{"id":206,"date":"2023-11-08T16:10:25","date_gmt":"2023-11-08T16:10:25","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/16-4-1-complex-numbers\/"},"modified":"2025-03-31T23:59:31","modified_gmt":"2025-03-31T23:59:31","slug":"5-9-complex-numbers","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/5-9-complex-numbers\/","title":{"raw":"5.9 Imaginary and Complex Numbers","rendered":"5.9 Imaginary and Complex Numbers"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Rewrite square roots with negative radicands in terms of [latex]i.[\/latex]<\/li>\r\n \t<li>State the real and imaginary parts of a complex number.<\/li>\r\n \t<li>Multiply radical expressions with negative radicands.<\/li>\r\n \t<li>Add and\/or subtract complex numbers, giving the result in the form [latex]a+bi.[\/latex]<\/li>\r\n \t<li>Multiply complex numbers, giving the result in the form [latex]a+bi.[\/latex]<\/li>\r\n \t<li>Simplify whole number powers of [latex]i.[\/latex]<\/li>\r\n \t<li>Divide complex numbers including using complex conjugates, giving the result in the form [latex]a+bi.[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\nWe have noted many times in this chapter that square roots of negative numbers do not exist as real numbers. However, these \"non-real numbers\" do still have applications in many areas of math and also in the real world in fields such as electrical engineering.\r\n\r\nYou really need only one new number to start working with the square roots of negative numbers. That number is the square root of [latex]\u22121,\\sqrt{-1}[\/latex]. The <i>real numbers<\/i> are those that can be shown on a number line.\u00a0When something is not real, we often say it is <i>imaginary<\/i>. So let us call this new number [latex]i[\/latex]\u00a0and use it<i> <\/i>to represent the square root of [latex]\u22121[\/latex].\r\n<p style=\"text-align: center;\">[latex] i=\\sqrt{-1}[\/latex]<\/p>\r\nBecause [latex] \\sqrt{x}\\,\\cdot \\,\\sqrt{x}=x[\/latex], we can also see that [latex] \\sqrt{-1}\\,\\cdot \\,\\sqrt{-1}=-1,[\/latex] or [latex] i^2=-1[\/latex]. Another way to say this is that [latex]i[\/latex] is a solution to the equation [latex]x^2=-1[\/latex] which previously did not have solutions.\r\n\r\nThe number [latex]i[\/latex]<i>\u00a0<\/i>allows us to work with roots of all negative numbers, not just [latex] \\sqrt{-1}[\/latex]. We will be using our previous Product Rule for Radicals,\u00a0[latex] \\sqrt{ab}=\\sqrt{a}\\sqrt{b}[\/latex]. <strong>Be very careful using this rule with imaginary numbers though!<\/strong> Remember when we introduced the rule, we required that both [latex]\\sqrt{a}[\/latex] and [latex]\\sqrt{b}[\/latex] exist, which is not the case in this section! For this section we redefine the product rule to work for imaginary numbers.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>product rule for imaginary numbers<\/h3>\r\nFor any positive number [latex]a[\/latex], define\r\n<p style=\"text-align: center;\">[latex]\\sqrt{-a}=\\sqrt{-1\\cdot a}=\\sqrt{-1}\\cdot\\sqrt{a}=i\\cdot\\sqrt{a}[\/latex]<\/p>\r\n\r\n<\/div>\r\nLet us try an example of using this rule. <strong>If a radicand of a square root is negative, always apply this rule first before any other steps.<\/strong>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] \\sqrt{-4}[\/latex]\r\n\r\n[reveal-answer q=\"793555\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"793555\"]\r\n\r\nUse the new product rule to rewrite the radical.\r\n<p style=\"text-align: center;\">[latex] \\sqrt{-4}=\\sqrt{-1\\cdot 4}=\\sqrt{-1}\\sqrt{4}=\\sqrt{-1}\\cdot 2[\/latex]<\/p>\r\nUse the definition of [latex]i[\/latex]\u00a0to rewrite [latex] \\sqrt{-1}[\/latex] as [latex]i[\/latex]<i>.<\/i>\r\n<p style=\"text-align: center;\">[latex] \\sqrt{-1}\\cdot 2=2i[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] \\sqrt{-18}[\/latex]\r\n\r\n[reveal-answer q=\"760057\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"760057\"]\r\n\r\nUse the new product rule to rewrite the radical.\r\n\r\n[latex] \\sqrt{-18}=\\sqrt{-1\\cdot 18}=\\sqrt{-1}\\sqrt{18}[\/latex]\r\n\r\nSince\u00a0[latex]18[\/latex] is <i>not<\/i> a perfect square, simplify the radical as done previously in this chapter.\r\n\r\n[latex] \\sqrt{-1}\\sqrt{18}=\\sqrt{-1}\\sqrt{9}\\sqrt{2}=\\sqrt{-1}3\\sqrt{2}=3\\sqrt{2} \\; i[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show more examples of how to use imaginary numbers to simplify a square root with a negative radicand.\r\n\r\nhttps:\/\/youtu.be\/LSp7yNP6Xxc\r\n<div class=\"textbox shaded\">\r\n<h3>Rewriting the Square Root of a Negative Number<\/h3>\r\n<ul>\r\n \t<li>Rewrite the radical using the rule [latex] \\sqrt{-a}=\\sqrt{-1}\\cdot \\sqrt{a}[\/latex].<\/li>\r\n \t<li>Rewrite [latex] \\sqrt{-1}[\/latex] as [latex]i[\/latex] and simplify the other radical if possible.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Complex Numbers<\/h2>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>COMPLEX NUMBERS<\/h3>\r\nA <strong>complex number<\/strong>\u00a0is any number that can be expressed in the standard form [latex]a+ bi[\/latex]\u00a0where [latex]a[\/latex] is the real part and [latex]b[\/latex] is the imaginary part. We can say that a complex number is the sum of a real number\u00a0[latex]a[\/latex] and a <strong>pure imaginary number<\/strong>\u00a0[latex]bi.[\/latex]\r\n\r\n<\/div>\r\nFor example, [latex]5+2i[\/latex] is a complex number. So, too, is [latex]-3+4i\\sqrt{3}[\/latex]. Here are some examples of complex numbers, with real part and imaginary part labeled. Note that [latex]i[\/latex] is not included in the imaginary part.