{"id":239,"date":"2019-08-15T17:06:52","date_gmt":"2019-08-15T17:06:52","guid":{"rendered":"http:\/\/104.42.120.246.xip.io\/calculus-tutorials\/?page_id=239"},"modified":"2020-06-17T18:39:30","modified_gmt":"2020-06-17T18:39:30","slug":"complex-numbers","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/calculus\/hmc-mathematics-calculus-online-tutorials\/precalculus\/complex-numbers\/","title":{"rendered":"Complex Numbers"},"content":{"rendered":"\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n\n\n\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<meta http-equiv=\"X-UA-Compatible\" content=\"IE=EmulateIE7\">\n\n\n\n<title>Complex Numbers &#8211; HMC Calculus Tutorial<\/title>\n\n\n\n<p> The complex numbers are an extension of the real numbers containing  all  roots of quadratic equations.  If we define $i$ to be a solution of   the equation $x^{2}= -1$, them the set $\\mathbb{C}$ of complex numbers   is represented in <strong>standard form<\/strong> as $$ \\left\\{ a+bi |  a,b \\in  R\\right\\}. $$ We often use the variable $z=a+bi$ to represent a  complex  number.  The number $a$ is called the <strong>real part<\/strong> of $z$: Re $z$ while $b$ is called the <strong>imaginary part<\/strong> of $z$: Im $z$.  Two complex numbers are <strong>equal<\/strong> if and only if their real parts are equal and their imaginary parts are equal. <\/p>\n\n\n\n<p>We represent complex numbers graphically by associating $z=a+bi$ with the point $(a,b)$ on the <strong>complex plane<\/strong>.  <\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"234\" height=\"195\" src=\"https:\/\/i0.wp.com\/math.hmc.edu\/calculus\/wp-content\/uploads\/sites\/3\/2019\/08\/complex1.gif?resize=234%2C195&#038;ssl=1\" alt=\"Follow the image link for a complete description of the image\" class=\"wp-image-326\"\/><\/figure><\/div>\n\n\n\n<h4 class=\"wp-block-heading\">Basic Operations<\/h4>\n\n\n\n<p>\nThe basic operations on complex numbers are defined as follows: \n\\begin{eqnarray*} (a+bi) + (c+di) &amp; = &amp; (a+c) + (b+d)i \\\\ (a+bi)\n &#8211; (c+di) &amp; = &amp; (a-c) + (b-d)i \\\\ (a+bi)(c+di) &amp; = &amp; ac +\n adi + bci + bdi^2 \\\\ &amp; = &amp; (ac-bd) + (bc+ad)i \\end{eqnarray*} \n$$ \\frac{a+bi}{c+di} = \\frac{a+bi}{c+di}\\cdot\\frac{c-di}{c-di} = \n\\frac{ac+bd}{c^2+d^2} + \\frac{bc-ad}{c^2+d^2}i $$  In dividing $a+bi$ by\n $c+di$, we rationalized the denominator using the fact that \n$(c+di)(c-di) = c^2 -cdi +cdi -d^2i^2 = c^2 + d^2$.  The complex numbers\n $c+di$ and $c-di$ are called <strong>complex conjugates<\/strong>.  If $z=c+di$, \nwe use $\\overline{z}$ to denote $c-di$. Viewing $z=a+bi$ as a vector in \nthe complex plane, it has magnitude $$ |z| = \\sqrt{a^2+b^2}, $$ which we\n call the <strong>modulus<\/strong> or <strong>absolute value<\/strong> of $z$. \n\n<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"132\" height=\"168\" src=\"https:\/\/i0.wp.com\/math.hmc.edu\/calculus\/wp-content\/uploads\/sites\/3\/2019\/08\/complex2.gif?resize=132%2C168&#038;ssl=1\" alt=\"Follow the image link for a complete description of the image\" class=\"wp-image-327\"\/><\/figure><\/div>\n\n\n\n<h6 class=\"wp-block-heading\">Examples<\/h6>\n\n\n\n<ul class=\"wp-block-list\"><li> $(2+3i)(2-3i)=4-6i+6i-9i^2=4+9=13.$\n\n\t<\/li><li> $|2+3i| = |2-3i| = \\sqrt{4+9}=\\sqrt{13}.