{"id":190,"date":"2023-06-21T13:22:42","date_gmt":"2023-06-21T13:22:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/divide-complex-numbers\/"},"modified":"2023-07-04T03:45:44","modified_gmt":"2023-07-04T03:45:44","slug":"divide-complex-numbers","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/divide-complex-numbers\/","title":{"raw":"\u25aa   Divide Complex Numbers","rendered":"\u25aa   Divide Complex Numbers"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify and write the complex conjugate of a complex number.<\/li>\r\n \t<li>Divide complex numbers.<\/li>\r\n \t<li>Simplify powers of [latex]i[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>recall a technique: rationalizing the denominator<\/h3>\r\nIn mathematics, a technique that works well in one situation often works equally as well in a different, but stylistically similar situation.\r\n\r\nDividing complex numbers uses a technique you used when rewriting\u00a0a fraction that contains a radical in the denominator. We called it <em>rationalizing a denominator.\u00a0<\/em>To accomplish it, you multiplied the fraction in the numerator and denominator by a number that would clear the radical in the denominator. We use the same general idea when dividing complex numbers. This time, instead of multiplying a radical by a specially chosen radical though, we'll multiply a complex number by a specially chosen complex number.\r\n\r\nWhen rationalizing the denominator, we took advantage of the fact that [latex]\\left(\\sqrt{a}\\right)^2=a[\/latex].\r\n\r\nWhen dividing complex numbers, we'll take advantage of the fact that [latex]i^2 = -1[\/latex].\r\n\r\n<\/div>\r\n<h2>Dividing Complex Numbers<\/h2>\r\nDivision of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the <strong>complex conjugate<\/strong> of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of [latex]a+bi[\/latex] is [latex]a-bi[\/latex].\r\n\r\nNote that complex conjugates have a reciprocal relationship: The complex conjugate of [latex]a+bi[\/latex] is [latex]a-bi[\/latex], and the complex conjugate of [latex]a-bi[\/latex] is [latex]a+bi[\/latex]. Importantly, complex conjugate pairs have a special property. Their product is always real.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}(a+bi)(a-bi)&amp;=a^2-abi+abi-b^2i^2\\\\[2mm]&amp;=a^2-b^2(-1)\\\\[2mm]&amp;=a^2+b^2\\end{align}[\/latex]<\/p>\r\nSuppose we want to divide [latex]c+di[\/latex] by [latex]a+bi[\/latex], where neither [latex]a[\/latex] nor [latex]b[\/latex] equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{c+di}{a+bi}[\/latex] where [latex]a\\ne 0[\/latex] and [latex]b\\ne 0[\/latex].<\/p>\r\nMultiply the numerator and denominator by the complex conjugate of the denominator.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{\\left(c+di\\right)}{\\left(a+bi\\right)}\\cdot \\dfrac{\\left(a-bi\\right)}{\\left(a-bi\\right)}=\\dfrac{\\left(c+di\\right)\\left(a-bi\\right)}{\\left(a+bi\\right)\\left(a-bi\\right)}[\/latex]<\/p>\r\nApply the distributive property.\r\n<p style=\"text-align: center;\">[latex]=\\dfrac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}}[\/latex]<\/p>\r\nSimplify, remembering that [latex]{i}^{2}=-1[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;=\\dfrac{ca-cbi+adi-bd\\left(-1\\right)}{{a}^{2}-abi+abi-{b}^{2}\\left(-1\\right)} \\\\[2mm] &amp;=\\dfrac{\\left(ca+bd\\right)+\\left(ad-cb\\right)i}{{a}^{2}+{b}^{2}}\\end{align}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Complex Conjugate<\/h3>\r\nThe <strong>complex conjugate<\/strong> of a complex number [latex]a+bi[\/latex] is [latex]a-bi[\/latex]. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.\r\n<ul>\r\n \t<li>When a complex number is multiplied by its complex conjugate, the result is a real number.<\/li>\r\n \t<li>When a complex number is added to its complex conjugate, the result is a real number.