\r\n<table cellspacing=\"0\" cellpadding=\"0\">\r\n<thead>\r\n<tr>\r\n<th>Complex Number<\/th>\r\n<th>Real Part<\/th>\r\n<th>Imaginary Part<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]3+7i[\/latex]<\/td>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]18\u201332i[\/latex]<\/td>\r\n<td>[latex]18[\/latex]<\/td>\r\n<td>[latex]\u221232[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] -\\frac{3}{5}+i\\sqrt{2}[\/latex]<\/td>\r\n<td>[latex] -\\frac{3}{5}[\/latex]<\/td>\r\n<td>[latex] \\sqrt{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] \\frac{\\sqrt{2}}{2}-\\frac{1}{2}i[\/latex]<\/td>\r\n<td>[latex] \\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]-\\frac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIn a number with a radical as part of [latex]b[\/latex], such as [latex]-\\frac{3}{5}+i\\sqrt{2}[\/latex]\u00a0above, it is acceptable to write the imaginary unit [latex]i[\/latex] in front of the radical. Though writing this number as [latex] -\\frac{3}{5}+\\sqrt{2} \\; i[\/latex] is technically correct, it makes it more difficult to tell whether\u00a0[latex]i[\/latex]\u00a0is inside or outside of the radical. Putting it before the radical, as in [latex] -\\frac{3}{5}+i\\sqrt{2}[\/latex], clears up any confusion. While usually we will require answers to be in the form of a complex number [latex]a+bi[\/latex], we make an exception if [latex]b[\/latex] is a radical. Look at these last two examples.\r\n<table cellspacing=\"0\" cellpadding=\"0\">\r\n<thead>\r\n<tr>\r\n<th>Number<\/th>\r\n<th>Complex Form:\r\n[latex]a+bi[\/latex]<\/th>\r\n<th>Real Part<\/th>\r\n<th>Imaginary Part<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]17[\/latex]<\/td>\r\n<td>[latex]17+0i[\/latex]<\/td>\r\n<td>[latex]17[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u22123i[\/latex]<\/td>\r\n<td>[latex]0\u20133i[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]\u22123[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nBy making [latex]b=0[\/latex], any real number can be expressed as a complex number. The real number [latex]a[\/latex] is written as [latex]a+0i[\/latex] in complex form. Similarly, any pure imaginary number can be expressed as a complex number. By making [latex]a=0[\/latex], any pure imaginary number [latex]bi[\/latex] can be written as [latex]0+bi[\/latex] in complex form.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite [latex]83.6[\/latex] in the form of a complex number.\r\n\r\n[reveal-answer q=\"704457\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"704457\"]\r\n\r\nRemember that a complex number has the form [latex]a+bi[\/latex]. You need to figure out what\u00a0[latex]a[\/latex]\u00a0and [latex]b[\/latex] need to be. Since [latex]83.6[\/latex] is a real number, it\u00a0does not contain any imaginary parts, so the value of [latex]b[\/latex] is\u00a0[latex]0[\/latex].\r\n\r\nThe answer is [latex]83.6+0i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite [latex]\u22123i[\/latex] in the form of a complex number.\r\n\r\n[reveal-answer q=\"451549\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"451549\"]\r\n\r\nRemember that a complex number has the form [latex]a+bi[\/latex].\u00a0You need to figure out what [latex]a[\/latex] and [latex]b[\/latex] need to be. Since [latex]\u22123i[\/latex] is a pure imaginary number, the value of [latex]a[\/latex] is [latex]0[\/latex].\r\n\r\nThe answer is\u00a0[latex]0-3i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn practice if the real or imaginary part of a complex number is [latex]0[\/latex] we will not write that part at all (so we would leave it as [latex]-3i[\/latex] rather than writing the real part of [latex]0[\/latex]).\r\n<h2>Adding and Subtracting Complex Numbers<\/h2>\r\nOur goal for the rest of this section is to prove the claim that <strong>the complex numbers form a number system just like the real numbers<\/strong> - they can be added, subtracted, multiplied, and divided. We will need some of these skills in the next chapter. In order for this claim to be true, we must be able to write each final answer as a complex number, in the form [latex]a+bi.[\/latex]\r\n\r\nFirst, consider the following expression.\r\n<p style=\"text-align: center;\">[latex](7+6i)+(2+4i)[\/latex]<\/p>\r\nIf we think of [latex]i[\/latex] as a variable for a moment, we just combine the like terms,\r\n<p style=\"text-align: center;\">[latex](7+6i)+(2+4i)=9+10i[\/latex]<\/p>\r\nThe same rules apply to adding and subtracting complex numbers. You combine the corresponding parts and write the final answer again in the form [latex]a+bi.[\/latex]\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nAdd.\u00a0[latex](\u22123+3i)+(7\u20132i)[\/latex]\r\n\r\n[reveal-answer q=\"929105\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"929105\"]\r\n\r\nRearrange the sums to put like terms together.\r\n<p style=\"text-align: center;\">[latex]\u22123+3i+7\u20132i=\u22123+7+3i\u20132i[\/latex]<\/p>\r\nCombine like terms.\r\n<p style=\"text-align: center;\">[latex]\u22123+7=4[\/latex]<\/p>\r\n<p style=\"text-align: center;\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]3i\u20132i=(3\u20132)i=i[\/latex]<\/p>\r\nThe answer is [latex]4+i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSubtract.\u00a0[latex](\u22123+3i)\u2013(7\u20132i)[\/latex]\r\n\r\n[reveal-answer q=\"203125\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"203125\"]\r\n\r\nBe sure to distribute the subtraction sign to all terms in the set of parentheses that follows.\r\n<p style=\"text-align: center;\">[latex](\u22123+3i)\u2013(7\u20132i)=\u22123+3i\u20137+2i[\/latex]<i>\u00a0<\/i><\/p>\r\nRearrange the terms to put like terms together.\r\n<p style=\"text-align: center;\">[latex]\u22123\u20137+3i+2i[\/latex]<\/p>\r\nCombine like terms.\r\n<p style=\"text-align: center;\">[latex]\u22123\u20137=\u221210[\/latex]<\/p>\r\n<p style=\"text-align: center;\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]3i+2i=(3+2)i=5i[\/latex]<\/p>\r\nThe answer is [latex]-10+5i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show more examples of how to add and subtract complex numbers (Note in this video that it incorrectly calls [latex]bi[\/latex] the imaginary part instead of just [latex]b[\/latex]).\r\n\r\nhttps:\/\/youtu.be\/SGhTjioGqqA\r\n<h2>Multiplying Radicals with Negative Radicands<\/h2>\r\nWe have to be careful when multiplying radicals with negative radicands. Remember to use the product rule for imaginary numbers first before any other steps.