$\n\n<\/li><\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Polar Form<\/h4>\n\n\n\n<p>\nFor $z=a+bi$, let \\begin{eqnarray*} a &amp; = &amp; r\\cos\\theta \\\\ b \n&amp; = &amp; r\\sin\\theta  \\end{eqnarray*} from which we can also obtain\n\n<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"184\" height=\"167\" src=\"https:\/\/i0.wp.com\/math.hmc.edu\/calculus\/wp-content\/uploads\/sites\/3\/2019\/08\/complex3.gif?resize=184%2C167&#038;ssl=1\" alt=\"Follow the image link for a complete description of the image\" class=\"wp-image-328\"\/><\/figure><\/div>\n\n\n\n<p>\n\t\t$\\qquad$\\begin{eqnarray*}\n\t\t\tr &amp; = &amp; \\sqrt{a^2+b^2} = |z| \\\\\t\n\t\t\t\\tan\\theta &amp; = &amp; \\frac{b}{a}. \n\t\t\\end{eqnarray*}\n<\/p>\n\n\n\n<p>\n\t\t\tIf you write \n\t\t\t$$\n\t\t\t\\theta = \\tan^{-1}\\frac{y}{x},\n\t\t\t$$\n\t\t\tbe careful to choose the value for $\\theta$ in the correct quadrant.\n<\/p>\n\n\n\n<p>\n\t\tThen\n\t\t$$\n\t\tz=r\\cos\\theta + ir\\sin\\theta\n\t\t$$\n\t\tand so, by Euler&#8217;s Equation, we obtain the <b>polar form<\/b>\n\t\t$$\n\t\tz=re^{i\\theta}.\n\t\t$$\n<\/p>\n\n\n\n<p>\n\t\t\t<b>Euler&#8217;s Equation:<\/b>\n\t\t\t$$\n\t\t\te^{i\\theta}=\\cos\\theta + i\\sin\\theta\n\t\t\t$$\n<\/p>\n\n\n\n<p> Here, $r$ is the magnitude of $z$ and $\\theta$ is called the <b>argument<\/b>  of $z$: arg $z$.  The argument is not unique; we can add multiples of $2\\pi$ to $\\theta$ without changing $z$.  We define Arg $z$, the principal value of the argument, to be in $(-\\pi,\\pi]$.  The principal value is unique for each $z$ but creates unavoidable, yet interesting, complications due to its discontinuity across the negative real axis where it jumps from $\\pi$ to $-\\pi$. This jump is called a <b>branch cut<\/b>. <\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Examples<\/h6>\n\n\n\n<ul class=\"wp-block-list\"><li> $e^{i\\pi} = \\cos\\pi + i\\sin\\pi = -1$\n\n\t<\/li><li> $3e^{i\\pi\/2} = 3(\\cos\\frac{\\pi}{2} + i\\sin\\frac{\\pi}{2}) = 3i$\n\n\t<\/li><li> $2e^{i\\pi\/6} = 2(\\cos\\frac{\\pi}{6} + i\\sin\\frac{\\pi}{6}) =\t\\sqrt{3} + i$\n\n<\/li><\/ul>\n\n\n\n<p> Multiplication and division of complex numbers is amazingly simple in polar form!  If $z_1 = r_1e^{i\\theta_1}$ and $z_2 = r_2e^{i\\theta_2}$, then \\begin{eqnarray*} z_1z_2 &amp; = &amp; r_1r_2e^{i(\\theta_1 + \\theta_2)} \\\\ \\frac{z_1}{z_2} &amp; = &amp; \\frac{r_1}{r_2}e^{i(\\theta_1-\\theta_2)}  \\end{eqnarray*} If $z=re^{i\\theta}$, then $\\overline{z}=re^{-i\\theta}$. Do you see why? So $z\\overline{z} = (re^{i\\theta})(re^{-i\\theta}) = r^{2}$. <\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\t\tTo calculate $(1+i)^{8}$, we can first rewrite $1+i$ as\n\t\t$\\sqrt{2}e^{i\\pi\/4}$.  Then\n\t\t\\begin{eqnarray*}\n\t\t\t\\left(\\sqrt{2}e^{i\\pi\/4}\\right)^{8} &amp; = &amp; (\\sqrt{2})^{8}e^{i8\\pi\/4} \\\\\n\t\t\t &amp; = &amp; 16e^{2\\pi i} \\\\\n\t\t\t &amp; = &amp; 16.  \n\t\t\\end{eqnarray*}\n<\/p>\n\n\n\n<p>\n\t\t\t\\begin{eqnarray*}\n\t\t\t\t\\sqrt{1^2+1^2} &amp; = &amp; \\sqrt{2} \\\\\n\t\t\t\t\\tan^{-1} \\left( \\frac{1}{1} \\right) &amp; = &amp; \\frac{\\pi}{4} \n\t\t\t\\end{eqnarray*}\n<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Roots of Unity<\/h4>\n\n\n\n<p>\n\nThe equation\n$$\nz^{n}=1\n$$\nhas $n$ complex-valued solutions, called the $n^{th}$ roots of unity.\nSince we know each root has magnitude 1, let $z=e^{i\\theta}$.  Then\n\n<\/p>\n\n\n\n<p>\n\t\t\\begin{eqnarray*}\n\t\t\t(e^{i\\theta})^{n} &amp; = &amp; 1 \\\\\n\t\t\te^{in\\theta} &amp; = &amp; e^{i(2\\pi k)} \\\\\n\t\t\tn\\theta &amp; = &amp; 2\\pi k \\\\\n\t\t\t\\theta &amp; = &amp; \\frac{2 \\pi k}{n} \n\t\t\\end{eqnarray*}\n\t\tso the $n^{th}$ roots of unity are of the form\n\t\t$$\n\t\tz=e^{i\\frac{2 \\pi k}{n}}.