\hfill \thepage} %} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section{Ma 116 Lecture 4/17/00} \vspace{1pt}Recall that last time we had introduced the concept of the roots of a complex number. \vspace{1pt} \subsection{\protect\vspace{1pt}Roots and Fractional Powers} \vspace{1pt}Definition. If \begin{equation*} w^{n}=z, \end{equation*} we say $w$ is an $n$ th root of $z.$ We denote this by $w=z^{\frac{1}{n}},$ \ $n=1,2,\ldots $ \vspace{1pt}We derived the formula% \begin{equation*} w=z^{\frac{1}{n}}=r^{\frac{1}{n}}\left[ \cos \left( \frac{\theta +2k\pi }{n}% \right) +i\sin \left( \frac{\theta +2k\pi }{n}\right) \right] \text{ \ }% k=0,1,\ldots ,n-1 \end{equation*} for the $n$th roots of the complex number $z.$ \vspace{1pt} To find $z^{\frac{m}{n}}$ we find $(z^{m})^{\frac{1}{n}}$. \paragraph{Example:} Find all possible values of $(2-2i)^{\frac{3}{5}}$. \begin{center} \begin{equation*} (2-2i)^{3}=(4-8i-4)(2-2i)=-16-16i. \end{equation*} \end{center} \vspace{1pt} Therefore we need the fifth roots of $-16-16i$. Now \begin{equation*} \tan \theta =\frac{y}{x}=\frac{-16}{-16}=+1. \end{equation*} \vspace{1pt} Since $-16-16i$ is in the third quadrant $\theta =\pi +\frac{\pi }{4}=\frac{% 5\pi }{4}$. Thus the fifth roots are: \begin{equation*} (\sqrt{512})^{\frac{1}{5}}\left[ \cos \left( \frac{\frac{5\pi }{4}}{5}% \right) +i\sin \left( \frac{5\pi }{20}\right) \right] \qquad \qquad \qquad k=0 \end{equation*} \begin{eqnarray*} (\sqrt{512})^{\frac{1}{5}}\left[ \cos \left( \frac{\frac{5\pi }{4}+2\pi }{5}% \right) +i\sin \left( \frac{5\pi }{20}+\frac{2\pi }{5}\right) \right] \qquad \ \ \ k &=&1 \\ &=&(\sqrt{512})^{\frac{1}{5}}\left[ \cos \left( \frac{13\pi }{20}\right) +i\sin \left( \frac{13\pi }{20}\right) \right] \end{eqnarray*} $\qquad $ \begin{equation*} (\sqrt{512})^{\frac{1}{5}}\left[ \cos \left( \frac{5\pi }{20}+\frac{4\pi }{5}% \right) +i\sin \left( \frac{21\pi }{20}\right) \right] \qquad \qquad \ \ k=2 \end{equation*} \begin{equation*} (\sqrt{512})^{\frac{1}{5}}\left[ \cos \left( \frac{5\pi }{20}+\frac{6\pi }{5}% \right) +i\sin \left( \frac{29\pi }{20}\right) \right] \qquad \qquad \ \ k=3 \end{equation*} \begin{equation*} (\sqrt{512})^{\frac{1}{5}}\left[ \cos \left( \frac{5\pi }{20}+\frac{8\pi }{5}% \right) +i\sin \left( \frac{37\pi }{20}\right) \right] \qquad \qquad \ \ k=4 \end{equation*} \vspace{1pt} \subsection{Complex Functions} A complex-valued function $f$ of a real variable $x$ is a function of the form $f(x)=u(x)+i$ $v(x)$ \ where $u(x)$ and $v(x)$ are real functions and $% i=\sqrt{-1}$. Definition. \ If $f=u+iv$ , $u\;,v$ real functions, then $f$ is continuous if $u$ and $v$ are continuous; $f$ is differentiable if $u$ and $v$ are differentiable and \begin{equation*} f^{\prime }(x)=u^{\prime }(x)+iv^{\prime }(x) \end{equation*} . \paragraph{Examples:} \begin{enumerate} \item \vspace{1pt}Let $f(x)=3x+ix^{2},$the $f(4)=3\left( 4\right) +i\left( 4\right) ^{2}=\allowbreak 12+16i.$ Also $f^{\prime }(x)=3+2ix.