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nMultiply and simplify [latex]\\sqrt{-8}\\cdot\\sqrt{-12}.[\/latex] Write answers in the form [latex]a+bi[\/latex]\r\n\r\n[reveal-answer q=\"676845\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"676845\"]\r\n\r\nFirst rewrite each radical using the product rule for imaginary numbers.\r\n\r\n[latex]\\begin{align}&amp;\\quad\\sqrt{-8}\\cdot\\sqrt{-12}\\\\\r\n=&amp;\\quad\\sqrt{-1}\\cdot\\sqrt{8}\\cdot\\sqrt{-1}\\cdot\\sqrt{12}\\\\\r\n=&amp;\\quad i\\cdot i\\cdot\\sqrt{8\\cdot 12}\\\\\r\n=&amp;\\quad i^2\\sqrt{96}\\\\\r\n=&amp;\\quad -\\sqrt{96}\\end{align}[\/latex]\r\n\r\nThe radical can now be simplified normally.\r\n\r\n[latex]\\begin{align}=&amp;\\quad -\\sqrt{16}\\sqrt{6}\\\\\r\n=&amp;\\quad -4\\sqrt{6}\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n\r\n<img class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01182614\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"37\" height=\"33\" \/>Notice in the previous example, it would be <span style=\"color: #ff0000;\">incorrect<\/span>\u00a0to use our previous product rule for radicals to multiply the radicands:\u00a0 [latex]\\sqrt{-8}\\cdot\\sqrt{-12}\\color{red}{\\neq \\sqrt{-8\\cdot -12}=\\sqrt{96}}.[\/latex] Doing so would cause us to miss the negative sign that should have appeared upon simplifying [latex]i^2.[\/latex]\r\n\r\n<\/div>\r\n<h2>Multiplying Complex Numbers<\/h2>\r\n<section id=\"fs-id1165137417169\">\r\n<p id=\"fs-id1165137832911\">The process of multiplying complex numbers is again similar to the corresponding process for polynomials. The only difference is that we should replace [latex]i^2=-1[\/latex] if it appears.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nMultiply and simplify [latex]3i(-6+5i).[\/latex] Write answers in the form [latex]a+bi.[\/latex]\r\n\r\n[reveal-answer q=\"603325\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"603325\"]\r\n<p style=\"text-align: center;\">We can use the distributive property.\r\n[latex]\\begin{align}&amp;\\quad 3i(-6+5i)\\\\\r\n=&amp;\\quad 3i\\cdot (-6) + 3i\\cdot 5i&amp;&amp; \\color{blue}{\\textsf{distribute}}\\\\\r\n=&amp;\\quad -18i+15i^2\\\\\r\n=&amp;\\quad -18i+15(-1) &amp;&amp; \\color{blue}{\\textsf{substitute }i^2=-1}\\\\\r\n=&amp;\\quad -15-18i&amp;&amp; \\color{blue}{\\textsf{write in the form }a+bi} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137650841\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nMultiply and simplify [latex]\\left(4+3i\\right)\\left(2 - 5i\\right).[\/latex] Write answers in the form [latex]a+bi.[\/latex]\r\n[reveal-answer q=\"188458\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"188458\"]\r\n<p id=\"fs-id1165137459488\">Multiply (using FOIL, for example).<\/p>\r\n\r\n<div id=\"eip-id1165137762412\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll}\\left(4+3i\\right)\\left(2 - 5i\\right) &amp; =\\left(4\\cdot 2\\right)+\\left(4\\cdot \\left(-5i\\right)\\right)+\\left(3i\\cdot 2\\right)+\\left(\\left(3i\\right)\\cdot \\left(-5i\\right)\\right) &amp; \\\\ &amp; =8-20i+6i-15i^2 &amp; \\\\ &amp; = 8-20i+6i-15\\left(-1\\right) &amp; \\color{blue}{\\textsf{substitute } i^2=-1} \\\\ &amp; = 8-20i+6i+15 &amp; \\\\ &amp; = \\left(8+15\\right)+\\left(-20+6\\right)i &amp; \\color{blue}{\\textsf{group like terms together}} \\\\ &amp; =23-14i &amp; \\end{array}[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the first video, we show more examples of multiplying complex numbers.\r\n\r\nhttps:\/\/youtu.be\/Fmr3o2zkwLM\r\n<h2>Simplifying Powers of\u00a0[latex]i[\/latex]<\/h2>\r\n<p id=\"fs-id1165132919554\">There is a pattern to evaluating powers of [latex]i[\/latex]. Let us look at what happens when we raise\u00a0[latex]i[\/latex]\u00a0to increasing powers.<\/p>\r\n\r\n<div id=\"eip-783\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{rl}{i}^{1}&amp;=\\quad i\\\\ {i}^{2}&amp;=\\quad -1\\\\ {i}^{3}&amp;=\\quad {i}^{2}\\cdot i=-1\\cdot i=-i\\\\ {i}^{4}&amp;=\\quad {i}^{3}\\cdot i=-i\\cdot i=-{i}^{2}=-\\left(-1\\right)=1\\\\ {i}^{5}&amp;=\\quad {i}^{4}\\cdot i=1\\cdot i=i\\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n<div class=\"equation unnumbered\">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\u00a0[latex]i[\/latex]\u00a0by itself for increasing powers, we will see a cycle of\u00a0four. Here are the next four powers of [latex]i[\/latex].<\/div>\r\n<div><\/div>\r\n<div id=\"eip-477\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{rl}{i}^{6}&amp;=\\quad {i}^{5}\\cdot i=i\\cdot i={i}^{2}=-1\\\\ {i}^{7}&amp;= \\quad{i}^{6}\\cdot i={i}^{2}\\cdot i={i}^{3}=-i\\\\ {i}^{8}&amp;=\\quad {i}^{7}\\cdot i={i}^{3}\\cdot i={i}^{4}=1\\\\ {i}^{9}&amp;=\\quad {i}^{8}\\cdot i={i}^{4}\\cdot i={i}^{5}=i\\end{array}[\/latex]<\/div>\r\n<div class=\"equation unnumbered\"><\/div>\r\n<div class=\"equation unnumbered\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nEvaluate [latex]{i}^{35}[\/latex].\r\n[reveal-answer q=\"295805\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"295805\"]\r\n<p id=\"fs-id1165137728290\">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\u00a0[latex]4[\/latex] goes into\u00a0[latex]35[\/latex]: [latex]35=4\\cdot 8+3[\/latex]. The remainder of [latex]3[\/latex] is the most important part.<\/p>\r\n\r\n<div id=\"eip-id1165134069265\" class=\"equation unnumbered\">[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]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nThe previous example illustrates the easiest way to simplify a large power of [latex]i[\/latex]. Rewrite your old exponent as a multiple of [latex]4[\/latex] plus a remainder. After some simplifying, the final power of\u00a0[latex]i[\/latex] is just the remainder. Then you just need to know the first four powers of\u00a0[latex]i.[\/latex]\r\n\r\nIn the following video, you will see more examples of how to simplify powers of [latex]i.