=2\\left(\\cos \\left(30^\\circ \\right)+i\\sin \\left(30^\\circ \\right)\\right)[\/latex]\r\n\r\n[latex]{z}_{1}=2\\left(\\cos \\left(120^\\circ \\right)+i\\sin \\left(120^\\circ \\right)\\right)[\/latex]\r\n\r\n[latex]{z}_{2}=2\\left(\\cos \\left(210^\\circ \\right)+i\\sin \\left(210^\\circ \\right)\\right)[\/latex]\r\n\r\n[latex]{z}_{3}=2\\left(\\cos \\left(300^\\circ \\right)+i\\sin \\left(300^\\circ \\right)\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>Complex numbers in the form [latex]a+bi[\/latex] are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Label the <em>x-<\/em>axis as the <em>real <\/em>axis and the <em>y-<\/em>axis as the <em>imaginary<\/em> axis.<\/li>\r\n \t<li>The absolute value of a complex number is the same as its magnitude. It is the distance from the origin to the point: [latex]|z|=\\sqrt{{a}^{2}+{b}^{2}}[\/latex].<\/li>\r\n \t<li>To write complex numbers in polar form, we use the formulas [latex]x=r\\cos \\theta ,y=r\\sin \\theta [\/latex], and [latex]r=\\sqrt{{x}^{2}+{y}^{2}}[\/latex]. Then, [latex]z=r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex].<\/li>\r\n \t<li>To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through by [latex]r[\/latex].<\/li>\r\n \t<li>To find the product of two complex numbers, multiply the two moduli and add the two angles. Evaluate the trigonometric functions, and multiply using the distributive property.<\/li>\r\n \t<li>To find the quotient of two complex numbers in polar form, find the quotient of the two moduli and the difference of the two angles.<\/li>\r\n \t<li>To find the power of a complex number [latex]{z}^{n}[\/latex], raise [latex]r[\/latex] to the power [latex]n[\/latex], and multiply [latex]\\theta [\/latex] by [latex]n[\/latex].<\/li>\r\n \t<li>Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135344835\" class=\"definition\">\r\n \t<dt>argument<\/dt>\r\n \t<dd id=\"fs-id1165135344841\">the angle associated with a complex number; the angle between the line from the origin to the point and the positive real axis<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135344846\" class=\"definition\">\r\n \t<dt>De Moivre\u2019s Theorem<\/dt>\r\n \t<dd id=\"fs-id1165134357535\">formula used to find the [latex]n\\text{th}[\/latex] power or <em>n<\/em>th roots of a complex number; states that, for a positive integer [latex]n,{z}^{n}[\/latex] is found by raising the modulus to the [latex]n\\text{th}[\/latex] power and multiplying the angles by [latex]n[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134279306\" class=\"definition\">\r\n \t<dt>modulus<\/dt>\r\n \t<dd id=\"fs-id1165134279312\">the absolute value of a complex number, or the distance from the origin to the point [latex]\\left(x,y\\right)[\/latex]; also called the amplitude<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165133162992\" class=\"definition\">\r\n \t<dt>polar form of a complex number<\/dt>\r\n \t<dd id=\"fs-id1165133162998\">a complex number expressed in terms of an angle [latex]\\theta [\/latex] and its distance from the origin [latex]r[\/latex]; can be found by using conversion formulas [latex]x=r\\cos \\theta ,y=r\\sin \\theta [\/latex], and [latex]r=\\sqrt{{x}^{2}+{y}^{2}}[\/latex]<\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Plot complex numbers in the complex plane.<\/li>\n<li>Find the absolute value of a complex number.<\/li>\n<li>Write complex numbers in polar form.<\/li>\n<li>Convert a complex number from polar to rectangular form.<\/li>\n<li>Find products of complex numbers in polar form.<\/li>\n<li>Find quotients of complex numbers in polar form.<\/li>\n<li>Find powers and roots of complex numbers in polar form.<\/li>\n<\/ul>\n<\/div>\n<p>&#8220;God made the integers; all else is the work of man.&#8221; This rather famous quote by nineteenth-century German mathematician Leopold <strong>Kronecker<\/strong> sets the stage for this section on the polar form of a complex number. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as <strong>Pythagoras<\/strong>, <strong>Descartes<\/strong>, De Moivre, <strong>Euler<\/strong>, <strong>Gauss<\/strong>, and others. Complex numbers answered questions that for centuries had puzzled the greatest minds in science.