{"id":237,"date":"2016-06-01T20:49:55","date_gmt":"2016-06-01T20:49:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=237"},"modified":"2018-05-17T00:02:03","modified_gmt":"2018-05-17T00:02:03","slug":"16-4-1-complex-numbers","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/16-4-1-complex-numbers\/","title":{"raw":"Imaginary and Complex Numbers","rendered":"Imaginary and Complex Numbers"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Define imaginary and complex numbers\r\n<ul>\r\n \t<li>Express roots of negative numbers in terms of <em>i<\/em>.<\/li>\r\n \t<li>Express imaginary numbers as bi and complex numbers as [latex]a+bi[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Algebraic operations on complex numbers\r\n<ul>\r\n \t<li>Add complex numbers.<\/li>\r\n \t<li>Subtract complex numbers<\/li>\r\n \t<li>Multiply complex numbers.<\/li>\r\n \t<li>Find conjugates of complex numbers.<\/li>\r\n \t<li>Divide complex numbers.<\/li>\r\n \t<li>Simplify\u00a0powers of\u00a0<em>i<\/em><\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\nUp to now, you\u2019ve known it was impossible to take a square root of a negative number. This is true, using only the <i>real numbers<\/i>. But here you will learn about a new kind of number that lets you work with square roots of negative numbers! Complex numbers are made from both real and imaginary numbers. \u00a0Imaginary numbers are called imaginary because they are impossible and, therefore, exist only in the world of ideas and pure imagination. Imaginary numbers result from taking the square root of a negative number.\r\n\r\nHere we will first define and perform algebraic operations on complex numbers, then we will provide examples of quadratic equations that have solutions that are complex numbers.\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\u2014they seem pretty <i>real<\/i> to us! When something\u2019s not real, you often say it is <i>imaginary<\/i>. So let\u2019s call this new number <i>i<\/i> and 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\\,\\cdot \\,i=-1[\/latex]. We also know that [latex] i\\,\\cdot \\,i={{i}^{2}}[\/latex], so we can conclude that [latex] {{i}^{2}}=-1[\/latex].\r\n<p style=\"text-align: center\">[latex] {{i}^{2}}=-1[\/latex]<\/p>\r\nThe number <i>i <\/i>allows us to work with roots of all negative numbers, not just [latex] \\sqrt{-1}[\/latex]. There are two important rules to remember: [latex] \\sqrt{-1}=i[\/latex], and [latex] \\sqrt{ab}=\\sqrt{a}\\sqrt{b}[\/latex]. You will use these rules to rewrite the square root of a negative number as the square root of a positive number times [latex] \\sqrt{-1}[\/latex]. Next you will simplify the square root and rewrite [latex] \\sqrt{-1}[\/latex] as <i>i. <\/i>Let\u2019s try an example.\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\"]Use the rule [latex] \\sqrt{ab}=\\sqrt{a}\\sqrt{b}[\/latex] to rewrite this as a product using [latex] \\sqrt{-1}[\/latex].\r\n\r\n[latex] \\sqrt{-4}=\\sqrt{4\\cdot -1}=\\sqrt{4}\\sqrt{-1}[\/latex]\r\n\r\nSince 4 is a perfect square\u00a0[latex](4=2^{2})[\/latex], you can simplify the square root of 4.\r\n\r\n[latex] \\sqrt{4}\\sqrt{-1}=2\\sqrt{-1}[\/latex]\r\n\r\nUse the definition of <i>i<\/i> to rewrite [latex] \\sqrt{-1}[\/latex] as <i>i.<\/i>\r\n\r\n[latex] 2\\sqrt{-1}=2i[\/latex]\r\n<h4>Answer<\/h4>\r\n[latex] \\sqrt{-4}=2i[\/latex]\r\n\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\"]Use the rule [latex] \\sqrt{ab}=\\sqrt{a}\\sqrt{b}[\/latex] to rewrite this as a product using [latex] \\sqrt{-1}[\/latex].\r\n\r\n[latex] \\sqrt{-18}=\\sqrt{18\\cdot -1}=\\sqrt{18}\\sqrt{-1}[\/latex]\r\n\r\nSince 18 is <i>not<\/i> a perfect square, use the same rule to rewrite it using factors that are perfect squares. In this case, 9 is the only perfect square factor, and the square root of 9 is 3.\r\n\r\n[latex] \\sqrt{18}\\sqrt{-1}=\\sqrt{9}\\sqrt{2}\\sqrt{-1}=3\\sqrt{2}\\sqrt{-1}[\/latex]\r\n\r\nUse the definition of <i>i<\/i> to rewrite [latex] \\sqrt{-1}[\/latex] as <i>i.<\/i>\r\n\r\n[latex] 3\\sqrt{2}\\sqrt{-1}=3\\sqrt{2}i=3i\\sqrt{2}[\/latex]\r\n\r\nRemember to write <i>i<\/i> in front of the radical.\r\n<h4>Answer<\/h4>\r\n[latex] \\sqrt{18}=3i\\sqrt[{}]{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] -\\sqrt{-72}[\/latex]\r\n\r\n[reveal-answer q=\"503996\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"503996\"]Use the rule [latex] \\sqrt{ab}=\\sqrt{a}\\sqrt{b}[\/latex] to rewrite this as a product using [latex] \\sqrt{-1}[\/latex].\r\n\r\n[latex] -\\sqrt{-72}=-\\sqrt{72\\cdot -1}=-\\sqrt{72}\\sqrt{-1}[\/latex]\r\n\r\nSince 72 is <i>not<\/i> a perfect square, use the same rule to rewrite it using factors that are perfect squares. Notice that 72 has three perfect squares as factors: 4, 9, and 36. It\u2019s easiest to use the largest factor that is a perfect square.\r\n\r\n[latex] -\\sqrt{72}\\sqrt{-1}=-\\sqrt{36}\\sqrt{2}\\sqrt{-1}=-6\\sqrt{2}\\sqrt{-1}[\/latex]\r\n\r\nUse the definition of <i>i<\/i> to rewrite [latex] \\sqrt{-1}[\/latex] as <i>i.<\/i>\r\n\r\n[latex] -6\\sqrt{2}\\sqrt{-1}=-6\\sqrt{2}i=-6i\\sqrt{2}[\/latex]\r\n\r\nRemember to write <i>i<\/i> in front of the radical.\r\n<h4>Answer<\/h4>\r\n[latex] -\\sqrt{-}72=-6i\\sqrt[{}]{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou may have wanted to simplify [latex] -\\sqrt{-72}[\/latex] using different factors. Some may have thought of rewriting this radical as [latex] -\\sqrt{-9}\\sqrt{8}[\/latex], or [latex] -\\sqrt{-4}\\sqrt{18}[\/latex], or [latex] -\\sqrt{-6}\\sqrt{12}[\/latex] for instance. Each of these radicals would have eventually yielded the same answer of [latex] -6i\\sqrt{2}[\/latex].\r\n\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>Find perfect squares within the radical.<\/li>\r\n \t<li>Rewrite the radical using the rule [latex] \\sqrt{ab}=\\sqrt{a}\\cdot \\sqrt{b}[\/latex].