[\/latex]\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=XyksedEntD4\r\n<h2>Dividing Complex Numbers<\/h2>\r\n<p id=\"fs-id1165137612241\">If we write a division of complex numbers in radical form, it should look similar to something we've done earlier in this chapter.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{2+3i}{4-i}=\\frac{2+3\\sqrt{-1}}{4-\\sqrt{-1}}[\/latex]<\/p>\r\nWe have previously removed radicals from the denominator by rationalizing the denominator using the conjugate. To eliminate the imaginary number [latex]i[\/latex] in the denominator, you multiply by\u00a0the <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<div class=\"textbox key-takeaways\">\r\n<h3>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. When a complex number is multiplied by its complex conjugate, the result is a real number.\r\n\r\n<\/div>\r\nIn our previous example, the denominator is [latex]4-i.[\/latex] The complex conjugate is [latex]4+i.[\/latex] Multiplying, we get\u00a0[latex](4-i)(4+i)=16+4i-4i-i^2=16-(-1)=17.[\/latex] Let's use this to finish our division problem.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\quad\\frac{2+3i}{4-i}\\\\\r\n=&amp;\\quad\\frac{2+3i}{4-i}\\cdot\\frac{4+i}{4+i}\\\\\r\n=&amp;\\quad\\frac{(2+3i)(4+i)}{(4-i)(4+i)}\\\\\r\n=&amp;\\quad\\frac{8+2i+12i+3i^2}{17}&amp;&amp;\\color{blue}{\\textsf{FOIL numerator and denominator}}\\\\\r\n=&amp;\\quad\\frac{8+2i+12i+3(-1)}{17}\\\\\r\n=&amp;\\quad\\frac{5+14i}{17}&amp;&amp;\\color{blue}{\\textsf{simplify the numerator}}\\\\\r\n=&amp;\\quad\\frac{5}{17}+\\frac{14}{17}i&amp;&amp;\\color{blue}{\\textsf{write in the form }a+bi}\\end{align}[\/latex]<\/p>\r\nWe are now done, and most importantly the final answer is in the form of a complex number, [latex]a+bi.[\/latex]\r\n<div class=\"textbox shaded\">\r\n<h3>DIVIDING COMPLEX NUMBERS<\/h3>\r\n<ul>\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 and simplify.<\/li>\r\n \t<li>Write the answer as [latex]a+bi.[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nDivide [latex]\\left(6-5i\\right)[\/latex] by [latex]\\left(1+3i\\right)[\/latex]. Write answers in the form [latex]a+bi.[\/latex]\r\n[reveal-answer q=\"665746\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"665746\"]\r\n<p id=\"fs-id1165137605861\">We begin by writing the problem as a fraction.<\/p>\r\n\r\n<div id=\"eip-id1165134234232\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Large\\frac{\\left(6-5i\\right)}{\\left(1+3i\\right)}[\/latex]<\/div>\r\n<p id=\"fs-id1165137639613\">Then we multiply the numerator and denominator by the complex conjugate of the denominator.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcll}\\Large{=\\frac{(6-5i)}{(1+3i)}}\\cdot\\frac{(1-3i)}{(1-3i)}&amp;=&amp;\\Large{\\frac{6-18i-5i+15{i}^{2}}{1-3i+3i-9{i}^{2}}}&amp; \\\\&amp;=&amp;\\Large{\\frac{6-18i-5i+15(-1)}{1-3i+3i-9(-1)}}&amp;\\quad\\color{blue}{\\textsf{substitute }i^2=-1}\\\\&amp;=&amp;\\Large{\\frac{-9-23i}{10}}&amp;\\quad\\color{blue}{\\textsf{combine like terms}} \\\\&amp;=&amp;\\Large{-\\frac{9}{10}-\\frac{23}{10}i}&amp;\\quad\\color{blue}{\\textsf{write in the form }a+bi}\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nDivide [latex]\\dfrac{6-8i}{3i}[\/latex]. Write answers in the form [latex]a+bi.[\/latex]\r\n\r\n[reveal-answer q=\"634438\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"634438\"]\r\n<div id=\"eip-id1165134234232\" class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\r\n<p id=\"fs-id1165137639613\">We could multiply the numerator and denominator by the complex conjugate, but since the denominator is a monomial it is easier to just multiply by [latex]-i.[\/latex] You will see this still results in the denominator being a real number.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{6-8i}{3i}&amp;=\\frac{\\left(6-8i\\right)}{3i}\\cdot\\frac{-i}{-i}\\\\\r\n&amp;=\\frac{-6i+8i^2}{-3i^2}\\\\\r\n&amp;=\\frac{-6i+8(-1)}{-3(-1)}\\\\\r\n&amp;=\\frac{-6i-8}{3}\\\\\r\n&amp;=\\frac{-6i}{3}+\\frac{-8}{3}\\\\\r\n&amp;=-\\frac{8}{3}-2i\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the last video, you will see more examples of dividing complex numbers.\r\n\r\nhttps:\/\/youtu.be\/XBJjbJAwM1c\r\n\r\n<\/section>\r\n<h2>Summary<\/h2>\r\nComplex numbers have the form [latex]a+bi[\/latex], where [latex]a[\/latex] and [latex]b[\/latex] are real numbers and [latex]i[\/latex] is the square root of [latex]\u22121[\/latex]. All real numbers can be written as complex numbers by setting [latex]b=0[\/latex]. Square roots of negative numbers can be simplified using [latex] \\sqrt{-a}=i\\sqrt{a}[\/latex] where [latex]a&gt;0.[\/latex] We can perform arithmetic operations on complex numbers in much the same way as working with polynomials and combining like terms, making sure to simplify [latex]i^2=-1[\/latex] when appropriate. Powers of\u00a0[latex]i[\/latex] form a cycle that repeats every four powers. Division of complex numbers is done by first making the denominator a real number using the complex conjugate multiplied on the top and bottom of the fraction.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Rewrite square roots with negative radicands in terms of [latex]i.[\/latex]<\/li>\n<li>State the real and imaginary parts of a complex number.<\/li>\n<li>Multiply radical expressions with negative radicands.<\/li>\n<li>Add and\/or subtract complex numbers, giving the result in the form [latex]a+bi.[\/latex]<\/li>\n<li>Multiply complex numbers, giving the result in the form [latex]a+bi.[\/latex]<\/li>\n<li>Simplify whole number powers of [latex]i.[\/latex]<\/li>\n<li>Divide complex numbers including using complex conjugates, giving the result in the form [latex]a+bi.[\/latex]<\/li>\n<\/ul>\n<\/div>\n<p>We have noted many times in this chapter that square roots of negative numbers do not exist as real numbers. However, these &#8220;non-real numbers&#8221; do still have applications in many areas of math and also in the real world in fields such as electrical engineering.