<\/p>\n<p>We first encountered complex numbers in Precalculus I. In this section, we will focus on the mechanics of working with complex numbers: translation of complex numbers from polar form to rectangular form and vice versa, interpretation of complex numbers in the scheme of applications, and application of De Moivre\u2019s Theorem.<\/p>\n<h2>Plotting Complex Numbers in the Complex Plane<\/h2>\n<p>Plotting a <strong>complex number<\/strong> [latex]a+bi[\/latex] is similar to plotting a real number, except that the horizontal axis represents the real part of the number, [latex]a[\/latex], and the vertical axis represents the imaginary part of the number, [latex]bi[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a complex number [latex]a+bi[\/latex], plot it in the complex plane.<\/h3>\n<ol>\n<li>Label the horizontal axis as the <em>real<\/em> axis and the vertical axis as the <em>imaginary axis.<\/em><\/li>\n<li>Plot the point in the complex plane by moving [latex]a[\/latex] units in the horizontal direction and [latex]b[\/latex] units in the vertical direction.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Plotting a Complex Number in the Complex Plane<\/h3>\n<p>Plot the complex number [latex]2 - 3i[\/latex] in the <strong>complex plane<\/strong>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q821515\">Show Solution<\/span><\/p>\n<div id=\"q821515\" class=\"hidden-answer\" style=\"display: none\">\n<p>From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180842\/CNX_Precalc_Figure_08_05_0012.jpg\" alt=\"Plot of 2-3i in the complex plane (2 along the real axis, -3 along the imaginary axis).\" width=\"487\" height=\"331\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Plot the point [latex]1+5i[\/latex] in the complex plane.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q522113\">Show Solution<\/span><\/p>\n<div id=\"q522113\" class=\"hidden-answer\" style=\"display: none\">\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180857\/CNX_Precalc_Figure_08_05_0022.jpg\" alt=\"Plot of 1+5i in the complex plane (1 along the real axis, 5 along the imaginary axis).\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Finding the Absolute Value of a Complex Number<\/h2>\n<p>The first step toward working with a complex number in <strong>polar form<\/strong> is to find the absolute value. The absolute value of a complex number is the same as its <strong>magnitude<\/strong>, or [latex]|z|[\/latex]. It measures the distance from the origin to a point in the plane. For example, the graph of [latex]z=2+4i[\/latex], in Figure 2, shows [latex]|z|[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180845\/CNX_Precalc_Figure_08_05_0032.jpg\" alt=\"Plot of 2+4i in the complex plane and its magnitude, |z| = rad 20.\" width=\"487\" height=\"368\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Absolute Value of a Complex Number<\/h3>\n<p>Given [latex]z=x+yi[\/latex], a complex number, the absolute value of [latex]z[\/latex] is defined as<\/p>\n<p style=\"text-align: center;\">[latex]|z|=\\sqrt{{x}^{2}+{y}^{2}}[\/latex]<\/p>\n<p>It is the distance from the origin to the point [latex]\\left(x,y\\right)[\/latex].<\/p>\n<p>Notice that the absolute value of a real number gives the distance of the number from 0, while the absolute value of a complex number gives the distance of the number from the origin, [latex]\\left(0,\\text{ }0\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Finding the Absolute Value of a Complex Number with a Radical<\/h3>\n<p>Find the absolute value of [latex]z=\\sqrt{5}-i[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q219333\">Show Solution<\/span><\/p>\n<div id=\"q219333\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the formula, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&|z|=\\sqrt{{x}^{2}+{y}^{2}}\\\\ &|z|=\\sqrt{{\\left(\\sqrt{5}\\right)}^{2}+{\\left(-1\\right)}^{2}} \\\\ &|z|=\\sqrt{5+1} \\\\ &|z|=\\sqrt{6} \\end{align}[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180847\/CNX_Precalc_Figure_08_05_0042.jpg\" alt=\"Plot of z=(rad5 - i) in the complex plane and its magnitude rad6.