<\/li>\r\n \t<li>Rewrite [latex] \\sqrt{-1}[\/latex] as <i>i<\/i>.<\/li>\r\n<\/ul>\r\nExample: [latex] \\sqrt{-18}=\\sqrt{9}\\sqrt{-2}=\\sqrt{9}\\sqrt{2}\\sqrt{-1}=3i\\sqrt{2}[\/latex]\r\n\r\n<\/div>\r\n<h2>Complex Numbers<\/h2>\r\n<img class=\"wp-image-2527 size-full aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2016\/06\/22231825\/CNX_Precalc_Figure_03_01_0012.jpg\" alt=\"Showing the real and imaginary parts of 5 + 2i. In this complex number, 5 is the real part and 2i is the complex part.\" width=\"487\" height=\"72\" \/>\r\n<p id=\"fs-id1165135500790\">A <strong>complex number<\/strong> is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written <em>a\u00a0<\/em>+ <em>bi<\/em>\u00a0where <em>a<\/em>\u00a0is the real part and <em>bi<\/em>\u00a0is the imaginary part. For example, [latex]5+2i[\/latex] is a complex number. So, too, is [latex]3+4\\sqrt{3}i[\/latex].<span id=\"fs-id1165137832295\">\r\n<\/span><\/p>\r\n<p id=\"fs-id1165137892327\">Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative\u00a0real number. Recall, when a positive real number is squared, the result is a positive real number and when a negative real number is squared, again, the result is a positive real number. Complex numbers are a combination of real and imaginary numbers.<\/p>\r\nYou can use the usual operations (addition, subtraction, multiplication, and so on) with imaginary numbers. You\u2019ll see more of that, later. When you <i>add<\/i> a real number to an imaginary number, however, you get a complex number. A complex number is any number in the form [latex]a+bi[\/latex], where <i>a<\/i> is a real number and <i>bi<\/i> is an imaginary number. The number <i>a<\/i> is sometimes called the real part of the complex number, and <i>bi<\/i> is sometimes called 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>3<\/td>\r\n<td>[latex]7i[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]18\u201332i[\/latex]<\/td>\r\n<td>18<\/td>\r\n<td>[latex]\u221232i[\/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] i\\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}i[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIn a number with a radical as part of <i>b<\/i>, such as [latex]-\\frac{3}{5}+i\\sqrt{2}[\/latex]\u00a0above, the imaginary <i>i<\/i> should be written in front of the radical. Though writing this number as [latex] -\\frac{3}{5}+\\sqrt{2}i[\/latex] is technically correct, it makes it much more difficult to tell whether <i>i<\/i> is 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. 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>Number in 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>17<\/td>\r\n<td>[latex]17+0i[\/latex]<\/td>\r\n<td>17<\/td>\r\n<td>[latex]0i[\/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>0<\/td>\r\n<td>[latex]\u22123i[\/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 <i>a<\/i> is written [latex]a+0i[\/latex] in complex form. Similarly, any imaginary number can be expressed as a complex number. By making [latex]a=0[\/latex], any imaginary number [latex]bi[\/latex] is written [latex]0+bi[\/latex] in complex form.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite 83.6 as a complex number.\r\n\r\n[reveal-answer q=\"704457\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"704457\"]Remember that a complex number has the form [latex]a+bi[\/latex]. You need to figure out what <i>a<\/i> and <i>b<\/i> need to be.\r\n\r\n[latex]a+bi[\/latex]\r\n\r\nSince 83.6 is a real number, it is the real part (a) of the complex number [latex]a+bi[\/latex]. A real number does not contain any imaginary parts, so the value of b is 0.<i>\r\n<\/i>\r\n\r\n[latex]83.6+bi[\/latex]\r\n<h4>Answer<\/h4>\r\n[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] as a complex number.\r\n\r\n[reveal-answer q=\"451549\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"451549\"]Remember that a complex number has the form [latex]a+bi[\/latex].\u00a0You need to figure out what <i>a<\/i> and <i>b<\/i> need to be.\r\n\r\n[latex]a+bi[\/latex]\r\n\r\nSince [latex]\u22123i[\/latex] is an imaginary number, it is the imaginary part (<em>bi<\/em>) of the complex number [latex]a+bi[\/latex]. This imaginary number has no real parts, so the value of a is 0.\r\n\r\n[latex]a\u20133i[\/latex]\r\n<h4>Answer<\/h4>\r\n[latex]0\u20133i[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next video we show more examples of how to write numbers as complex numbers.\r\n\r\nhttps:\/\/youtu.be\/mfoOYdDkuyY\r\n<h2>Add and Subtract Complex Numbers<\/h2>\r\nAny time new kinds of numbers are introduced, one of the first questions that needs to be addressed is, \u201cHow do you add them?\u201d In this topic, you\u2019ll learn how to add complex numbers and also how to subtract.\r\n\r\nFirst, consider the following expression.\r\n<p style=\"text-align: center\">[latex](6x+8)+(4x+2)[\/latex]<\/p>\r\nTo simplify this expression, you combine the like terms, [latex]6x[\/latex] and [latex]4x[\/latex].<i>\u00a0<\/i>These are like terms because they have the same variable with the same exponents. Similarly, 8 and 2 are like terms because they are both constants, with no variables.\r\n<p style=\"text-align: center\">[latex](6x+8)+(4x+2)=10x+10[\/latex]<\/p>\r\nIn the same way, you can simplify expressions with radicals.\r\n<p style=\"text-align: center\">[latex] (6\\sqrt{3}+8)+(4\\sqrt{3}+2)=10\\sqrt{3}+10[\/latex]<\/p>\r\nYou can add [latex] 6\\sqrt{3}[\/latex] to [latex] 4\\sqrt{3}[\/latex] because the two terms have the same radical, [latex] \\sqrt{3}[\/latex], just as 6<i>x<\/i> and 4<i>x<\/i> have the same variable and exponent.\r\n\r\nThe number <i>i <\/i>looks like a variable, but remember that it is equal to [latex]\\sqrt{-1}[\/latex]. The great thing is you have no new rules to worry about\u2014whether you treat it as a variable or a radical, the exact same rules apply to adding and subtracting <strong>complex numbers<\/strong>. You combine the imaginary parts (the terms with <i>i<\/i>),<i> <\/i>and you combine the real parts.