<\/p>\n<p>You really need only one new number to start working with the square roots of negative numbers. That number is the square root of [latex]\u22121,\\sqrt{-1}[\/latex]. The <i>real numbers<\/i> are those that can be shown on a number line.\u00a0When something is not real, we often say it is <i>imaginary<\/i>. So let us call this new number [latex]i[\/latex]\u00a0and use it<i> <\/i>to represent the square root of [latex]\u22121[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]i=\\sqrt{-1}[\/latex]<\/p>\n<p>Because [latex]\\sqrt{x}\\,\\cdot \\,\\sqrt{x}=x[\/latex], we can also see that [latex]\\sqrt{-1}\\,\\cdot \\,\\sqrt{-1}=-1,[\/latex] or [latex]i^2=-1[\/latex]. Another way to say this is that [latex]i[\/latex] is a solution to the equation [latex]x^2=-1[\/latex] which previously did not have solutions.<\/p>\n<p>The number [latex]i[\/latex]<i>\u00a0<\/i>allows us to work with roots of all negative numbers, not just [latex]\\sqrt{-1}[\/latex]. We will be using our previous Product Rule for Radicals,\u00a0[latex]\\sqrt{ab}=\\sqrt{a}\\sqrt{b}[\/latex]. <strong>Be very careful using this rule with imaginary numbers though!<\/strong> Remember when we introduced the rule, we required that both [latex]\\sqrt{a}[\/latex] and [latex]\\sqrt{b}[\/latex] exist, which is not the case in this section! For this section we redefine the product rule to work for imaginary numbers.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>product rule for imaginary numbers<\/h3>\n<p>For any positive number [latex]a[\/latex], define<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{-a}=\\sqrt{-1\\cdot a}=\\sqrt{-1}\\cdot\\sqrt{a}=i\\cdot\\sqrt{a}[\/latex]<\/p>\n<\/div>\n<p>Let us try an example of using this rule. <strong>If a radicand of a square root is negative, always apply this rule first before any other steps.<\/strong><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\sqrt{-4}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q793555\">Show Solution<\/span><\/p>\n<div id=\"q793555\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the new product rule to rewrite the radical.<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{-4}=\\sqrt{-1\\cdot 4}=\\sqrt{-1}\\sqrt{4}=\\sqrt{-1}\\cdot 2[\/latex]<\/p>\n<p>Use the definition of [latex]i[\/latex]\u00a0to rewrite [latex]\\sqrt{-1}[\/latex] as [latex]i[\/latex]<i>.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{-1}\\cdot 2=2i[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\sqrt{-18}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q760057\">Show Solution<\/span><\/p>\n<div id=\"q760057\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the new product rule to rewrite the radical.<\/p>\n<p>[latex]\\sqrt{-18}=\\sqrt{-1\\cdot 18}=\\sqrt{-1}\\sqrt{18}[\/latex]<\/p>\n<p>Since\u00a0[latex]18[\/latex] is <i>not<\/i> a perfect square, simplify the radical as done previously in this chapter.<\/p>\n<p>[latex]\\sqrt{-1}\\sqrt{18}=\\sqrt{-1}\\sqrt{9}\\sqrt{2}=\\sqrt{-1}3\\sqrt{2}=3\\sqrt{2} \\; i[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show more examples of how to use imaginary numbers to simplify a square root with a negative radicand.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Simplify Square Roots to Imaginary Numbers\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/LSp7yNP6Xxc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox shaded\">\n<h3>Rewriting the Square Root of a Negative Number<\/h3>\n<ul>\n<li>Rewrite the radical using the rule [latex]\\sqrt{-a}=\\sqrt{-1}\\cdot \\sqrt{a}[\/latex].<\/li>\n<li>Rewrite [latex]\\sqrt{-1}[\/latex] as [latex]i[\/latex] and simplify the other radical if possible.<\/li>\n<\/ul>\n<\/div>\n<h2>Complex Numbers<\/h2>\n<div class=\"textbox key-takeaways\">\n<h3>COMPLEX NUMBERS<\/h3>\n<p>A <strong>complex number<\/strong>\u00a0is any number that can be expressed in the standard form [latex]a+ bi[\/latex]\u00a0where [latex]a[\/latex] is the real part and [latex]b[\/latex] is the imaginary part. We can say that a complex number is the sum of a real number\u00a0[latex]a[\/latex] and a <strong>pure imaginary number<\/strong>\u00a0[latex]bi.[\/latex]<\/p>\n<\/div>\n<p>For example, [latex]5+2i[\/latex] is a complex number. So, too, is [latex]-3+4i\\sqrt{3}[\/latex]. Here are some examples of complex numbers, with real part and imaginary part labeled. Note that [latex]i[\/latex] is not included in the imaginary part.<\/p>\n<table cellpadding=\"0\" style=\"border-spacing: 0px;\">\n<thead>\n<tr>\n<th>Complex Number<\/th>\n<th>Real Part<\/th>\n<th>Imaginary Part<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]3+7i[\/latex]<\/td>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]18\u201332i[\/latex]<\/td>\n<td>[latex]18[\/latex]<\/td>\n<td>[latex]\u221232[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-\\frac{3}{5}+i\\sqrt{2}[\/latex]<\/td>\n<td>[latex]-\\frac{3}{5}[\/latex]<\/td>\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\frac{\\sqrt{2}}{2}-\\frac{1}{2}i[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]-\\frac{1}{2}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>In a number with a radical as part of [latex]b[\/latex], such as [latex]-\\frac{3}{5}+i\\sqrt{2}[\/latex]\u00a0above, it is acceptable to write the imaginary unit [latex]i[\/latex] in front of the radical. Though writing this number as [latex]-\\frac{3}{5}+\\sqrt{2} \\; i[\/latex] is technically correct, it makes it more difficult to tell whether\u00a0[latex]i[\/latex]\u00a0is inside or outside of the radical. Putting it before the radical, as in [latex]-\\frac{3}{5}+i\\sqrt{2}[\/latex], clears up any confusion. While usually we will require answers to be in the form of a complex number [latex]a+bi[\/latex], we make an exception if [latex]b[\/latex] is a radical. Look at these last two examples.