\" width=\"487\" height=\"331\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the absolute value of the complex number [latex]z=12 - 5i[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q514025\">Show Solution<\/span><\/p>\n<div id=\"q514025\" class=\"hidden-answer\" style=\"display: none\">\n<p>13<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Finding the Absolute Value of a Complex Number<\/h3>\n<p>Given [latex]z=3 - 4i[\/latex], find [latex]|z|[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q748063\">Show Solution<\/span><\/p>\n<div id=\"q748063\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the formula, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&|z|=\\sqrt{{x}^{2}+{y}^{2}} \\\\ &|z|=\\sqrt{{\\left(3\\right)}^{2}+{\\left(-4\\right)}^{2}} \\\\ &|z|=\\sqrt{9+16} \\\\ &|z|=\\sqrt{25}\\\\ &|z|=5 \\end{align}[\/latex]<\/p>\n<p>The absolute value [latex]z[\/latex] is 5.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180849\/CNX_Precalc_Figure_08_05_0052-1.jpg\" alt=\"Plot of (3-4i) in the complex plane and its magnitude |z| =5.\" width=\"487\" height=\"331\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given [latex]z=1 - 7i[\/latex], find [latex]|z|[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q473571\">Show Solution<\/span><\/p>\n<div id=\"q473571\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]|z|=\\sqrt{50}=5\\sqrt{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173801\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173801&theme=oea&iframe_resize_id=ohm173801\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>\u00a0Writing Complex Numbers in Polar Form<\/h2>\n<p>The <strong>polar form of a complex number<\/strong> expresses a number in terms of an angle [latex]\\theta[\/latex] and its distance from the origin [latex]r[\/latex]. Given a complex number in <strong>rectangular form<\/strong> expressed as [latex]z=x+yi[\/latex], we use the same conversion formulas as we do to write the number in trigonometric form:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}x=r\\cos \\theta \\\\ y=r\\sin \\theta \\\\ r=\\sqrt{{x}^{2}+{y}^{2}} \\end{gathered}[\/latex]<\/div>\n<p>We review these relationships in Figure 5.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180851\/CNX_Precalc_Figure_08_05_0062.jpg\" alt=\"Triangle plotted in the complex plane (x axis is real, y axis is imaginary). Base is along the x\/real axis, height is some y\/imaginary value in Q 1, and hypotenuse r extends from origin to that point (x+yi) in Q 1. The angle at the origin is theta. There is an arc going through (x+yi).\" width=\"487\" height=\"331\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<p>We use the term <strong>modulus<\/strong> to represent the absolute value of a complex number, or the distance from the origin to the point [latex]\\left(x,y\\right)[\/latex]. The modulus, then, is the same as [latex]r[\/latex], the radius in polar form. We use [latex]\\theta[\/latex] to indicate the angle of direction (just as with polar coordinates). Substituting, we have<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&z=x+yi \\\\ &z=r\\cos \\theta +\\left(r\\sin \\theta \\right)i \\\\ &z=r\\left(\\cos \\theta +i\\sin \\theta \\right) \\end{align}[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Polar Form of a Complex Number<\/h3>\n<p>Writing a complex number in polar form involves the following conversion formulas:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} x=r\\cos \\theta \\\\ y=r\\sin \\theta \\\\ r=\\sqrt{{x}^{2}+{y}^{2}} \\end{gathered}[\/latex]<\/p>\n<p>Making a direct substitution, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&z=x+yi \\\\ &z=\\left(r\\cos \\theta \\right)+i\\left(r\\sin \\theta \\right) \\\\ &z=r\\left(\\cos \\theta +i\\sin \\theta \\right) \\end{align}[\/latex]<\/p>\n<p>where [latex]r[\/latex] is the <strong>modulus<\/strong> and [latex]\\theta[\/latex] is the <strong>argument<\/strong>. We often use the abbreviation [latex]r\\text{cis}\\theta[\/latex] to represent [latex]r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Expressing a Complex Number Using Polar Coordinates<\/h3>\n<p>Express the complex number [latex]4i[\/latex] using polar coordinates.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3488\">Show Solution<\/span><\/p>\n<div id=\"q3488\" class=\"hidden-answer\" style=\"display: none\">\n<p>On the complex plane, the number [latex]z=4i[\/latex] is the same as [latex]z=0+4i[\/latex]. Writing it in polar form, we have to calculate [latex]r[\/latex] first.