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding Complex Conjugates<\/h3>\r\nFind the complex conjugate of each number.\r\n<ol>\r\n \t<li>[latex]2+i\\sqrt{5}[\/latex]<\/li>\r\n \t<li>[latex]-\\frac{1}{2}i[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"349660\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"349660\"]\r\n<ol>\r\n \t<li>The number is already in the form [latex]a+bi[\/latex]. The complex conjugate is [latex]a-bi[\/latex], or [latex]2-i\\sqrt{5}[\/latex].<\/li>\r\n \t<li>We can rewrite this number in the form [latex]a+bi[\/latex] as [latex]0-\\frac{1}{2}i[\/latex]. The complex conjugate is [latex]a-bi[\/latex], or [latex]0+\\frac{1}{2}i[\/latex]. This can be written simply as [latex]\\frac{1}{2}i[\/latex].<\/li>\r\n<\/ol>\r\n<h4>Analysis of the Solution<\/h4>\r\nAlthough we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by [latex]i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given two complex numbers, divide one by the other.<\/h3>\r\n<ol>\r\n \t<li>Write the division problem as a fraction.<\/li>\r\n \t<li>Determine the complex conjugate of the denominator.<\/li>\r\n \t<li>Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Dividing Complex Numbers<\/h3>\r\nDivide [latex]\\left(2+5i\\right)[\/latex] by [latex]\\left(4-i\\right)[\/latex].\r\n\r\n[reveal-answer q=\"932961\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"932961\"]\r\n\r\nWe begin by writing the problem as a fraction.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{\\left(2+5i\\right)}{\\left(4-i\\right)}[\/latex]<\/p>\r\nThen we multiply the numerator and denominator by the complex conjugate of the denominator.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{\\left(2+5i\\right)}{\\left(4-i\\right)}\\cdot \\dfrac{\\left(4+i\\right)}{\\left(4+i\\right)}[\/latex]<\/p>\r\nTo multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL).\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\dfrac{\\left(2+5i\\right)}{\\left(4-i\\right)}\\cdot \\dfrac{\\left(4+i\\right)}{\\left(4+i\\right)}&amp;=\\dfrac{8+2i+20i+5{i}^{2}}{16+4i - 4i-{i}^{2}}\\\\[2mm] &amp;=\\dfrac{8+2i+20i+5\\left(-1\\right)}{16+4i - 4i-\\left(-1\\right)} &amp;&amp; \\text{Because } {i}^{2}=-1 \\\\[2mm] &amp;=\\frac{3+22i}{17} \\\\[2mm] &amp;=\\dfrac{3}{17}+\\frac{22}{17}i &amp;&amp; \\text{Separate real and imaginary parts}.\\end{align}[\/latex]<\/p>\r\nNote that this expresses the quotient in standard form.\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]61715[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nWe have seen that we can evaluate functions for any real number or algebraic expression we choose. For example,\r\n\r\ngiven the function [latex]f(x)=2x-7[\/latex],\r\n\r\n[latex]f(-1) = 2(-1) -7 = -9[\/latex]\r\n\r\n[latex]f(5+h)=2(5+h)-7 = 10 +2h - 7 = 3 +2h[\/latex].\r\n\r\nWe may also evaluate a function for a complex number if we wish, as shown in the Example and Try It boxes below.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Substituting a Complex Number into a Polynomial Function<\/h3>\r\nLet [latex]f\\left(x\\right)={x}^{2}-5x+2[\/latex]. Evaluate [latex]f\\left(3+i\\right)[\/latex].\r\n\r\n[reveal-answer q=\"323196\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"323196\"]\r\n\r\nSubstitute [latex]x=3+i[\/latex] into the function [latex]f\\left(x\\right)={x}^{2}-5x+2[\/latex] and simplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}f(3+i)&amp;=(3+i)^2-5(3+i)+2&amp;&amp;\\text{Substitute } 3+i \\text{ for }x\\\\[2mm]&amp;=(3+6i+i^2)-(15+5i)+2&amp;&amp;\\text{Multiply}\\\\[2mm]&amp;=9+6i+(-1)-15-5i+2&amp;&amp;\\text{Substitute }-1\\text{ for }i^2 \\\\[2mm]&amp;=-5+i&amp;&amp;\\text{Combine like terms}\\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWe write [latex]f\\left(3+i\\right)=-5+i[\/latex]. Notice that the input is [latex]3+i[\/latex] and the output is [latex]-5+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\nLet [latex]f\\left(x\\right)=2{x}^{2}-3x[\/latex]. Evaluate [latex]f\\left(8-i\\right)[\/latex].