<\/p>\n<table cellpadding=\"0\" style=\"border-spacing: 0px;\">\n<thead>\n<tr>\n<th>Number<\/th>\n<th>Complex Form:<br \/>\n[latex]a+bi[\/latex]<\/th>\n<th>Real Part<\/th>\n<th>Imaginary Part<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]17[\/latex]<\/td>\n<td>[latex]17+0i[\/latex]<\/td>\n<td>[latex]17[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u22123i[\/latex]<\/td>\n<td>[latex]0\u20133i[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\u22123[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>By making [latex]b=0[\/latex], any real number can be expressed as a complex number. The real number [latex]a[\/latex] is written as [latex]a+0i[\/latex] in complex form. Similarly, any pure imaginary number can be expressed as a complex number. By making [latex]a=0[\/latex], any pure imaginary number [latex]bi[\/latex] can be written as [latex]0+bi[\/latex] in complex form.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write [latex]83.6[\/latex] in the form of a complex number.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q704457\">Show Solution<\/span><\/p>\n<div id=\"q704457\" class=\"hidden-answer\" style=\"display: none\">\n<p>Remember that a complex number has the form [latex]a+bi[\/latex]. You need to figure out what\u00a0[latex]a[\/latex]\u00a0and [latex]b[\/latex] need to be. Since [latex]83.6[\/latex] is a real number, it\u00a0does not contain any imaginary parts, so the value of [latex]b[\/latex] is\u00a0[latex]0[\/latex].<\/p>\n<p>The answer is [latex]83.6+0i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write [latex]\u22123i[\/latex] in the form of a complex number.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q451549\">Show Solution<\/span><\/p>\n<div id=\"q451549\" class=\"hidden-answer\" style=\"display: none\">\n<p>Remember that a complex number has the form [latex]a+bi[\/latex].\u00a0You need to figure out what [latex]a[\/latex] and [latex]b[\/latex] need to be. Since [latex]\u22123i[\/latex] is a pure imaginary number, the value of [latex]a[\/latex] is [latex]0[\/latex].<\/p>\n<p>The answer is\u00a0[latex]0-3i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In practice if the real or imaginary part of a complex number is [latex]0[\/latex] we will not write that part at all (so we would leave it as [latex]-3i[\/latex] rather than writing the real part of [latex]0[\/latex]).<\/p>\n<h2>Adding and Subtracting Complex Numbers<\/h2>\n<p>Our goal for the rest of this section is to prove the claim that <strong>the complex numbers form a number system just like the real numbers<\/strong> &#8211; they can be added, subtracted, multiplied, and divided. We will need some of these skills in the next chapter. In order for this claim to be true, we must be able to write each final answer as a complex number, in the form [latex]a+bi.[\/latex]<\/p>\n<p>First, consider the following expression.<\/p>\n<p style=\"text-align: center;\">[latex](7+6i)+(2+4i)[\/latex]<\/p>\n<p>If we think of [latex]i[\/latex] as a variable for a moment, we just combine the like terms,<\/p>\n<p style=\"text-align: center;\">[latex](7+6i)+(2+4i)=9+10i[\/latex]<\/p>\n<p>The same rules apply to adding and subtracting complex numbers. You combine the corresponding parts and write the final answer again in the form [latex]a+bi.[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Add.\u00a0[latex](\u22123+3i)+(7\u20132i)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q929105\">Show Solution<\/span><\/p>\n<div id=\"q929105\" class=\"hidden-answer\" style=\"display: none\">\n<p>Rearrange the sums to put like terms together.<\/p>\n<p style=\"text-align: center;\">[latex]\u22123+3i+7\u20132i=\u22123+7+3i\u20132i[\/latex]<\/p>\n<p>Combine like terms.<\/p>\n<p style=\"text-align: center;\">[latex]\u22123+7=4[\/latex]<\/p>\n<p style=\"text-align: center;\">and<\/p>\n<p style=\"text-align: center;\">[latex]3i\u20132i=(3\u20132)i=i[\/latex]<\/p>\n<p>The answer is [latex]4+i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Subtract.\u00a0[latex](\u22123+3i)\u2013(7\u20132i)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q203125\">Show Solution<\/span><\/p>\n<div id=\"q203125\" class=\"hidden-answer\" style=\"display: none\">\n<p>Be sure to distribute the subtraction sign to all terms in the set of parentheses that follows.<\/p>\n<p style=\"text-align: center;\">[latex](\u22123+3i)\u2013(7\u20132i)=\u22123+3i\u20137+2i[\/latex]<i>\u00a0<\/i><\/p>\n<p>Rearrange the terms to put like terms together.<\/p>\n<p style=\"text-align: center;\">[latex]\u22123\u20137+3i+2i[\/latex]<\/p>\n<p>Combine like terms.<\/p>\n<p style=\"text-align: center;\">[latex]\u22123\u20137=\u221210[\/latex]<\/p>\n<p style=\"text-align: center;\">and<\/p>\n<p style=\"text-align: center;\">[latex]3i+2i=(3+2)i=5i[\/latex]<\/p>\n<p>The answer is [latex]-10+5i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show more examples of how to add and subtract complex numbers (Note in this video that it incorrectly calls [latex]bi[\/latex] the imaginary part instead of just [latex]b[\/latex]).<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 1:  Adding and Subtracting Complex Numbers\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/SGhTjioGqqA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Multiplying Radicals with Negative Radicands<\/h2>\n<p>We have to be careful when multiplying radicals with negative radicands. Remember to use the product rule for imaginary numbers first before any other steps.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Multiply and simplify [latex]\\sqrt{-8}\\cdot\\sqrt{-12}.[\/latex] Write answers in the form [latex]a+bi[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q676845\">Show Solution<\/span><\/p>\n<div id=\"q676845\" class=\"hidden-answer\" style=\"display: none\">\n<p>First rewrite each radical using the product rule for imaginary numbers.<\/p>\n<p>[latex]\\begin{align}&\\quad\\sqrt{-8}\\cdot\\sqrt{-12}\\\\  =&\\quad\\sqrt{-1}\\cdot\\sqrt{8}\\cdot\\sqrt{-1}\\cdot\\sqrt{12}\\\\  =&\\quad i\\cdot i\\cdot\\sqrt{8\\cdot 12}\\\\  =&\\quad i^2\\sqrt{96}\\\\  =&\\quad -\\sqrt{96}\\end{align}[\/latex]<\/p>\n<p>The radical can now be simplified normally.