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&r=\\sqrt{{x}^{2}+{y}^{2}} \\\\ &r=\\sqrt{{0}^{2}+{4}^{2}} \\\\ &r=\\sqrt{16} \\\\ &r=4 \\end{align}[\/latex]<\/p>\n<p>Next, we look at [latex]x[\/latex]. If [latex]x=r\\cos \\theta[\/latex], and [latex]x=0[\/latex], then [latex]\\theta =\\frac{\\pi }{2}[\/latex]. In polar coordinates, the complex number [latex]z=0+4i[\/latex] can be written as [latex]z=4\\left(\\cos \\left(\\frac{\\pi }{2}\\right)+i\\sin \\left(\\frac{\\pi }{2}\\right)\\right)[\/latex] or [latex]4\\text{cis}\\left(\\frac{\\pi }{2}\\right)[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180854\/CNX_Precalc_Figure_08_05_0072.jpg\" alt=\"Plot of z=4i in the complex plane, also shows that the in polar coordinate it would be (4,pi\/2).\" width=\"487\" height=\"294\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Express [latex]z=3i[\/latex] as [latex]r\\text{cis}\\theta[\/latex] in polar form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q364839\">Show Solution<\/span><\/p>\n<div id=\"q364839\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]z=3\\left(\\cos \\left(\\frac{\\pi }{2}\\right)+i\\sin \\left(\\frac{\\pi }{2}\\right)\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Finding the Polar Form of a Complex Number<\/h3>\n<p>Find the polar form of [latex]-4+4i[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q871564\">Show Solution<\/span><\/p>\n<div id=\"q871564\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, find the value of [latex]r[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&r=\\sqrt{{x}^{2}+{y}^{2}} \\\\ &r=\\sqrt{{\\left(-4\\right)}^{2}+\\left({4}^{2}\\right)} \\\\ &r=\\sqrt{32} \\\\ &r=4\\sqrt{2} \\end{align}[\/latex]<\/p>\n<p>Find the angle [latex]\\theta[\/latex] using the formula:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\cos \\theta =\\frac{x}{r} \\\\ &\\cos \\theta =\\frac{-4}{4\\sqrt{2}} \\\\ &\\cos \\theta =-\\frac{1}{\\sqrt{2}} \\\\ &\\theta ={\\cos }^{-1}\\left(-\\frac{1}{\\sqrt{2}}\\right)=\\frac{3\\pi }{4} \\end{align}[\/latex]<\/p>\n<p>Thus, the solution is [latex]4\\sqrt{2}\\text{cis}\\left(\\frac{3\\pi }{4}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Write [latex]z=\\sqrt{3}+i[\/latex] in polar form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q164997\">Show Solution<\/span><\/p>\n<div id=\"q164997\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]z=2\\left(\\cos \\left(\\frac{\\pi }{6}\\right)+i\\sin \\left(\\frac{\\pi }{6}\\right)\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173845\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173845&theme=oea&iframe_resize_id=ohm173845\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>\u00a0Converting a Complex Number from Polar to Rectangular Form<\/h2>\n<p>Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. In other words, given [latex]z=r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex], first evaluate the trigonometric functions [latex]\\cos \\theta[\/latex] and [latex]\\sin \\theta[\/latex]. Then, multiply through by [latex]r[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 6: Converting from Polar to Rectangular Form<\/h3>\n<p>Convert the polar form of the given complex number to rectangular form:<\/p>\n<p style=\"text-align: center;\">[latex]z=12\\left(\\cos \\left(\\frac{\\pi }{6}\\right)+i\\sin \\left(\\frac{\\pi }{6}\\right)\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q145161\">Show Solution<\/span><\/p>\n<div id=\"q145161\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by evaluating the trigonometric expressions.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(\\frac{\\pi }{6}\\right)=\\frac{\\sqrt{3}}{2}\\\\\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}\\end{gathered}[\/latex]<\/p>\n<p>After substitution, the complex number is<\/p>\n<p style=\"text-align: center;\">[latex]z=12\\left(\\frac{\\sqrt{3}}{2}+\\frac{1}{2}i\\right)[\/latex]<\/p>\n<p>We apply the distributive property:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}z&=12\\left(\\frac{\\sqrt{3}}{2}+\\frac{1}{2}i\\right) \\\\ &=\\left(12\\right)\\frac{\\sqrt{3}}{2}+\\left(12\\right)\\frac{1}{2}i \\\\ &=6\\sqrt{3}+6i \\end{align}[\/latex]<\/p>\n<p>The rectangular form of the given point in complex form is [latex]6\\sqrt{3}+6i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Finding the Rectangular Form of a Complex Number<\/h3>\n<p>Find the rectangular form of the complex number given [latex]r=13[\/latex] and [latex]\\tan \\theta =\\frac{5}{12}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q11326\">Show Solution<\/span><\/p>\n<div id=\"q11326\" class=\"hidden-answer\" style=\"display: none\">\n<p>If [latex]\\tan \\theta =\\frac{5}{12}[\/latex], and [latex]\\tan \\theta =\\frac{y}{x}[\/latex], we first determine [latex]r=\\sqrt{{x}^{2}+{y}^{2}}=\\sqrt{{12}^{2}+{5}^{2}}=13\\text{.