$ \end{enumerate} $\vspace{1pt}$ \begin{enumerate} \item[2.] $\frac{d}{dx}(3x+ix^{2})^{2}$ $% =2(3x+ix^{2})(3+2ix)=2(9x-2x^{3}+9ix^{2})$ \ \ \end{enumerate} \ \ \ \subsection{\protect\vspace{1pt}The Complex Exponential} Consider the function \begin{equation*} E(x)=e^{ax}(\cos bx+i\sin bx). \end{equation*} Differentiating $E\left( x\right) ,$ we get \begin{center} \ \begin{eqnarray*} E^{\prime }(x) &=&ae^{ax}(\cos bx+i\sin bx)+e^{ax}(-b\sin bx+bi\cos bx) \\ &=&e^{ax}[a(\cos bx+i\sin bx)+bi(\cos bx+i\sin bx) \\ &=&e^{ax}[a+bi](\cos bx+i\sin bx) \\ &=&(a+bi)E(x) \end{eqnarray*} $\ \ $ . \end{center} Hence $E^{\prime }(x)=(a+bi)$ $E(x)$. Recalling that $\left( e^{cx}\right) ^{\prime }=ce^{cx}$ we define the complex exponential via \begin{center} \begin{equation*} e^{(a+bi)x}=e^{ax}\cos bx+ie^{ax}\sin bx \end{equation*} \end{center} If we let $a=0$ in this last expression, then we get Euler's formula, namely, \vspace{1pt} \begin{equation*} e^{bix}=\cos bx+i\sin bx \end{equation*} Hence \begin{center} \begin{equation*} e^{(a+bi)x}=e^{ax}\cdot e^{bix} \end{equation*} \vspace{1pt} \end{center} Based on the polar representation of a complex number $z=r\left( \cos \theta +i\sin \theta \right) $ and Euler's formula, we may write $z$ in \emph{% exponential form }as \vspace{1pt} \begin{equation*} z=\left| z\right| e^{i\theta }=re^{i\theta } \end{equation*} \vspace{1pt}Note that $\theta $ should be in radians. \paragraph{Example:} We shall write $z=1+\sqrt{3}i$ in exponential form. Now $r=\left| z\right| =% \sqrt{\left( 1\right) ^{2}+\left( \sqrt{3}\right) ^{2}}=2.$ Also $\theta =\arctan \frac{\sqrt{3}}{1}=\allowbreak \frac{1}{3}\pi .$ Thus \begin{equation*} 1+\sqrt{3}i=2e^{i\frac{\pi }{3}}.\vspace{1pt} \end{equation*} \vspace{1pt} \paragraph{Example:} Express $\left( -1,2\right) $ in exponential form. $r=\sqrt{\left( -1\right) ^{2}+\left( 2\right) ^{2}}=\allowbreak \sqrt{5}$ and $\arctan \dfrac{2}{-1}=\allowbreak -\arctan 2\approx \allowbreak -1.\,\allowbreak 107\,1.$ (Note that this is in radians.) Thus \begin{equation*} \left( -1,2\right) =\sqrt{5}e^{-1.\,\allowbreak 107\,1i} \end{equation*} \vspace{1pt} Remark: \begin{equation*} \left( re^{i\theta }\right) ^{n}=r^{n}e^{in\theta } \end{equation*} for $n=1,2,....$ \vspace{1pt} \subsection{Power Series for Sine and Cosine} \vspace{1pt} Recall that if $f\left( x\right) $ has a power series expansion near $x=a,$ then \begin{equation*} f\left( x\right) =\sum_{n=0}^{\infty }\dfrac{f^{\left( n\right) }\left( a\right) }{n!}\left( x-a\right) ^{n} \end{equation*} \vspace{1pt} When we discussed power series, we showed that near $t=0$ \begin{equation*} e^{t}=\sum_{n=0}^{\infty }\dfrac{t^{n}}{n!} \end{equation*} \vspace{1pt} Thus \begin{equation*} e^{i\theta }=\sum_{n=0}^{\infty }\dfrac{\left( i\theta \right) ^{n}}{n!}% =\sum_{n=0}^{\infty }\dfrac{\left( i^{n}\theta ^{n}\right) }{n!} \end{equation*} \vspace{1pt} But \begin{equation*} e^{i\theta }=\cos \theta +i\sin \theta \end{equation*} \vspace{1pt} so that \begin{equation*} \cos \theta +i\sin \theta =\sum_{n=0}^{\infty }\dfrac{\left( i^{n}\theta ^{n}\right) }{n!