\n\t\t$$\n<\/p>\n\n\n\n<p>\n\t\t\t\t$(e^{i\\theta})^{n} = e^{in\\theta}$\n\t\t\t\ttogether with Euler&#8217;s Equation, gives us <b>deMoivre&#8217;s Formula:<\/b>\n\t\t\t\t$$\n\t\t\t\t(\\cos\\theta + i\\sin\\theta)^{n} = \\cos n\\theta + i\\sin n\\theta\n\t\t\t\t$$\n<\/p>\n\n\n\n<p>\n\t\t\t\t$$\n\t\t\t\t1=e^{0i}=e^{2\\pi ki}\n\t\t\t\t$$\n\t\t\t\tfor $k=0,\\pm 1,\\pm2,\\ldots$\n<\/p>\n\n\n\n<p>\n\t\tThere are $n$ distinct roots, after which we start dumplicating roots\n\t\talready found.\n<\/p>\n\n\n\n<p>\n\t\t\tThese are evenly spaced around the unit circle.\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nThe $3rd$ roots of unity are\n\\begin{eqnarray*}\n\t1 &amp; &amp;  \\\\\n\te^{i\\frac{2\\pi}{3}} &amp; = &amp; -\\frac{1}{2} + i\\frac{\\sqrt{3}}{2} \\\\\n\te^{-i\\frac{2\\pi}{3}} &amp; = &amp; -\\frac{1}{2} &#8211;  i\\frac{\\sqrt{3}}{2}\n\\end{eqnarray*}\nYou can verify that $(-\\frac{1}{2}+ i\\frac{\\sqrt{3}}{2})^{3} = 1$ and\n$(-\\frac{1}{2} &#8211;  i\\frac{\\sqrt{3}}{2})^{3} = 1$.\n\n<\/p>\n\n\n\n<p>\nThis tutorial has reviewed the basics of complex arithmetic.  The\nmethods of complex analysis, which build on this background, are both\nintriguing and powerful!\n\n<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<center>\n<h3>Key Concept<\/h3>\n<\/center>\n\n\n\n<p>\n\t\t<b>Standard Form<\/b><br>\n\t\t$\\begin{array}{rcl}\n\t\t\tz &amp; = &amp; a +bi \\\\\n\t\t\ta &amp; = &amp; \\mbox{Re } z \\\\\n\t\t\tb &amp; = &amp; \\mbox{Im } z \\\\\n\t\t\t|z| &amp; = &amp; \\sqrt{a^2+b^2} \\\\\n\t\t\t\\overline{z} &amp; = &amp; a-bi \n\t\t\\end{array}$\n<\/p>\n\n\n\n<p>\n\t\t\t$\\begin{array}{rcl}\n\t\t\t\ta &amp; = &amp; r\\cos\\theta \\\\\n\t\t\t\tb &amp; = &amp; r\\sin\\theta \\\\\n\t\t\t\tr &amp; = &amp; \\sqrt{a^2 + b^2} \\\\\n\t\t\t\t\\tan\\theta &amp; = &amp; \\frac{b}{a}\n\t\t\t\\end{array}$\n<\/p>\n\n\n\n<p>\n\t\t<b>Polar Form<\/b><br>\n\t\t$\\begin{array}{rcl}\n\t\t\tz &amp; = &amp; re^{i\\theta} \\\\\n\t\t\tr &amp; = &amp; |z| \\\\\n\t\t\t\\theta &amp; = &amp; \\mbox{arg } z \\\\\n\t\t\t\\overline{z} &amp; = &amp; re^{-i\\theta}  \n\t\t\\end{array}$\n<\/p>\n\n\n\n<p>\nEuler&#8217;s Equation,\n$$\ne^{i\\theta} = \\cos\\theta + i\\sin\\theta,\n$$\nprovides the connection between these two representations of complex\nnumbers.\n\n<!------------------------>\n\n\n<br>\n\n<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p>\n\n[<a href=\"https:\/\/physics.hmc.edu\/ct\/quiz\/QZ3610\/\">I&#8217;m ready to take the quiz.<\/a>]\n[<a href=\"#top\">I need to review more.<\/a>]<br>\n\n\n<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Complex Numbers &#8211; HMC Calculus Tutorial The complex numbers are an extension of the real numbers containing all roots of quadratic equations. If we define $i$ to be a solution of the equation $x^{2}= -1$, them the set $\\mathbb{C}$ of complex numbers is represented in standard form as $$ \\left\\{ a+bi | a,b \\in R\\right\\}.&hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"parent":55,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[],"class_list":["post-239","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/239","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/comments?post=239"}],"version-history":[{"count":24,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/239\/revisions"}],"predecessor-version":[{"id":1195,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/239\/revisions\/1195"}],"up":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/55"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/media?parent=239"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/tags?post=239"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}