\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\"]Rearrange 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] and [latex]3i\u20132i=(3\u20132)i=i[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex](\u22123+3i)+(7\u20132i)=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\"]Be sure to distribute the subtraction sign to all terms in the subtrahend.\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] and\u00a0[latex]3i+2i=(3+2)i=5i[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex](\u22123+3i)\u2013(7\u20132i)=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.\r\n\r\nhttps:\/\/youtu.be\/SGhTjioGqqA\r\n<h2>Multiply and Divide Complex Numbers<\/h2>\r\n<section id=\"fs-id1165137417169\">\r\n<p id=\"fs-id1165137832911\">Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.<\/p>\r\n\r\n<h3>\u00a0Multiplying a Complex Number by a Real Number<\/h3>\r\n<img class=\"wp-image-2535 size-full aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2016\/06\/23153423\/CNX_Precalc_Figure_03_01_0062.jpg\" alt=\"Showing how distribution works for complex numbers. For 3(6+2i), 3 is multiplied to both the real and imaginary parts. So we have (3)(6)+(3)(2i) = 18 + 6i. \" width=\"487\" height=\"87\" \/>\r\n\r\nLet\u2019s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So,for 3(6+2i), 3 is multiplied to both the real and imaginary parts. So we have (3)(6)+(3)(2i) = 18 + 6i.\r\n<div id=\"fs-id1165137745292\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137426118\">How To: Given a complex number and a real number, multiply to find the product.<\/h3>\r\n<ol id=\"fs-id1165137793647\">\r\n \t<li>Use the distributive property.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the product [latex]4\\left(2+5i\\right)[\/latex].\r\n[reveal-answer q=\"374377\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"374377\"]\r\n<p id=\"fs-id1165137804818\">Distribute the 4.<\/p>\r\n\r\n<div id=\"eip-id1165135436310\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{cc}4\\left(2+5i\\right)=\\left(4\\cdot 2\\right)+\\left(4\\cdot 5i\\right)\\hfill \\\\ =8+20i\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Multiplying Complex Numbers Together<\/h2>\r\n<\/section><section id=\"fs-id1165137650841\">\r\n<p id=\"fs-id1165137832483\">Now, let\u2019s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get<\/p>\r\n\r\n<div id=\"eip-586\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left(a+bi\\right)\\left(c+di\\right)=ac+adi+bci+bd{i}^{2}[\/latex]<\/div>\r\n<p id=\"fs-id1165137734803\">Because [latex]{i}^{2}=-1[\/latex], we have<\/p>\r\n\r\n<div id=\"eip-523\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left(a+bi\\right)\\left(c+di\\right)=ac+adi+bci-bd[\/latex]<\/div>\r\n<p id=\"fs-id1165135186757\">To simplify, we combine the real parts, and we combine the imaginary parts.<\/p>\r\n\r\n<div id=\"eip-794\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left(a+bi\\right)\\left(c+di\\right)=\\left(ac-bd\\right)+\\left(ad+bc\\right)i[\/latex]<\/div>\r\n<div id=\"fs-id1165137642817\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137724898\">How To: Given two complex numbers, multiply to find the product.<\/h3>\r\n<ol id=\"fs-id1165137561156\">\r\n \t<li>Use the distributive property or the FOIL method.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nMultiply [latex]\\left(4+3i\\right)\\left(2 - 5i\\right)[\/latex].\r\n[reveal-answer q=\"188458\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"188458\"]\r\n<p id=\"fs-id1165137459488\">Use [latex]\\left(a+bi\\right)\\left(c+di\\right)=\\left(ac-bd\\right)+\\left(ad+bc\\right)i[\/latex]<\/p>\r\n\r\n<div id=\"eip-id1165137762412\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{ccc}\\left(4+3i\\right)\\left(2 - 5i\\right)=\\left(4\\cdot 2 - 3\\cdot \\left(-5\\right)\\right)+\\left(4\\cdot \\left(-5\\right)+3\\cdot 2\\right)i\\hfill \\\\ \\text{ }=\\left(8+15\\right)+\\left(-20+6\\right)i\\hfill \\\\ \\text{ }=23 - 14i\\hfill \\end{array}[\/latex]<\/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 <em>i<\/em><\/h2>\r\n<p id=\"fs-id1165132919554\">The powers of <em>i<\/em>\u00a0are cyclic. Let\u2019s look at what happens when we raise <em>i<\/em>\u00a0to increasing powers.<\/p>\r\n\r\n<div id=\"eip-783\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}{i}^{1}=i\\\\ {i}^{2}=-1\\\\ {i}^{3}={i}^{2}\\cdot i=-1\\cdot i=-i\\\\ {i}^{4}={i}^{3}\\cdot i=-i\\cdot i=-{i}^{2}=-\\left(-1\\right)=1\\\\ {i}^{5}={i}^{4}\\cdot i=1\\cdot i=i\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137530297\">We can see that when we get to the fifth power of <em>i<\/em>, it is equal to the first power. As we continue to multiply <em>i<\/em>\u00a0by itself for increasing powers, we will see a cycle of 4. Let\u2019s examine the next 4 powers of <em>i<\/em>.<\/p>\r\n\r\n<div id=\"eip-477\" class=\"equation unnumbered\">[latex]\\begin{array}{i}^{6}={i}^{5}\\cdot i=i\\cdot i={i}^{2}=-1\\\\ {i}^{7}={i}^{6}\\cdot i={i}^{2}\\cdot i={i}^{3}=-i\\\\ {i}^{8}={i}^{7}\\cdot i={i}^{3}\\cdot i={i}^{4}=1\\\\ {i}^{9}={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 Answer[\/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 4 goes into 35: [latex]35=4\\cdot 8+3[\/latex].<\/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\n<div id=\"fs-id1165137758921\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"fs-id1165135186727\"><strong>Can we write [latex]{i}^{35}[\/latex] in other helpful ways?<\/strong><\/p>\r\n<p id=\"fs-id1165135444053\"><em>As we saw in Example 11, 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>\r\n\r\n<table>\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<p id=\"fs-id1165135255472\"><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>\r\n\r\n<\/div>\r\nIn the following video you will see more examples of how to simplify powers of\u00a0<em>i<\/em>.