\r\n\r\n[reveal-answer q=\"324665\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"324665\"]\r\n\r\n[latex]102 - 29i[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[ohm_question]120193[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Substituting an Imaginary Number in a Rational Function<\/h3>\r\nLet [latex]f\\left(x\\right)=\\dfrac{2+x}{x+3}[\/latex]. Evaluate [latex]f\\left(10i\\right)[\/latex].\r\n\r\n[reveal-answer q=\"462657\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"462657\"]\r\n\r\nSubstitute [latex]x=10i[\/latex] and simplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\dfrac{2+10i}{10i+3} &amp;&amp; \\text{Substitute }10i\\text{ for }x\\\\[2mm] &amp;\\dfrac{2+10i}{3+10i} &amp;&amp; \\text{Rewrite the denominator in standard form}\\\\[2mm] &amp;\\dfrac{2+10i}{3+10i}\\cdot \\dfrac{3 - 10i}{3 - 10i} &amp;&amp; \\text{Multiply the numerator and denominator by the complex conjugate of the denominator}\\\\[2mm] &amp;\\dfrac{6 - 20i+30i - 100{i}^{2}}{9 - 30i+30i - 100{i}^{2}} &amp;&amp; \\text{Multiply using the distributive property or the FOIL method} \\\\[2mm] &amp;\\dfrac{6 - 20i+30i - 100\\left(-1\\right)}{9 - 30i+30i - 100\\left(-1\\right)} &amp;&amp; \\text{Substitute }-1\\text{ for } {i}^{2} \\\\[2mm] &amp;\\dfrac{106+10i}{109} &amp;&amp; \\text{Simplify} \\\\[2mm] &amp;\\dfrac{106}{109}+\\dfrac{10}{109}i &amp;&amp; \\text{Separate the real and imaginary parts} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [latex]f\\left(x\\right)=\\dfrac{x+1}{x - 4}[\/latex]. Evaluate [latex]f\\left(-i\\right)[\/latex].\r\n\r\n[reveal-answer q=\"453508\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"453508\"]\r\n\r\n[latex]-\\dfrac{3}{17}+\\dfrac{5}{17}i[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/XBJjbJAwM1c\r\n<h2>Simplifying Powers of [latex]i[\/latex]<\/h2>\r\nThe powers of [latex]i[\/latex] are cyclic. Let\u2019s look at what happens when we raise [latex]i[\/latex] to increasing powers.\r\n\r\n[latex]{i}^{1}=i[\/latex]\r\n[latex]{i}^{2}=-1[\/latex]\r\n[latex]{i}^{3}={i}^{2}\\cdot i=-1\\cdot i=-i[\/latex]\r\n[latex]{i}^{4}={i}^{3}\\cdot i=-i\\cdot i=-{i}^{2}=-\\left(-1\\right)=1[\/latex]\r\n[latex]{i}^{5}={i}^{4}\\cdot i=1\\cdot i=i[\/latex]\r\n\r\nWe can see that when we get to the fifth power of [latex]i[\/latex], it is equal to the first power. As we continue to multiply [latex]i[\/latex]\u00a0by itself for increasing powers, we will see a cycle of 4. Let\u2019s examine the next 4 powers of [latex]i[\/latex].\r\n\r\n[latex]{i}^{6}={i}^{5}\\cdot i=i\\cdot i={i}^{2}=-1[\/latex]\r\n[latex]{i}^{7}={i}^{6}\\cdot i={i}^{2}\\cdot i={i}^{3}=-i[\/latex]\r\n[latex]{i}^{8}={i}^{7}\\cdot i={i}^{3}\\cdot i={i}^{4}=1[\/latex]\r\n[latex]{i}^{9}={i}^{8}\\cdot i={i}^{4}\\cdot i={i}^{5}=i[\/latex]\r\n<div class=\"textbox examples\">\r\n<h3>recall properties of exponents<\/h3>\r\nRecall that [latex]\\left(a^m\\right)^n=a^{mn}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Simplifying Powers of [latex]i[\/latex]<\/h3>\r\nEvaluate [latex]{i}^{35}[\/latex].\r\n\r\n[reveal-answer q=\"879646\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"879646\"]\r\n\r\nSince [latex]{i}^{4}=1[\/latex], we can simplify the problem by factoring out as many factors of [latex]{i}^{4}[\/latex] as possible. To do so, first determine how many times [latex]4[\/latex] goes into [latex]35[\/latex]: [latex]35=4\\cdot 8+3[\/latex].\r\n\r\n[latex]{i}^{35}={i}^{4\\cdot 8+3}={i}^{4\\cdot 8}\\cdot {i}^{3}={\\left({i}^{4}\\right)}^{8}\\cdot {i}^{3}={1}^{8}\\cdot {i}^{3}={i}^{3}=-i[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Can we write [latex]{i}^{35}[\/latex] in other helpful ways?