<\/p>\n<p>[latex]\\begin{align}=&\\quad -\\sqrt{16}\\sqrt{6}\\\\  =&\\quad -4\\sqrt{6}\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01182614\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"37\" height=\"33\" \/>Notice in the previous example, it would be <span style=\"color: #ff0000;\">incorrect<\/span>\u00a0to use our previous product rule for radicals to multiply the radicands:\u00a0 [latex]\\sqrt{-8}\\cdot\\sqrt{-12}\\color{red}{\\neq \\sqrt{-8\\cdot -12}=\\sqrt{96}}.[\/latex] Doing so would cause us to miss the negative sign that should have appeared upon simplifying [latex]i^2.[\/latex]<\/p>\n<\/div>\n<h2>Multiplying Complex Numbers<\/h2>\n<section id=\"fs-id1165137417169\">\n<p id=\"fs-id1165137832911\">The process of multiplying complex numbers is again similar to the corresponding process for polynomials. The only difference is that we should replace [latex]i^2=-1[\/latex] if it appears.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Multiply and simplify [latex]3i(-6+5i).[\/latex] Write answers in the form [latex]a+bi.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q603325\">Show Solution<\/span><\/p>\n<div id=\"q603325\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">We can use the distributive property.<br \/>\n[latex]\\begin{align}&\\quad 3i(-6+5i)\\\\  =&\\quad 3i\\cdot (-6) + 3i\\cdot 5i&& \\color{blue}{\\textsf{distribute}}\\\\  =&\\quad -18i+15i^2\\\\  =&\\quad -18i+15(-1) && \\color{blue}{\\textsf{substitute }i^2=-1}\\\\  =&\\quad -15-18i&& \\color{blue}{\\textsf{write in the form }a+bi} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137650841\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Multiply and simplify [latex]\\left(4+3i\\right)\\left(2 - 5i\\right).[\/latex] Write answers in the form [latex]a+bi.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q188458\">Show Solution<\/span><\/p>\n<div id=\"q188458\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137459488\">Multiply (using FOIL, for example).<\/p>\n<div id=\"eip-id1165137762412\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll}\\left(4+3i\\right)\\left(2 - 5i\\right) & =\\left(4\\cdot 2\\right)+\\left(4\\cdot \\left(-5i\\right)\\right)+\\left(3i\\cdot 2\\right)+\\left(\\left(3i\\right)\\cdot \\left(-5i\\right)\\right) & \\\\ & =8-20i+6i-15i^2 & \\\\ & = 8-20i+6i-15\\left(-1\\right) & \\color{blue}{\\textsf{substitute } i^2=-1} \\\\ & = 8-20i+6i+15 & \\\\ & = \\left(8+15\\right)+\\left(-20+6\\right)i & \\color{blue}{\\textsf{group like terms together}} \\\\ & =23-14i & \\end{array}[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In the first video, we show more examples of multiplying complex numbers.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 3:  Multiply Complex Numbers\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Fmr3o2zkwLM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Simplifying Powers of\u00a0[latex]i[\/latex]<\/h2>\n<p id=\"fs-id1165132919554\">There is a pattern to evaluating powers of [latex]i[\/latex]. Let us look at what happens when we raise\u00a0[latex]i[\/latex]\u00a0to increasing powers.<\/p>\n<div id=\"eip-783\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{rl}{i}^{1}&=\\quad i\\\\ {i}^{2}&=\\quad -1\\\\ {i}^{3}&=\\quad {i}^{2}\\cdot i=-1\\cdot i=-i\\\\ {i}^{4}&=\\quad {i}^{3}\\cdot i=-i\\cdot i=-{i}^{2}=-\\left(-1\\right)=1\\\\ {i}^{5}&=\\quad {i}^{4}\\cdot i=1\\cdot i=i\\end{array}[\/latex]<\/div>\n<div><\/div>\n<div class=\"equation unnumbered\">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\u00a0[latex]i[\/latex]\u00a0by itself for increasing powers, we will see a cycle of\u00a0four. Here are the next four powers of [latex]i[\/latex].<\/div>\n<div><\/div>\n<div id=\"eip-477\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{rl}{i}^{6}&=\\quad {i}^{5}\\cdot i=i\\cdot i={i}^{2}=-1\\\\ {i}^{7}&= \\quad{i}^{6}\\cdot i={i}^{2}\\cdot i={i}^{3}=-i\\\\ {i}^{8}&=\\quad {i}^{7}\\cdot i={i}^{3}\\cdot i={i}^{4}=1\\\\ {i}^{9}&=\\quad {i}^{8}\\cdot i={i}^{4}\\cdot i={i}^{5}=i\\end{array}[\/latex]<\/div>\n<div class=\"equation unnumbered\"><\/div>\n<div class=\"equation unnumbered\">\n<div class=\"textbox exercises\">\n<h3>Example<\/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=\"q295805\">Show Solution<\/span><\/p>\n<div id=\"q295805\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137728290\">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\u00a0[latex]4[\/latex] goes into\u00a0[latex]35[\/latex]: [latex]35=4\\cdot 8+3[\/latex]. The remainder of [latex]3[\/latex] is the most important part.<\/p>\n<div id=\"eip-id1165134069265\" class=\"equation unnumbered\">[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]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>The previous example illustrates the easiest way to simplify a large power of [latex]i[\/latex]. Rewrite your old exponent as a multiple of [latex]4[\/latex] plus a remainder. After some simplifying, the final power of\u00a0[latex]i[\/latex] is just the remainder. Then you just need to know the first four powers of\u00a0[latex]i.[\/latex]<\/p>\n<p>In the following video, you will see more examples of how to simplify powers of [latex]i.[\/latex]<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Simplifying Complex Numbers\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/XyksedEntD4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Dividing Complex Numbers<\/h2>\n<p id=\"fs-id1165137612241\">If we write a division of complex numbers in radical form, it should look similar to something we&#8217;ve done earlier in this chapter.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{2+3i}{4-i}=\\frac{2+3\\sqrt{-1}}{4-\\sqrt{-1}}[\/latex]<\/p>\n<p>We have previously removed radicals from the denominator by rationalizing the denominator using the conjugate. To eliminate the imaginary number [latex]i[\/latex] in the denominator, you multiply by\u00a0the <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<div class=\"textbox key-takeaways\">\n<h3>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. When a complex number is multiplied by its complex conjugate, the result is a real number.