\r\n\r\nhttps:\/\/youtu.be\/sfP6SmEYHRw\r\n<h2>Dividing Complex Numbers<\/h2>\r\n<p id=\"fs-id1165137612241\">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. This idea is similar to rationalizing the denominator of a fraction that contains a radical.\u00a0To eliminate the complex or imaginary number 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>\r\n<p id=\"fs-id1165137435064\">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]. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.<\/p>\r\n<p id=\"fs-id1165137611741\">Suppose we want to divide [latex]c+di[\/latex] by [latex]a+bi[\/latex], where neither <em>a<\/em>\u00a0nor <em>b<\/em>\u00a0equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.<\/p>\r\n\r\n<div id=\"eip-225\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\displaystyle \\frac{c+di}{a+bi}\\text{ where }a\\ne 0\\text{ and }b\\ne 0[\/latex]<\/div>\r\n<p id=\"fs-id1165134148263\">Multiply the numerator and denominator by the complex conjugate of the denominator.<\/p>\r\n\r\n<div id=\"eip-32\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\displaystyle \\frac{\\left(c+di\\right)}{\\left(a+bi\\right)}\\cdot \\frac{\\left(a-bi\\right)}{\\left(a-bi\\right)}=\\frac{\\left(c+di\\right)\\left(a-bi\\right)}{\\left(a+bi\\right)\\left(a-bi\\right)}[\/latex]<\/div>\r\n<p id=\"fs-id1165135260723\">Apply the distributive property.<\/p>\r\n\r\n<div id=\"eip-736\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\displaystyle =\\frac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}}[\/latex]<\/div>\r\n<p id=\"fs-id1165137871668\">Simplify, remembering that [latex]{i}^{2}=-1[\/latex].<\/p>\r\n\r\n<div id=\"eip-64\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}=\\frac{ca-cbi+adi-bd\\left(-1\\right)}{{a}^{2}-abi+abi-{b}^{2}\\left(-1\\right)}\\hfill \\\\ =\\frac{\\left(ca+bd\\right)+\\left(ad-cb\\right)i}{{a}^{2}+{b}^{2}}\\hfill \\end{array}[\/latex]<\/div>\r\n<div id=\"fs-id1165135203870\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Complex Conjugate<\/h3>\r\n<p id=\"fs-id1165137793758\">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>\r\n\r\n<ul id=\"fs-id1165135487089\">\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<\/h3>\r\n<p id=\"fs-id1165137896182\">Find the complex conjugate of each number.<\/p>\r\n\r\n<ol id=\"fs-id1165137896185\">\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=\"426623\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"426623\"]\r\n<ol id=\"fs-id1165137742669\">\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[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Analysis of the Solution<\/h3>\r\n<div id=\"Example_03_01_06\" class=\"example\">\r\n<div id=\"fs-id1165134032261\" class=\"exercise\">\r\n<div id=\"fs-id1165137772480\" class=\"commentary\">\r\n<p id=\"fs-id1165137762415\">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 <em>i<\/em>.<\/p>\r\nIn the last video you will see more examples of dividing complex numbers.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nhttps:\/\/youtu.be\/XBJjbJAwM1c\r\n<div id=\"fs-id1165137409413\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135471100\">How To: Given two complex numbers, divide one by the other.<\/h3>\r\n<ol id=\"fs-id1165135471104\">\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<\/h3>\r\nDivide [latex]\\left(2+5i\\right)[\/latex] by [latex]\\left(4-i\\right)[\/latex].\r\n[reveal-answer q=\"665746\"]Show Answer[\/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]\\displaystyle \\frac{\\left(2+5i\\right)}{\\left(4-i\\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\r\n<div id=\"eip-id1165137400110\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\displaystyle \\frac{\\left(2+5i\\right)}{\\left(4-i\\right)}\\cdot \\frac{\\left(4+i\\right)}{\\left(4+i\\right)}[\/latex]<\/div>\r\n<p id=\"fs-id1165137474228\">To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL).<\/p>\r\n\r\n<div id=\"eip-id1165131986989\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}\\frac{\\left(2+5i\\right)}{\\left(4-i\\right)}\\cdot \\frac{\\left(4+i\\right)}{\\left(4+i\\right)}=\\frac{8+2i+20i+5{i}^{2}}{16+4i - 4i-{i}^{2}}\\hfill &amp; \\hfill \\\\ \\text{ }=\\frac{8+2i+20i+5\\left(-1\\right)}{16+4i - 4i-\\left(-1\\right)}\\hfill &amp; \\text{Because } {i}^{2}=-1\\hfill \\\\ \\text{ }=\\frac{3+22i}{17}\\hfill &amp; \\hfill \\\\ \\text{ }=\\frac{3}{17}+\\frac{22}{17}i\\hfill &amp; \\text{Separate real and imaginary parts}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137930346\">Note that this expresses the quotient in standard form.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137650841\"><\/section>\r\n<h2><\/h2>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Define imaginary and complex numbers\n<ul>\n<li>Express roots of negative numbers in terms of <em>i<\/em>.<\/li>\n<li>Express imaginary numbers as bi and complex numbers as [latex]a+bi[\/latex].<\/li>\n<\/ul>\n<\/li>\n<li>Algebraic operations on complex numbers\n<ul>\n<li>Add complex numbers.<\/li>\n<li>Subtract complex numbers<\/li>\n<li>Multiply complex numbers.<\/li>\n<li>Find conjugates of complex numbers.<\/li>\n<li>Divide complex numbers.<\/li>\n<li>Simplify\u00a0powers of\u00a0<em>i<\/em><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<p>Up to now, you\u2019ve known it was impossible to take a square root of a negative number. This is true, using only the <i>real numbers<\/i>. But here you will learn about a new kind of number that lets you work with square roots of negative numbers! Complex numbers are made from both real and imaginary numbers. \u00a0Imaginary numbers are called imaginary because they are impossible and, therefore, exist only in the world of ideas and pure imagination. Imaginary numbers result from taking the square root of a negative number.