<\/strong>\r\n\r\n<em>As we saw in Example: Simplifying Powers of <\/em>[latex]i[\/latex]<em>, we reduced [latex]{i}^{35}[\/latex] to [latex]{i}^{3}[\/latex] by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of [latex]{i}^{35}[\/latex] may be more useful. The table below\u00a0shows some other possible factorizations.<\/em>\r\n<table summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Factorization of [latex]{i}^{35}[\/latex]<\/strong><\/td>\r\n<td>[latex]{i}^{34}\\cdot i[\/latex]<\/td>\r\n<td>[latex]{i}^{33}\\cdot {i}^{2}[\/latex]<\/td>\r\n<td>[latex]{i}^{31}\\cdot {i}^{4}[\/latex]<\/td>\r\n<td>[latex]{i}^{19}\\cdot {i}^{16}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Reduced form<\/strong><\/td>\r\n<td>[latex]{\\left({i}^{2}\\right)}^{17}\\cdot i[\/latex]<\/td>\r\n<td>[latex]{i}^{33}\\cdot \\left(-1\\right)[\/latex]<\/td>\r\n<td>[latex]{i}^{31}\\cdot 1[\/latex]<\/td>\r\n<td>[latex]{i}^{19}\\cdot {\\left({i}^{4}\\right)}^{4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Simplified form<\/strong><\/td>\r\n<td>[latex]{\\left(-1\\right)}^{17}\\cdot i[\/latex]<\/td>\r\n<td>[latex]-{i}^{33}[\/latex]<\/td>\r\n<td>[latex]{i}^{31}[\/latex]<\/td>\r\n<td>[latex]{i}^{19}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<em>Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.<\/em>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify and write the complex conjugate of a complex number.<\/li>\n<li>Divide complex numbers.<\/li>\n<li>Simplify powers of [latex]i[\/latex].<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>recall a technique: rationalizing the denominator<\/h3>\n<p>In mathematics, a technique that works well in one situation often works equally as well in a different, but stylistically similar situation.<\/p>\n<p>Dividing complex numbers uses a technique you used when rewriting\u00a0a fraction that contains a radical in the denominator. We called it <em>rationalizing a denominator.\u00a0<\/em>To accomplish it, you multiplied the fraction in the numerator and denominator by a number that would clear the radical in the denominator. We use the same general idea when dividing complex numbers. This time, instead of multiplying a radical by a specially chosen radical though, we&#8217;ll multiply a complex number by a specially chosen complex number.<\/p>\n<p>When rationalizing the denominator, we took advantage of the fact that [latex]\\left(\\sqrt{a}\\right)^2=a[\/latex].<\/p>\n<p>When dividing complex numbers, we&#8217;ll take advantage of the fact that [latex]i^2 = -1[\/latex].<\/p>\n<\/div>\n<h2>Dividing Complex Numbers<\/h2>\n<p>Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the <strong>complex conjugate<\/strong> of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of [latex]a+bi[\/latex] is [latex]a-bi[\/latex].<\/p>\n<p>Note that complex conjugates have a reciprocal relationship: The complex conjugate of [latex]a+bi[\/latex] is [latex]a-bi[\/latex], and the complex conjugate of [latex]a-bi[\/latex] is [latex]a+bi[\/latex]. Importantly, complex conjugate pairs have a special property. Their product is always real.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}(a+bi)(a-bi)&=a^2-abi+abi-b^2i^2\\\\[2mm]&=a^2-b^2(-1)\\\\[2mm]&=a^2+b^2\\end{align}[\/latex]<\/p>\n<p>Suppose we want to divide [latex]c+di[\/latex] by [latex]a+bi[\/latex], where neither [latex]a[\/latex] nor [latex]b[\/latex] equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{c+di}{a+bi}[\/latex] where [latex]a\\ne 0[\/latex] and [latex]b\\ne 0[\/latex].<\/p>\n<p>Multiply the numerator and denominator by the complex conjugate of the denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{\\left(c+di\\right)}{\\left(a+bi\\right)}\\cdot \\dfrac{\\left(a-bi\\right)}{\\left(a-bi\\right)}=\\dfrac{\\left(c+di\\right)\\left(a-bi\\right)}{\\left(a+bi\\right)\\left(a-bi\\right)}[\/latex]<\/p>\n<p>Apply the distributive property.