}[\/latex] We then find [latex]\\cos \\theta =\\frac{x}{r}[\/latex] and [latex]\\sin \\theta =\\frac{y}{r}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}z&=13\\left(\\cos \\theta +i\\sin \\theta \\right) \\\\ &=13\\left(\\frac{12}{13}+\\frac{5}{13}i\\right) \\\\ &=12+5i \\end{align}[\/latex]<\/p>\n<p>The rectangular form of the given number in complex form is [latex]12+5i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Convert the complex number to rectangular form:<\/p>\n<p style=\"text-align: center;\">[latex]z=4\\left(\\cos \\frac{11\\pi }{6}+i\\sin \\frac{11\\pi }{6}\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q9609\">Show Solution<\/span><\/p>\n<div id=\"q9609\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">[latex]z=2\\sqrt{3}-2i[\/latex]<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173840\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173840&theme=oea&iframe_resize_id=ohm173840\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>\u00a0Finding Products and Quotients of Complex Numbers in Polar Form<\/h2>\n<p>Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham <strong>de Moivre<\/strong> (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Products of Complex Numbers in Polar Form<\/h3>\n<p>If [latex]{z}_{1}={r}_{1}\\left(\\cos {\\theta }_{1}+i\\sin {\\theta }_{1}\\right)[\/latex] and [latex]{z}_{2}={r}_{2}\\left(\\cos {\\theta }_{2}+i\\sin {\\theta }_{2}\\right)[\/latex], then the product of these numbers is given as:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{z}_{1}{z}_{2}&={r}_{1}{r}_{2}\\left[\\cos \\left({\\theta }_{1}+{\\theta }_{2}\\right)+i\\sin \\left({\\theta }_{1}+{\\theta }_{2}\\right)\\right] \\\\ {z}_{1}{z}_{2}&={r}_{1}{r}_{2}\\text{cis}\\left({\\theta }_{1}+{\\theta }_{2}\\right) \\end{align}[\/latex]<\/p>\n<p>Notice that the product calls for multiplying the moduli and adding the angles.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Finding the Product of Two Complex Numbers in Polar Form<\/h3>\n<p>Find the product of [latex]{z}_{1}{z}_{2}[\/latex], given [latex]{z}_{1}=4\\left(\\cos \\left(80^\\circ \\right)+i\\sin \\left(80^\\circ \\right)\\right)[\/latex] and [latex]{z}_{2}=2\\left(\\cos \\left(145^\\circ \\right)+i\\sin \\left(145^\\circ \\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q385516\">Show Solution<\/span><\/p>\n<div id=\"q385516\" class=\"hidden-answer\" style=\"display: none\">\n<p>Follow the formula<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{z}_{1}{z}_{2}=4\\cdot 2\\left[\\cos \\left(80^\\circ +145^\\circ \\right)+i\\sin \\left(80^\\circ +145^\\circ \\right)\\right] \\\\ &{z}_{1}{z}_{2}=8\\left[\\cos \\left(225^\\circ \\right)+i\\sin \\left(225^\\circ \\right)\\right] \\\\ &{z}_{1}{z}_{2}=8\\left[\\cos \\left(\\frac{5\\pi }{4}\\right)+i\\sin \\left(\\frac{5\\pi }{4}\\right)\\right] \\\\ {z}_{1}{z}_{2}=8\\left[-\\frac{\\sqrt{2}}{2}+i\\left(-\\frac{\\sqrt{2}}{2}\\right)\\right] \\\\ &{z}_{1}{z}_{2}=-4\\sqrt{2}-4i\\sqrt{2} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding Quotients of Complex Numbers in Polar Form<\/h2>\n<p>The quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Quotients of Complex Numbers in Polar Form<\/h3>\n<p>If [latex]{z}_{1}={r}_{1}\\left(\\cos {\\theta }_{1}+i\\sin {\\theta }_{1}\\right)[\/latex] and [latex]{z}_{2}={r}_{2}\\left(\\cos {\\theta }_{2}+i\\sin {\\theta }_{2}\\right)[\/latex], then the quotient of these numbers is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\frac{{z}_{1}}{{z}_{2}}=\\frac{{r}_{1}}{{r}_{2}}\\left[\\cos \\left({\\theta }_{1}-{\\theta }_{2}\\right)+i\\sin \\left({\\theta }_{1}-{\\theta }_{2}\\right)\\right],{z}_{2}\\ne 0\\\\ &\\frac{{z}_{1}}{{z}_{2}}=\\frac{{r}_{1}}{{r}_{2}}\\text{cis}\\left({\\theta }_{1}-{\\theta }_{2}\\right),{z}_{2}\\ne 0\\end{align}[\/latex]<\/p>\n<p>Notice that the moduli are divided, and the angles are subtracted.