<\/p>\n<p>Here we will first define and perform algebraic operations on complex numbers, then we will provide examples of quadratic equations that have solutions that are complex numbers.<\/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\u2014they seem pretty <i>real<\/i> to us! When something\u2019s not real, you often say it is <i>imaginary<\/i>. So let\u2019s call this new number <i>i<\/i> and 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\\,\\cdot \\,i=-1[\/latex]. We also know that [latex]i\\,\\cdot \\,i={{i}^{2}}[\/latex], so we can conclude that [latex]{{i}^{2}}=-1[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]{{i}^{2}}=-1[\/latex]<\/p>\n<p>The number <i>i <\/i>allows us to work with roots of all negative numbers, not just [latex]\\sqrt{-1}[\/latex]. There are two important rules to remember: [latex]\\sqrt{-1}=i[\/latex], and [latex]\\sqrt{ab}=\\sqrt{a}\\sqrt{b}[\/latex]. You will use these rules to rewrite the square root of a negative number as the square root of a positive number times [latex]\\sqrt{-1}[\/latex]. Next you will simplify the square root and rewrite [latex]\\sqrt{-1}[\/latex] as <i>i. <\/i>Let\u2019s try an example.<\/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\">Use the rule [latex]\\sqrt{ab}=\\sqrt{a}\\sqrt{b}[\/latex] to rewrite this as a product using [latex]\\sqrt{-1}[\/latex].<\/p>\n<p>[latex]\\sqrt{-4}=\\sqrt{4\\cdot -1}=\\sqrt{4}\\sqrt{-1}[\/latex]<\/p>\n<p>Since 4 is a perfect square\u00a0[latex](4=2^{2})[\/latex], you can simplify the square root of 4.<\/p>\n<p>[latex]\\sqrt{4}\\sqrt{-1}=2\\sqrt{-1}[\/latex]<\/p>\n<p>Use the definition of <i>i<\/i> to rewrite [latex]\\sqrt{-1}[\/latex] as <i>i.<\/i><\/p>\n<p>[latex]2\\sqrt{-1}=2i[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\sqrt{-4}=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\">Use the rule [latex]\\sqrt{ab}=\\sqrt{a}\\sqrt{b}[\/latex] to rewrite this as a product using [latex]\\sqrt{-1}[\/latex].<\/p>\n<p>[latex]\\sqrt{-18}=\\sqrt{18\\cdot -1}=\\sqrt{18}\\sqrt{-1}[\/latex]<\/p>\n<p>Since 18 is <i>not<\/i> a perfect square, use the same rule to rewrite it using factors that are perfect squares. In this case, 9 is the only perfect square factor, and the square root of 9 is 3.<\/p>\n<p>[latex]\\sqrt{18}\\sqrt{-1}=\\sqrt{9}\\sqrt{2}\\sqrt{-1}=3\\sqrt{2}\\sqrt{-1}[\/latex]<\/p>\n<p>Use the definition of <i>i<\/i> to rewrite [latex]\\sqrt{-1}[\/latex] as <i>i.<\/i><\/p>\n<p>[latex]3\\sqrt{2}\\sqrt{-1}=3\\sqrt{2}i=3i\\sqrt{2}[\/latex]<\/p>\n<p>Remember to write <i>i<\/i> in front of the radical.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\sqrt{18}=3i\\sqrt[{}]{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]-\\sqrt{-72}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q503996\">Show Solution<\/span><\/p>\n<div id=\"q503996\" class=\"hidden-answer\" style=\"display: none\">Use the rule [latex]\\sqrt{ab}=\\sqrt{a}\\sqrt{b}[\/latex] to rewrite this as a product using [latex]\\sqrt{-1}[\/latex].<\/p>\n<p>[latex]-\\sqrt{-72}=-\\sqrt{72\\cdot -1}=-\\sqrt{72}\\sqrt{-1}[\/latex]<\/p>\n<p>Since 72 is <i>not<\/i> a perfect square, use the same rule to rewrite it using factors that are perfect squares. Notice that 72 has three perfect squares as factors: 4, 9, and 36. It\u2019s easiest to use the largest factor that is a perfect square.<\/p>\n<p>[latex]-\\sqrt{72}\\sqrt{-1}=-\\sqrt{36}\\sqrt{2}\\sqrt{-1}=-6\\sqrt{2}\\sqrt{-1}[\/latex]<\/p>\n<p>Use the definition of <i>i<\/i> to rewrite [latex]\\sqrt{-1}[\/latex] as <i>i.<\/i><\/p>\n<p>[latex]-6\\sqrt{2}\\sqrt{-1}=-6\\sqrt{2}i=-6i\\sqrt{2}[\/latex]<\/p>\n<p>Remember to write <i>i<\/i> in front of the radical.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]-\\sqrt{-}72=-6i\\sqrt[{}]{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>You may have wanted to simplify [latex]-\\sqrt{-72}[\/latex] using different factors. Some may have thought of rewriting this radical as [latex]-\\sqrt{-9}\\sqrt{8}[\/latex], or [latex]-\\sqrt{-4}\\sqrt{18}[\/latex], or [latex]-\\sqrt{-6}\\sqrt{12}[\/latex] for instance. Each of these radicals would have eventually yielded the same answer of [latex]-6i\\sqrt{2}[\/latex].<\/p>\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>Find perfect squares within the radical.<\/li>\n<li>Rewrite the radical using the rule [latex]\\sqrt{ab}=\\sqrt{a}\\cdot \\sqrt{b}[\/latex].<\/li>\n<li>Rewrite [latex]\\sqrt{-1}[\/latex] as <i>i<\/i>.<\/li>\n<\/ul>\n<p>Example: [latex]\\sqrt{-18}=\\sqrt{9}\\sqrt{-2}=\\sqrt{9}\\sqrt{2}\\sqrt{-1}=3i\\sqrt{2}[\/latex]<\/p>\n<\/div>\n<h2>Complex Numbers<\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2527 size-full aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2016\/06\/22231825\/CNX_Precalc_Figure_03_01_0012.jpg\" alt=\"Showing the real and imaginary parts of 5 + 2i. In this complex number, 5 is the real part and 2i is the complex part.\" width=\"487\" height=\"72\" \/><\/p>\n<p id=\"fs-id1165135500790\">A <strong>complex number<\/strong> is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written <em>a\u00a0<\/em>+ <em>bi<\/em>\u00a0where <em>a<\/em>\u00a0is the real part and <em>bi<\/em>\u00a0is the imaginary part. For example, [latex]5+2i[\/latex] is a complex number. So, too, is [latex]3+4\\sqrt{3}i[\/latex].<span id=\"fs-id1165137832295\"><br \/>\n<\/span><\/p>\n<p id=\"fs-id1165137892327\">Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative\u00a0real number. Recall, when a positive real number is squared, the result is a positive real number and when a negative real number is squared, again, the result is a positive real number. Complex numbers are a combination of real and imaginary numbers.