<\/p>\n<p style=\"text-align: center;\">[latex]=\\dfrac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}}[\/latex]<\/p>\n<p>Simplify, remembering that [latex]{i}^{2}=-1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&=\\dfrac{ca-cbi+adi-bd\\left(-1\\right)}{{a}^{2}-abi+abi-{b}^{2}\\left(-1\\right)} \\\\[2mm] &=\\dfrac{\\left(ca+bd\\right)+\\left(ad-cb\\right)i}{{a}^{2}+{b}^{2}}\\end{align}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Complex Conjugate<\/h3>\n<p>The <strong>complex conjugate<\/strong> of a complex number [latex]a+bi[\/latex] is [latex]a-bi[\/latex]. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.<\/p>\n<ul>\n<li>When a complex number is multiplied by its complex conjugate, the result is a real number.<\/li>\n<li>When a complex number is added to its complex conjugate, the result is a real number.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding Complex Conjugates<\/h3>\n<p>Find the complex conjugate of each number.<\/p>\n<ol>\n<li>[latex]2+i\\sqrt{5}[\/latex]<\/li>\n<li>[latex]-\\frac{1}{2}i[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q349660\">Show Solution<\/span><\/p>\n<div id=\"q349660\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The number is already in the form [latex]a+bi[\/latex]. The complex conjugate is [latex]a-bi[\/latex], or [latex]2-i\\sqrt{5}[\/latex].<\/li>\n<li>We can rewrite this number in the form [latex]a+bi[\/latex] as [latex]0-\\frac{1}{2}i[\/latex]. The complex conjugate is [latex]a-bi[\/latex], or [latex]0+\\frac{1}{2}i[\/latex]. This can be written simply as [latex]\\frac{1}{2}i[\/latex].<\/li>\n<\/ol>\n<h4>Analysis of the Solution<\/h4>\n<p>Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by [latex]i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given two complex numbers, divide one by the other.<\/h3>\n<ol>\n<li>Write the division problem as a fraction.<\/li>\n<li>Determine the complex conjugate of the denominator.<\/li>\n<li>Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Dividing Complex Numbers<\/h3>\n<p>Divide [latex]\\left(2+5i\\right)[\/latex] by [latex]\\left(4-i\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q932961\">Show Solution<\/span><\/p>\n<div id=\"q932961\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by writing the problem as a fraction.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{\\left(2+5i\\right)}{\\left(4-i\\right)}[\/latex]<\/p>\n<p>Then we multiply the numerator and denominator by the complex conjugate of the denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{\\left(2+5i\\right)}{\\left(4-i\\right)}\\cdot \\dfrac{\\left(4+i\\right)}{\\left(4+i\\right)}[\/latex]<\/p>\n<p>To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\dfrac{\\left(2+5i\\right)}{\\left(4-i\\right)}\\cdot \\dfrac{\\left(4+i\\right)}{\\left(4+i\\right)}&=\\dfrac{8+2i+20i+5{i}^{2}}{16+4i - 4i-{i}^{2}}\\\\[2mm] &=\\dfrac{8+2i+20i+5\\left(-1\\right)}{16+4i - 4i-\\left(-1\\right)} && \\text{Because } {i}^{2}=-1 \\\\[2mm] &=\\frac{3+22i}{17} \\\\[2mm] &=\\dfrac{3}{17}+\\frac{22}{17}i && \\text{Separate real and imaginary parts}.\\end{align}[\/latex]<\/p>\n<p>Note that this expresses the quotient in standard form.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm61715\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=61715&theme=oea&iframe_resize_id=ohm61715&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>We have seen that we can evaluate functions for any real number or algebraic expression we choose. For example,<\/p>\n<p>given the function [latex]f(x)=2x-7[\/latex],<\/p>\n<p>[latex]f(-1) = 2(-1) -7 = -9[\/latex]<\/p>\n<p>[latex]f(5+h)=2(5+h)-7 = 10 +2h - 7 = 3 +2h[\/latex].