} \end{equation*} We can get power series for $\sin \theta $ and $\cos \theta $ by equating the real and imaginary parts of the above equation. Note that \vspace{1pt} \begin{equation*} i^{2k}=\left( i^{2}\right) ^{k}=\left( -1\right) ^{k} \end{equation*} \vspace{1pt} and \begin{equation*} \left( i\right) ^{2k+1}=\left( i\right) ^{2k}i=\left( -1\right) ^{k}i \end{equation*} \vspace{1pt} Thus \begin{eqnarray*} \cos \theta +i\sin \theta &=&\sum_{n=0}^{\infty }\dfrac{\left( i^{n}\theta ^{n}\right) }{n!}=\sum_{k=0}^{\infty }\dfrac{i^{2k}\theta ^{2k}}{\left( 2k\right) !}+\sum_{k=0}^{\infty }\dfrac{i^{2k+1}\theta ^{2k+1}}{\left( 2k+1\right) !} \\ &=&\sum_{k=0}^{\infty }\left( -1\right) ^{k}\dfrac{\theta ^{2k}}{\left( 2k\right) !}+i\sum_{k=0}^{\infty }\left( -1\right) ^{k}\dfrac{\theta ^{2k+1}% }{\left( 2k+1\right) !} \end{eqnarray*} \vspace{1pt} Hence \begin{equation*} \cos \theta =\sum_{k=0}^{\infty }\left( -1\right) ^{k}\dfrac{\theta ^{2k}}{% \left( 2k\right) !} \end{equation*} \vspace{1pt} and \begin{equation*} \sin \theta =\sum_{k=0}^{\infty }\left( -1\right) ^{k}\dfrac{\theta ^{2k+1}}{% \left( 2k+1\right) !} \end{equation*} \vspace{1pt} These series are usually derived by taking successive derivatives of $\sin \theta $ and/or $\cos \theta $ and then evaluating them at $x=0.$ These series are valid for all $\theta .$ \vspace{1pt} \subsection{Complex Integrals} \vspace{1pt} Definition: If $u\left( x\right) =f\left( x\right) +ig\left( x\right) $ is a complex-values function of a real variable, then its indefinite integral \begin{equation*} \int u\left( x\right) dx \end{equation*} is an antiderivative of $u.$ \vspace{1pt} \paragraph{Example:} Evaluate \begin{equation*} \int e^{\left( 1+i\right) x}dx \end{equation*} \vspace{1pt} Solution:% \begin{equation*} \int e^{\left( 1+i\right) x}dx=\dfrac{1}{1+i}e^{\left( 1+i\right) x}+C \end{equation*} \vspace{1pt} \paragraph{Example:} Use the result of the previous example to evaluate% \begin{equation*} \int e^{x}\cos xdx\text{ \ \ and \ \ \ }\int e^{x}\sin xdx \end{equation*} Solution: \vspace{1pt} Note that% \begin{eqnarray*} \int e^{x}\cos xdx\text{ }+i\text{\ }\int e^{x}\sin xdx &=&\int e^{\left( 1+i\right) x}dx \\ &=&\left( \dfrac{1}{1+i}\right) e^{\left( 1+i\right) x}+C \end{eqnarray*} \vspace{1pt} Since \begin{equation*} \int e^{x}\cos xdx=\func{Re}\left[ \int e^{\left( 1+i\right) x}dx\right] \end{equation*} and% \begin{equation*} \int e^{x}\sin xdx=\func{Im}\left[ \int e^{\left( 1+i\right) x}dx\right] \end{equation*} Then we can find these integrals by finding the real and imaginary parts of $% \dfrac{1}{1+i}e^{\left( 1+i\right) x}.$ Now% \begin{eqnarray*} \left( \dfrac{1}{1+i}\right) e^{\left( 1+i\right) x} &=&e^{x}\left( \cos x+i\sin x\right) \left( \dfrac{1}{1+i}\right) \\ &=&e^{x}\left( \cos x+i\sin x\right) \left( \dfrac{1}{1+i}\right) \left( \dfrac{1-i}{1-i}\right) \\ &=&e^{x}\left( \cos x+i\sin x\right) \left( \dfrac{1-i}{2}\right) \end{eqnarray*} \vspace{1pt} Thus% \begin{equation*} \int e^{x}\cos xdx=\dfrac{e^{x}}{2}\left( \cos x+\sin x\right) \end{equation*} \vspace{1pt} and% \begin{equation*} \int e^{x}\sin xdx=\dfrac{e^{x}}{2}\left( \sin x-\cos x\right) \end{equation*} \vspace{1pt} This technique is an alternative to technique of Integration by parts which can also be used to evaluate such an integral. \end{document} %%%%%%%%%%%%%%%%%%%% End /document/lec_4_17a_00.tex %%%%%%%%%%%%%%%%%%%