<\/p>\n<p>You can use the usual operations (addition, subtraction, multiplication, and so on) with imaginary numbers. You\u2019ll see more of that, later. When you <i>add<\/i> a real number to an imaginary number, however, you get a complex number. A complex number is any number in the form [latex]a+bi[\/latex], where <i>a<\/i> is a real number and <i>bi<\/i> is an imaginary number. The number <i>a<\/i> is sometimes called the real part of the complex number, and <i>bi<\/i> is sometimes called 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>3<\/td>\n<td>[latex]7i[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]18\u201332i[\/latex]<\/td>\n<td>18<\/td>\n<td>[latex]\u221232i[\/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]i\\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}i[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>In a number with a radical as part of <i>b<\/i>, such as [latex]-\\frac{3}{5}+i\\sqrt{2}[\/latex]\u00a0above, the imaginary <i>i<\/i> should be written in front of the radical. Though writing this number as [latex]-\\frac{3}{5}+\\sqrt{2}i[\/latex] is technically correct, it makes it much more difficult to tell whether <i>i<\/i> is 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. 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>Number in 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>17<\/td>\n<td>[latex]17+0i[\/latex]<\/td>\n<td>17<\/td>\n<td>[latex]0i[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u22123i[\/latex]<\/td>\n<td>[latex]0\u20133i[\/latex]<\/td>\n<td>0<\/td>\n<td>[latex]\u22123i[\/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 <i>a<\/i> is written [latex]a+0i[\/latex] in complex form. Similarly, any imaginary number can be expressed as a complex number. By making [latex]a=0[\/latex], any imaginary number [latex]bi[\/latex] is written [latex]0+bi[\/latex] in complex form.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write 83.6 as 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\">Remember that a complex number has the form [latex]a+bi[\/latex]. You need to figure out what <i>a<\/i> and <i>b<\/i> need to be.<\/p>\n<p>[latex]a+bi[\/latex]<\/p>\n<p>Since 83.6 is a real number, it is the real part (a) of the complex number [latex]a+bi[\/latex]. A real number does not contain any imaginary parts, so the value of b is 0.<i><br \/>\n<\/i><\/p>\n<p>[latex]83.6+bi[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[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] as 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\">Remember that a complex number has the form [latex]a+bi[\/latex].\u00a0You need to figure out what <i>a<\/i> and <i>b<\/i> need to be.<\/p>\n<p>[latex]a+bi[\/latex]<\/p>\n<p>Since [latex]\u22123i[\/latex] is an imaginary number, it is the imaginary part (<em>bi<\/em>) of the complex number [latex]a+bi[\/latex]. This imaginary number has no real parts, so the value of a is 0.<\/p>\n<p>[latex]a\u20133i[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]0\u20133i[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next video we show more examples of how to write numbers as complex numbers.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Write Number in the Form of Complex Numbers\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/mfoOYdDkuyY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Add and Subtract Complex Numbers<\/h2>\n<p>Any time new kinds of numbers are introduced, one of the first questions that needs to be addressed is, \u201cHow do you add them?\u201d In this topic, you\u2019ll learn how to add complex numbers and also how to subtract.<\/p>\n<p>First, consider the following expression.<\/p>\n<p style=\"text-align: center\">[latex](6x+8)+(4x+2)[\/latex]<\/p>\n<p>To simplify this expression, you combine the like terms, [latex]6x[\/latex] and [latex]4x[\/latex].<i>\u00a0<\/i>These are like terms because they have the same variable with the same exponents. Similarly, 8 and 2 are like terms because they are both constants, with no variables.<\/p>\n<p style=\"text-align: center\">[latex](6x+8)+(4x+2)=10x+10[\/latex]<\/p>\n<p>In the same way, you can simplify expressions with radicals.<\/p>\n<p style=\"text-align: center\">[latex](6\\sqrt{3}+8)+(4\\sqrt{3}+2)=10\\sqrt{3}+10[\/latex]<\/p>\n<p>You can add [latex]6\\sqrt{3}[\/latex] to [latex]4\\sqrt{3}[\/latex] because the two terms have the same radical, [latex]\\sqrt{3}[\/latex], just as 6<i>x<\/i> and 4<i>x<\/i> have the same variable and exponent.<\/p>\n<p>The number <i>i <\/i>looks like a variable, but remember that it is equal to [latex]\\sqrt{-1}[\/latex]. The great thing is you have no new rules to worry about\u2014whether you treat it as a variable or a radical, the exact same rules apply to adding and subtracting <strong>complex numbers<\/strong>. You combine the imaginary parts (the terms with <i>i<\/i>),<i> <\/i>and you combine the real parts.<\/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\">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] and [latex]3i\u20132i=(3\u20132)i=i[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex](\u22123+3i)+(7\u20132i)=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\">Be sure to distribute the subtraction sign to all terms in the subtrahend.<\/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] and\u00a0[latex]3i+2i=(3+2)i=5i[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex](\u22123+3i)\u2013(7\u20132i)=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.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" 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>Multiply and Divide Complex Numbers<\/h2>\n<section id=\"fs-id1165137417169\">\n<p id=\"fs-id1165137832911\">Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.<\/p>\n<h3>\u00a0Multiplying a Complex Number by a Real Number<\/h3>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2535 size-full aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2016\/06\/23153423\/CNX_Precalc_Figure_03_01_0062.jpg\" alt=\"Showing how distribution works for complex numbers. For 3(6+2i), 3 is multiplied to both the real and imaginary parts. So we have (3)(6)+(3)(2i) = 18 + 6i.