<\/p>\n<p>We may also evaluate a function for a complex number if we wish, as shown in the Example and Try It boxes below.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Substituting a Complex Number into a Polynomial Function<\/h3>\n<p>Let [latex]f\\left(x\\right)={x}^{2}-5x+2[\/latex]. Evaluate [latex]f\\left(3+i\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q323196\">Show Solution<\/span><\/p>\n<div id=\"q323196\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute [latex]x=3+i[\/latex] into the function [latex]f\\left(x\\right)={x}^{2}-5x+2[\/latex] and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}f(3+i)&=(3+i)^2-5(3+i)+2&&\\text{Substitute } 3+i \\text{ for }x\\\\[2mm]&=(3+6i+i^2)-(15+5i)+2&&\\text{Multiply}\\\\[2mm]&=9+6i+(-1)-15-5i+2&&\\text{Substitute }-1\\text{ for }i^2 \\\\[2mm]&=-5+i&&\\text{Combine like terms}\\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We write [latex]f\\left(3+i\\right)=-5+i[\/latex]. Notice that the input is [latex]3+i[\/latex] and the output is [latex]-5+i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]f\\left(x\\right)=2{x}^{2}-3x[\/latex]. Evaluate [latex]f\\left(8-i\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q324665\">Show Solution<\/span><\/p>\n<div id=\"q324665\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]102 - 29i[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm120193\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=120193&theme=oea&iframe_resize_id=ohm120193&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Substituting an Imaginary Number in a Rational Function<\/h3>\n<p>Let [latex]f\\left(x\\right)=\\dfrac{2+x}{x+3}[\/latex]. Evaluate [latex]f\\left(10i\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q462657\">Show Solution<\/span><\/p>\n<div id=\"q462657\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute [latex]x=10i[\/latex] and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\dfrac{2+10i}{10i+3} && \\text{Substitute }10i\\text{ for }x\\\\[2mm] &\\dfrac{2+10i}{3+10i} && \\text{Rewrite the denominator in standard form}\\\\[2mm] &\\dfrac{2+10i}{3+10i}\\cdot \\dfrac{3 - 10i}{3 - 10i} && \\text{Multiply the numerator and denominator by the complex conjugate of the denominator}\\\\[2mm] &\\dfrac{6 - 20i+30i - 100{i}^{2}}{9 - 30i+30i - 100{i}^{2}} && \\text{Multiply using the distributive property or the FOIL method} \\\\[2mm] &\\dfrac{6 - 20i+30i - 100\\left(-1\\right)}{9 - 30i+30i - 100\\left(-1\\right)} && \\text{Substitute }-1\\text{ for } {i}^{2} \\\\[2mm] &\\dfrac{106+10i}{109} && \\text{Simplify} \\\\[2mm] &\\dfrac{106}{109}+\\dfrac{10}{109}i && \\text{Separate the real and imaginary parts} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]f\\left(x\\right)=\\dfrac{x+1}{x - 4}[\/latex]. Evaluate [latex]f\\left(-i\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q453508\">Show Solution<\/span><\/p>\n<div id=\"q453508\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-\\dfrac{3}{17}+\\dfrac{5}{17}i[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Dividing Complex Numbers\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/XBJjbJAwM1c?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Simplifying Powers of [latex]i[\/latex]<\/h2>\n<p>The powers of [latex]i[\/latex] are cyclic. Let\u2019s look at what happens when we raise [latex]i[\/latex] to increasing powers.<\/p>\n<p>[latex]{i}^{1}=i[\/latex]<br \/>\n[latex]{i}^{2}=-1[\/latex]<br \/>\n[latex]{i}^{3}={i}^{2}\\cdot i=-1\\cdot i=-i[\/latex]<br \/>\n[latex]{i}^{4}={i}^{3}\\cdot i=-i\\cdot i=-{i}^{2}=-\\left(-1\\right)=1[\/latex]<br \/>\n[latex]{i}^{5}={i}^{4}\\cdot i=1\\cdot i=i[\/latex]<\/p>\n<p>We can see that when we get to the fifth power of [latex]i[\/latex], it is equal to the first power. As we continue to multiply [latex]i[\/latex]\u00a0by itself for increasing powers, we will see a cycle of 4. Let\u2019s examine the next 4 powers of [latex]i[\/latex].