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given two complex numbers in polar form, find the quotient.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Divide [latex]\\frac{{r}_{1}}{{r}_{2}}[\/latex].<\/li>\n<li>Find [latex]{\\theta }_{1}-{\\theta }_{2}[\/latex].<\/li>\n<li>Substitute the results into the formula: [latex]z=r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex]. Replace [latex]r[\/latex] with [latex]\\frac{{r}_{1}}{{r}_{2}}[\/latex], and replace [latex]\\theta[\/latex] with [latex]{\\theta }_{1}-{\\theta }_{2}[\/latex].<\/li>\n<li>Calculate the new trigonometric expressions and multiply through by [latex]r[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 9: Finding the Quotient of Two Complex Numbers<\/h3>\n<p>Find the quotient of [latex]{z}_{1}=2\\left(\\cos \\left(213^\\circ \\right)+i\\sin \\left(213^\\circ \\right)\\right)[\/latex] and [latex]{z}_{2}=4\\left(\\cos \\left(33^\\circ \\right)+i\\sin \\left(33^\\circ \\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q244676\">Show Solution<\/span><\/p>\n<div id=\"q244676\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the formula, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\frac{{z}_{1}}{{z}_{2}}=\\frac{2}{4}\\left[\\cos \\left(213^\\circ -33^\\circ \\right)+i\\sin \\left(213^\\circ -33^\\circ \\right)\\right] \\\\ &\\frac{{z}_{1}}{{z}_{2}}=\\frac{1}{2}\\left[\\cos \\left(180^\\circ \\right)+i\\sin \\left(180^\\circ \\right)\\right] \\\\ &\\frac{{z}_{1}}{{z}_{2}}=\\frac{1}{2}\\left[-1+0i\\right] \\\\ &\\frac{{z}_{1}}{{z}_{2}}=-\\frac{1}{2}+0i \\\\ &\\frac{{z}_{1}}{{z}_{2}}=-\\frac{1}{2} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the product and the quotient of [latex]{z}_{1}=2\\sqrt{3}\\left(\\cos \\left(150^\\circ \\right)+i\\sin \\left(150^\\circ \\right)\\right)[\/latex] and [latex]{z}_{2}=2\\left(\\cos \\left(30^\\circ \\right)+i\\sin \\left(30^\\circ \\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q381268\">Show Solution<\/span><\/p>\n<div id=\"q381268\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{z}_{1}{z}_{2}=-4\\sqrt{3};\\frac{{z}_{1}}{{z}_{2}}=-\\frac{\\sqrt{3}}{2}+\\frac{3}{2}i[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173851\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173851&theme=oea&iframe_resize_id=ohm173851\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>\u00a0Finding Powers and Roots of Complex Numbers in Polar Form<\/h2>\n<p>Finding powers of complex numbers is greatly simplified using <strong>De Moivre\u2019s Theorem<\/strong>. It states that, for a positive integer [latex]n,{z}^{n}[\/latex] is found by raising the modulus to the [latex]n\\text{th}[\/latex] power and multiplying the argument by [latex]n[\/latex]. It is the standard method used in modern mathematics.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: De Moivre\u2019s Theorem<\/h3>\n<p>If [latex]z=r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex] is a complex number, then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{z}^{n}={r}^{n}\\left[\\cos \\left(n\\theta \\right)+i\\sin \\left(n\\theta \\right)\\right]\\\\ &{z}^{n}={r}^{n}\\text{cis}\\left(n\\theta \\right)\\end{align}[\/latex]<\/p>\n<p>where [latex]n[\/latex]\u00a0is a positive integer.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 10: Evaluating an Expression Using De Moivre\u2019s Theorem<\/h3>\n<p>Evaluate the expression [latex]{\\left(1+i\\right)}^{5}[\/latex] using De Moivre\u2019s Theorem.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q728625\">Show Solution<\/span><\/p>\n<div id=\"q728625\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since De Moivre\u2019s Theorem applies to complex numbers written in polar form, we must first write [latex]\\left(1+i\\right)[\/latex] in polar form. Let us find [latex]r[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&r=\\sqrt{{x}^{2}+{y}^{2}} \\\\ &r=\\sqrt{{\\left(1\\right)}^{2}+{\\left(1\\right)}^{2}} \\\\ &r=\\sqrt{2} \\end{align}[\/latex]<\/p>\n<p>Then we find [latex]\\theta[\/latex]. Using the formula [latex]\\tan \\theta =\\frac{y}{x}[\/latex] gives<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\tan \\theta =\\frac{1}{1} \\\\ &\\tan \\theta =1 \\\\ &\\theta =\\frac{\\pi }{4} \\end{align}[\/latex]<\/p>\n<p>Use De Moivre\u2019s Theorem to evaluate the expression.