<\/p>\n<\/div>\n<p>In our previous example, the denominator is [latex]4-i.[\/latex] The complex conjugate is [latex]4+i.[\/latex] Multiplying, we get\u00a0[latex](4-i)(4+i)=16+4i-4i-i^2=16-(-1)=17.[\/latex] Let&#8217;s use this to finish our division problem.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\quad\\frac{2+3i}{4-i}\\\\  =&\\quad\\frac{2+3i}{4-i}\\cdot\\frac{4+i}{4+i}\\\\  =&\\quad\\frac{(2+3i)(4+i)}{(4-i)(4+i)}\\\\  =&\\quad\\frac{8+2i+12i+3i^2}{17}&&\\color{blue}{\\textsf{FOIL numerator and denominator}}\\\\  =&\\quad\\frac{8+2i+12i+3(-1)}{17}\\\\  =&\\quad\\frac{5+14i}{17}&&\\color{blue}{\\textsf{simplify the numerator}}\\\\  =&\\quad\\frac{5}{17}+\\frac{14}{17}i&&\\color{blue}{\\textsf{write in the form }a+bi}\\end{align}[\/latex]<\/p>\n<p>We are now done, and most importantly the final answer is in the form of a complex number, [latex]a+bi.[\/latex]<\/p>\n<div class=\"textbox shaded\">\n<h3>DIVIDING COMPLEX NUMBERS<\/h3>\n<ul>\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 and simplify.<\/li>\n<li>Write the answer as [latex]a+bi.[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Divide [latex]\\left(6-5i\\right)[\/latex] by [latex]\\left(1+3i\\right)[\/latex]. Write answers in the form [latex]a+bi.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q665746\">Show Solution<\/span><\/p>\n<div id=\"q665746\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137605861\">We begin by writing the problem as a fraction.<\/p>\n<div id=\"eip-id1165134234232\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Large\\frac{\\left(6-5i\\right)}{\\left(1+3i\\right)}[\/latex]<\/div>\n<p id=\"fs-id1165137639613\">Then we multiply the numerator and denominator by the complex conjugate of the denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcll}\\Large{=\\frac{(6-5i)}{(1+3i)}}\\cdot\\frac{(1-3i)}{(1-3i)}&=&\\Large{\\frac{6-18i-5i+15{i}^{2}}{1-3i+3i-9{i}^{2}}}& \\\\&=&\\Large{\\frac{6-18i-5i+15(-1)}{1-3i+3i-9(-1)}}&\\quad\\color{blue}{\\textsf{substitute }i^2=-1}\\\\&=&\\Large{\\frac{-9-23i}{10}}&\\quad\\color{blue}{\\textsf{combine like terms}} \\\\&=&\\Large{-\\frac{9}{10}-\\frac{23}{10}i}&\\quad\\color{blue}{\\textsf{write in the form }a+bi}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Divide [latex]\\dfrac{6-8i}{3i}[\/latex]. Write answers in the form [latex]a+bi.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q634438\">Show Solution<\/span><\/p>\n<div id=\"q634438\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"eip-id1165134234232\" class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<p id=\"fs-id1165137639613\">We could multiply the numerator and denominator by the complex conjugate, but since the denominator is a monomial it is easier to just multiply by [latex]-i.[\/latex] You will see this still results in the denominator being a real number.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{6-8i}{3i}&=\\frac{\\left(6-8i\\right)}{3i}\\cdot\\frac{-i}{-i}\\\\  &=\\frac{-6i+8i^2}{-3i^2}\\\\  &=\\frac{-6i+8(-1)}{-3(-1)}\\\\  &=\\frac{-6i-8}{3}\\\\  &=\\frac{-6i}{3}+\\frac{-8}{3}\\\\  &=-\\frac{8}{3}-2i\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the last video, you will see more examples of dividing complex numbers.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" 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<\/section>\n<h2>Summary<\/h2>\n<p>Complex numbers have the form [latex]a+bi[\/latex], where [latex]a[\/latex] and [latex]b[\/latex] are real numbers and [latex]i[\/latex] is the square root of [latex]\u22121[\/latex]. All real numbers can be written as complex numbers by setting [latex]b=0[\/latex]. Square roots of negative numbers can be simplified using [latex]\\sqrt{-a}=i\\sqrt{a}[\/latex] where [latex]a>0.[\/latex] We can perform arithmetic operations on complex numbers in much the same way as working with polynomials and combining like terms, making sure to simplify [latex]i^2=-1[\/latex] when appropriate. Powers of\u00a0[latex]i[\/latex] form a cycle that repeats every four powers. Division of complex numbers is done by first making the denominator a real number using the complex conjugate multiplied on the top and bottom of the fraction.<\/p>\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-206\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Write Number in the Form of Complex Numbers. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/mfoOYdDkuyY\">https:\/\/youtu.be\/mfoOYdDkuyY<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Simplify Square Roots to Imaginary Numbers. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/LSp7yNP6Xxc\">https:\/\/youtu.be\/LSp7yNP6Xxc<\/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":9,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Write Number in the Form of Complex Numbers\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/mfoOYdDkuyY\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Simplify Square Roots to Imaginary Numbers\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/LSp7yNP6Xxc\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"08e5cf72-4d52-4382-ac86-188a39ac7ad3","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-206","chapter","type-chapter","status-publish","hentry"],"part":184,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/206","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":15,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/206\/revisions"}],"predecessor-version":[{"id":2084,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/206\/revisions\/2084"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/parts\/184"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/206\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/media?parent=206"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=206"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/contributor?post=206"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/license?post=206"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}