<\/p>\n<p>[latex]{i}^{6}={i}^{5}\\cdot i=i\\cdot i={i}^{2}=-1[\/latex]<br \/>\n[latex]{i}^{7}={i}^{6}\\cdot i={i}^{2}\\cdot i={i}^{3}=-i[\/latex]<br \/>\n[latex]{i}^{8}={i}^{7}\\cdot i={i}^{3}\\cdot i={i}^{4}=1[\/latex]<br \/>\n[latex]{i}^{9}={i}^{8}\\cdot i={i}^{4}\\cdot i={i}^{5}=i[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>recall properties of exponents<\/h3>\n<p>Recall that [latex]\\left(a^m\\right)^n=a^{mn}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Simplifying Powers of [latex]i[\/latex]<\/h3>\n<p>Evaluate [latex]{i}^{35}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q879646\">Show Solution<\/span><\/p>\n<div id=\"q879646\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since [latex]{i}^{4}=1[\/latex], we can simplify the problem by factoring out as many factors of [latex]{i}^{4}[\/latex] as possible. To do so, first determine how many times [latex]4[\/latex] goes into [latex]35[\/latex]: [latex]35=4\\cdot 8+3[\/latex].<\/p>\n<p>[latex]{i}^{35}={i}^{4\\cdot 8+3}={i}^{4\\cdot 8}\\cdot {i}^{3}={\\left({i}^{4}\\right)}^{8}\\cdot {i}^{3}={1}^{8}\\cdot {i}^{3}={i}^{3}=-i[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Can we write [latex]{i}^{35}[\/latex] in other helpful ways?<\/strong><\/p>\n<p><em>As we saw in Example: Simplifying Powers of <\/em>[latex]i[\/latex]<em>, we reduced [latex]{i}^{35}[\/latex] to [latex]{i}^{3}[\/latex] by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of [latex]{i}^{35}[\/latex] may be more useful. The table below\u00a0shows some other possible factorizations.<\/em><\/p>\n<table summary=\"..\">\n<tbody>\n<tr>\n<td><strong>Factorization of [latex]{i}^{35}[\/latex]<\/strong><\/td>\n<td>[latex]{i}^{34}\\cdot i[\/latex]<\/td>\n<td>[latex]{i}^{33}\\cdot {i}^{2}[\/latex]<\/td>\n<td>[latex]{i}^{31}\\cdot {i}^{4}[\/latex]<\/td>\n<td>[latex]{i}^{19}\\cdot {i}^{16}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Reduced form<\/strong><\/td>\n<td>[latex]{\\left({i}^{2}\\right)}^{17}\\cdot i[\/latex]<\/td>\n<td>[latex]{i}^{33}\\cdot \\left(-1\\right)[\/latex]<\/td>\n<td>[latex]{i}^{31}\\cdot 1[\/latex]<\/td>\n<td>[latex]{i}^{19}\\cdot {\\left({i}^{4}\\right)}^{4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Simplified form<\/strong><\/td>\n<td>[latex]{\\left(-1\\right)}^{17}\\cdot i[\/latex]<\/td>\n<td>[latex]-{i}^{33}[\/latex]<\/td>\n<td>[latex]{i}^{31}[\/latex]<\/td>\n<td>[latex]{i}^{19}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><em>Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.<\/em><\/p>\n<\/div>\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-190\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex: Dividing Complex Numbers. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/XBJjbJAwM1c\">https:\/\/youtu.be\/XBJjbJAwM1c<\/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":395986,"menu_order":27,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex: Dividing Complex Numbers\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/XBJjbJAwM1c\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"9b67846d-4636-49f6-8785-1f1f09c24de5","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-190","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/190","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/190\/revisions"}],"predecessor-version":[{"id":761,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/190\/revisions\/761"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/190\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=190"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=190"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=190"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=190"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}