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{\\left(a+bi\\right)}^{n}={r}^{n}\\left[\\cos \\left(n\\theta \\right)+i\\sin \\left(n\\theta \\right)\\right]\\\\ &{\\left(1+i\\right)}^{5}={\\left(\\sqrt{2}\\right)}^{5}\\left[\\cos \\left(5\\cdot \\frac{\\pi }{4}\\right)+i\\sin \\left(5\\cdot \\frac{\\pi }{4}\\right)\\right] \\\\ &{\\left(1+i\\right)}^{5}=4\\sqrt{2}\\left[\\cos \\left(\\frac{5\\pi }{4}\\right)+i\\sin \\left(\\frac{5\\pi }{4}\\right)\\right] \\\\ &{\\left(1+i\\right)}^{5}=4\\sqrt{2}\\left[-\\frac{\\sqrt{2}}{2}+i\\left(-\\frac{\\sqrt{2}}{2}\\right)\\right] \\\\ &{\\left(1+i\\right)}^{5}=-4 - 4i \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding Roots of Complex Numbers in Polar Form<\/h2>\n<p>To find the <strong><em>n<\/em>th root of a complex number<\/strong> in polar form, we use the [latex]n\\text{th}[\/latex] Root Theorem or <strong>De Moivre\u2019s Theorem<\/strong> and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding [latex]n\\text{th}[\/latex] roots of complex numbers in polar form.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The <em>n<\/em>th Root Theorem<\/h3>\n<p>To find the [latex]n\\text{th}[\/latex] root of a complex number in polar form, use the formula given as<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{z}^{\\frac{1}{n}}={r}^{\\frac{1}{n}}\\left[\\cos \\left(\\frac{\\theta }{n}+\\frac{2k\\pi }{n}\\right)+i\\sin \\left(\\frac{\\theta }{n}+\\frac{2k\\pi }{n}\\right)\\right]\\end{align}[\/latex]<\/p>\n<p>where [latex]k=0,1,2,3,...,n - 1[\/latex]. We add [latex]\\frac{2k\\pi }{n}[\/latex] to [latex]\\frac{\\theta }{n}[\/latex] in order to obtain the periodic roots.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 11: Finding the <em>n<\/em>th Root of a Complex Number<\/h3>\n<p>Evaluate the cube roots of [latex]z=8\\left(\\cos \\left(\\frac{2\\pi }{3}\\right)+i\\sin \\left(\\frac{2\\pi }{3}\\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q791675\">Show Solution<\/span><\/p>\n<div id=\"q791675\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{z}^{\\frac{1}{3}}={8}^{\\frac{1}{3}}\\left[\\cos \\left(\\frac{\\frac{2\\pi }{3}}{3}+\\frac{2k\\pi }{3}\\right)+i\\sin \\left(\\frac{\\frac{2\\pi }{3}}{3}+\\frac{2k\\pi }{3}\\right)\\right] \\\\ &{z}^{\\frac{1}{3}}=2\\left[\\cos \\left(\\frac{2\\pi }{9}+\\frac{2k\\pi }{3}\\right)+i\\sin \\left(\\frac{2\\pi }{9}+\\frac{2k\\pi }{3}\\right)\\right] \\end{align}[\/latex]<\/p>\n<p>There will be three roots: [latex]k=0,1,2[\/latex]. When [latex]k=0[\/latex], we have<\/p>\n<p style=\"text-align: center;\">[latex]{z}^{\\frac{1}{3}}=2\\left(\\cos \\left(\\frac{2\\pi }{9}\\right)+i\\sin \\left(\\frac{2\\pi }{9}\\right)\\right)[\/latex]<\/p>\n<p>When [latex]k=1[\/latex], we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{z}^{\\frac{1}{3}}=2\\left[\\cos \\left(\\frac{2\\pi }{9}+\\frac{6\\pi }{9}\\right)+i\\sin \\left(\\frac{2\\pi }{9}+\\frac{6\\pi }{9}\\right)\\right] && \\text{ Add }\\frac{2\\left(1\\right)\\pi }{3}\\text{ to each angle.} \\\\ &{z}^{\\frac{1}{3}}=2\\left(\\cos \\left(\\frac{8\\pi }{9}\\right)+i\\sin \\left(\\frac{8\\pi }{9}\\right)\\right) \\end{align}[\/latex]<\/p>\n<p>When [latex]k=2[\/latex], we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{z}^{\\frac{1}{3}}=2\\left[\\cos \\left(\\frac{2\\pi }{9}+\\frac{12\\pi }{9}\\right)+i\\sin \\left(\\frac{2\\pi }{9}+\\frac{12\\pi }{9}\\right)\\right]&& \\text{Add }\\frac{2\\left(2\\right)\\pi }{3}\\text{ to each angle.