\" width=\"487\" height=\"87\" \/><\/p>\n<p>Let\u2019s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So,for 3(6+2i), 3 is multiplied to both the real and imaginary parts. So we have (3)(6)+(3)(2i) = 18 + 6i.<\/p>\n<div id=\"fs-id1165137745292\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137426118\">How To: Given a complex number and a real number, multiply to find the product.<\/h3>\n<ol id=\"fs-id1165137793647\">\n<li>Use the distributive property.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the product [latex]4\\left(2+5i\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q374377\">Show Answer<\/span><\/p>\n<div id=\"q374377\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137804818\">Distribute the 4.<\/p>\n<div id=\"eip-id1165135436310\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{cc}4\\left(2+5i\\right)=\\left(4\\cdot 2\\right)+\\left(4\\cdot 5i\\right)\\hfill \\\\ =8+20i\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Multiplying Complex Numbers Together<\/h2>\n<\/section>\n<section id=\"fs-id1165137650841\">\n<p id=\"fs-id1165137832483\">Now, let\u2019s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get<\/p>\n<div id=\"eip-586\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left(a+bi\\right)\\left(c+di\\right)=ac+adi+bci+bd{i}^{2}[\/latex]<\/div>\n<p id=\"fs-id1165137734803\">Because [latex]{i}^{2}=-1[\/latex], we have<\/p>\n<div id=\"eip-523\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left(a+bi\\right)\\left(c+di\\right)=ac+adi+bci-bd[\/latex]<\/div>\n<p id=\"fs-id1165135186757\">To simplify, we combine the real parts, and we combine the imaginary parts.<\/p>\n<div id=\"eip-794\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\left(a+bi\\right)\\left(c+di\\right)=\\left(ac-bd\\right)+\\left(ad+bc\\right)i[\/latex]<\/div>\n<div id=\"fs-id1165137642817\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137724898\">How To: Given two complex numbers, multiply to find the product.<\/h3>\n<ol id=\"fs-id1165137561156\">\n<li>Use the distributive property or the FOIL method.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Multiply [latex]\\left(4+3i\\right)\\left(2 - 5i\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q188458\">Show Answer<\/span><\/p>\n<div id=\"q188458\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137459488\">Use [latex]\\left(a+bi\\right)\\left(c+di\\right)=\\left(ac-bd\\right)+\\left(ad+bc\\right)i[\/latex]<\/p>\n<div id=\"eip-id1165137762412\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}{ccc}\\left(4+3i\\right)\\left(2 - 5i\\right)=\\left(4\\cdot 2 - 3\\cdot \\left(-5\\right)\\right)+\\left(4\\cdot \\left(-5\\right)+3\\cdot 2\\right)i\\hfill \\\\ \\text{ }=\\left(8+15\\right)+\\left(-20+6\\right)i\\hfill \\\\ \\text{ }=23 - 14i\\hfill \\end{array}[\/latex]<\/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-4\" 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 <em>i<\/em><\/h2>\n<p id=\"fs-id1165132919554\">The powers of <em>i<\/em>\u00a0are cyclic. Let\u2019s look at what happens when we raise <em>i<\/em>\u00a0to increasing powers.<\/p>\n<div id=\"eip-783\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}{i}^{1}=i\\\\ {i}^{2}=-1\\\\ {i}^{3}={i}^{2}\\cdot i=-1\\cdot i=-i\\\\ {i}^{4}={i}^{3}\\cdot i=-i\\cdot i=-{i}^{2}=-\\left(-1\\right)=1\\\\ {i}^{5}={i}^{4}\\cdot i=1\\cdot i=i\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137530297\">We can see that when we get to the fifth power of <em>i<\/em>, it is equal to the first power. As we continue to multiply <em>i<\/em>\u00a0by itself for increasing powers, we will see a cycle of 4. Let\u2019s examine the next 4 powers of <em>i<\/em>.<\/p>\n<div id=\"eip-477\" class=\"equation unnumbered\">[latex]\\begin{array}{i}^{6}={i}^{5}\\cdot i=i\\cdot i={i}^{2}=-1\\\\ {i}^{7}={i}^{6}\\cdot i={i}^{2}\\cdot i={i}^{3}=-i\\\\ {i}^{8}={i}^{7}\\cdot i={i}^{3}\\cdot i={i}^{4}=1\\\\ {i}^{9}={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 Answer<\/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 4 goes into 35: [latex]35=4\\cdot 8+3[\/latex].<\/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<div id=\"fs-id1165137758921\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"fs-id1165135186727\"><strong>Can we write [latex]{i}^{35}[\/latex] in other helpful ways?<\/strong><\/p>\n<p id=\"fs-id1165135444053\"><em>As we saw in Example 11, 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>\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 id=\"fs-id1165135255472\"><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<p>In the following video you will see more examples of how to simplify powers of\u00a0<em>i<\/em>.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex:  Raising the imaginary unit i to powers\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/sfP6SmEYHRw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Dividing Complex Numbers<\/h2>\n<p id=\"fs-id1165137612241\">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. This idea is similar to rationalizing the denominator of a fraction that contains a radical.\u00a0To eliminate the complex or imaginary number 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<p id=\"fs-id1165137435064\">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]. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.<\/p>\n<p id=\"fs-id1165137611741\">Suppose we want to divide [latex]c+di[\/latex] by [latex]a+bi[\/latex], where neither <em>a<\/em>\u00a0nor <em>b<\/em>\u00a0equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.<\/p>\n<div id=\"eip-225\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\displaystyle \\frac{c+di}{a+bi}\\text{ where }a\\ne 0\\text{ and }b\\ne 0[\/latex]<\/div>\n<p id=\"fs-id1165134148263\">Multiply the numerator and denominator by the complex conjugate of the denominator.