} \\\\ &{z}^{\\frac{1}{3}}=2\\left(\\cos \\left(\\frac{14\\pi }{9}\\right)+i\\sin \\left(\\frac{14\\pi }{9}\\right)\\right)\\end{align}[\/latex]<\/p>\n<p>Remember to find the common denominator to simplify fractions in situations like this one. For [latex]k=1[\/latex], the angle simplification is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{\\frac{2\\pi }{3}}{3}+\\frac{2\\left(1\\right)\\pi }{3}&=\\frac{2\\pi }{3}\\left(\\frac{1}{3}\\right)+\\frac{2\\left(1\\right)\\pi }{3}\\left(\\frac{3}{3}\\right)\\\\ &=\\frac{2\\pi }{9}+\\frac{6\\pi }{9} \\\\ &=\\frac{8\\pi }{9} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the four fourth roots of [latex]16\\left(\\cos \\left(120^\\circ \\right)+i\\sin \\left(120^\\circ \\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q929013\">Show Solution<\/span><\/p>\n<div id=\"q929013\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{z}_{0}=2\\left(\\cos \\left(30^\\circ \\right)+i\\sin \\left(30^\\circ \\right)\\right)[\/latex]<\/p>\n<p>[latex]{z}_{1}=2\\left(\\cos \\left(120^\\circ \\right)+i\\sin \\left(120^\\circ \\right)\\right)[\/latex]<\/p>\n<p>[latex]{z}_{2}=2\\left(\\cos \\left(210^\\circ \\right)+i\\sin \\left(210^\\circ \\right)\\right)[\/latex]<\/p>\n<p>[latex]{z}_{3}=2\\left(\\cos \\left(300^\\circ \\right)+i\\sin \\left(300^\\circ \\right)\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>Complex numbers in the form [latex]a+bi[\/latex] are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Label the <em>x-<\/em>axis as the <em>real <\/em>axis and the <em>y-<\/em>axis as the <em>imaginary<\/em> axis.<\/li>\n<li>The absolute value of a complex number is the same as its magnitude. It is the distance from the origin to the point: [latex]|z|=\\sqrt{{a}^{2}+{b}^{2}}[\/latex].<\/li>\n<li>To write complex numbers in polar form, we use the formulas [latex]x=r\\cos \\theta ,y=r\\sin \\theta[\/latex], and [latex]r=\\sqrt{{x}^{2}+{y}^{2}}[\/latex]. Then, [latex]z=r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex].<\/li>\n<li>To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through by [latex]r[\/latex].<\/li>\n<li>To find the product of two complex numbers, multiply the two moduli and add the two angles. Evaluate the trigonometric functions, and multiply using the distributive property.<\/li>\n<li>To find the quotient of two complex numbers in polar form, find the quotient of the two moduli and the difference of the two angles.<\/li>\n<li>To find the power of a complex number [latex]{z}^{n}[\/latex], raise [latex]r[\/latex] to the power [latex]n[\/latex], and multiply [latex]\\theta[\/latex] by [latex]n[\/latex].<\/li>\n<li>Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135344835\" class=\"definition\">\n<dt>argument<\/dt>\n<dd id=\"fs-id1165135344841\">the angle associated with a complex number; the angle between the line from the origin to the point and the positive real axis<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135344846\" class=\"definition\">\n<dt>De Moivre\u2019s Theorem<\/dt>\n<dd id=\"fs-id1165134357535\">formula used to find the [latex]n\\text{th}[\/latex] power or <em>n<\/em>th roots of a complex number; states that, for a positive integer [latex]n,{z}^{n}[\/latex] is found by raising the modulus to the [latex]n\\text{th}[\/latex] power and multiplying the angles by [latex]n[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134279306\" class=\"definition\">\n<dt>modulus<\/dt>\n<dd id=\"fs-id1165134279312\">the absolute value of a complex number, or the distance from the origin to the point [latex]\\left(x,y\\right)[\/latex]; also called the amplitude<\/dd>\n<\/dl>\n<dl id=\"fs-id1165133162992\" class=\"definition\">\n<dt>polar form of a complex number<\/dt>\n<dd id=\"fs-id1165133162998\">a complex number expressed in terms of an angle [latex]\\theta[\/latex] and its distance from the origin [latex]r[\/latex]; can be found by using conversion formulas [latex]x=r\\cos \\theta ,y=r\\sin \\theta[\/latex], and [latex]r=\\sqrt{{x}^{2}+{y}^{2}}[\/latex]<\/dd>\n<\/dl>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-14449\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-14449","chapter","type-chapter","status-publish","hentry"],"part":14256,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14449","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14449\/revisions"}],"predecessor-version":[{"id":16185,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14449\/revisions\/16185"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/14256"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14449\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=14449"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=14449"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=14449"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=14449"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}