<\/p>\n<div id=\"eip-32\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\displaystyle \\frac{\\left(c+di\\right)}{\\left(a+bi\\right)}\\cdot \\frac{\\left(a-bi\\right)}{\\left(a-bi\\right)}=\\frac{\\left(c+di\\right)\\left(a-bi\\right)}{\\left(a+bi\\right)\\left(a-bi\\right)}[\/latex]<\/div>\n<p id=\"fs-id1165135260723\">Apply the distributive property.<\/p>\n<div id=\"eip-736\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\displaystyle =\\frac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}}[\/latex]<\/div>\n<p id=\"fs-id1165137871668\">Simplify, remembering that [latex]{i}^{2}=-1[\/latex].<\/p>\n<div id=\"eip-64\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}=\\frac{ca-cbi+adi-bd\\left(-1\\right)}{{a}^{2}-abi+abi-{b}^{2}\\left(-1\\right)}\\hfill \\\\ =\\frac{\\left(ca+bd\\right)+\\left(ad-cb\\right)i}{{a}^{2}+{b}^{2}}\\hfill \\end{array}[\/latex]<\/div>\n<div id=\"fs-id1165135203870\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Complex Conjugate<\/h3>\n<p id=\"fs-id1165137793758\">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 id=\"fs-id1165135487089\">\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<\/h3>\n<p id=\"fs-id1165137896182\">Find the complex conjugate of each number.<\/p>\n<ol id=\"fs-id1165137896185\">\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=\"q426623\">Show Answer<\/span><\/p>\n<div id=\"q426623\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165137742669\">\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<\/div>\n<\/div>\n<\/div>\n<h3>Analysis of the Solution<\/h3>\n<div id=\"Example_03_01_06\" class=\"example\">\n<div id=\"fs-id1165134032261\" class=\"exercise\">\n<div id=\"fs-id1165137772480\" class=\"commentary\">\n<p id=\"fs-id1165137762415\">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 <em>i<\/em>.<\/p>\n<p>In the last video you will see more examples of dividing complex numbers.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" 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<div id=\"fs-id1165137409413\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135471100\">How To: Given two complex numbers, divide one by the other.<\/h3>\n<ol id=\"fs-id1165135471104\">\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<\/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=\"q665746\">Show Answer<\/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]\\displaystyle \\frac{\\left(2+5i\\right)}{\\left(4-i\\right)}[\/latex]<\/div>\n<p id=\"fs-id1165137639613\">Then we multiply the numerator and denominator by the complex conjugate of the denominator.<\/p>\n<div id=\"eip-id1165137400110\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\displaystyle \\frac{\\left(2+5i\\right)}{\\left(4-i\\right)}\\cdot \\frac{\\left(4+i\\right)}{\\left(4+i\\right)}[\/latex]<\/div>\n<p id=\"fs-id1165137474228\">To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL).<\/p>\n<div id=\"eip-id1165131986989\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}\\frac{\\left(2+5i\\right)}{\\left(4-i\\right)}\\cdot \\frac{\\left(4+i\\right)}{\\left(4+i\\right)}=\\frac{8+2i+20i+5{i}^{2}}{16+4i - 4i-{i}^{2}}\\hfill & \\hfill \\\\ \\text{ }=\\frac{8+2i+20i+5\\left(-1\\right)}{16+4i - 4i-\\left(-1\\right)}\\hfill & \\text{Because } {i}^{2}=-1\\hfill \\\\ \\text{ }=\\frac{3+22i}{17}\\hfill & \\hfill \\\\ \\text{ }=\\frac{3}{17}+\\frac{22}{17}i\\hfill & \\text{Separate real and imaginary parts}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137930346\">Note that this expresses the quotient in standard form.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137650841\"><\/section>\n<h2><\/h2>\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-237\">\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><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 1: Adding and Subtracting Complex Numbers. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/SGhTjioGqqA\">https:\/\/youtu.be\/SGhTjioGqqA<\/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>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download fro free at:  http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/li><li>Ex: Raising the imaginary unit i to powers. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/sfP6SmEYHRw\">https:\/\/youtu.be\/sfP6SmEYHRw<\/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>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":21,"menu_order":4,"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\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Adding and Subtracting Complex Numbers\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/SGhTjioGqqA\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download fro free at:  http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Raising the imaginary unit i to powers\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/sfP6SmEYHRw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"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":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-237","chapter","type-chapter","status-publish","hentry"],"part":774,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/237","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":20,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/237\/revisions"}],"predecessor-version":[{"id":5062,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/237\/revisions\/5062"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/774"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/237\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/media?parent=237"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=237"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/contributor?post=237"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/license?post=237"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}