\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 4/10/00} \begin{center} \vspace{1pt} \end{center} \subsection{Complex Numbers} Consider a number of the form $z=x+iy,$ where $x$ and $y$ are real and $% i^{2}=-1.$ We call $x$ the \emph{real part} of $z$ and $y$ is called the \emph{imaginary part.} We write $x=\func{Re}(z)$ and $y=\func{Im}\left( z\right) .$ Let $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$, then \begin{equation*} z_{1}=z_{2}\Leftrightarrow x_{1}=x_{2}\text{ and }y_{1}=y_{2} \end{equation*} \vspace{1pt} \vspace{1pt}Complex numbers are added and subtracted according to the rule \vspace{1pt} \begin{equation*} z_{1}\pm z_{2}=\left( x_{1}\pm x_{2}\right) +i\left( y_{1}\pm y_{2}\right) \end{equation*} and are multiplied according to the rule \vspace{1pt} \begin{equation*} z_{1}\cdot z_{2}=\left( x_{1}+iy_{1}\right) \left( x_{2}+iy_{2}\right) =\left( x_{1}x_{2}-y_{1}y_{2}\right) +i\left( x_{1}y_{2}+x_{2}y_{1}\right) \end{equation*} \paragraph{Examples:} $\left( 2+3i\right) +(4-2i)=\allowbreak 6+i$ $\left( 2+3i\right) -(4-2i)=\allowbreak -2+5i$ $\left( 2+3i\right) (4-2i)=\allowbreak 14+8i$ \vspace{1pt} Note that $z=x+iy=0\Leftrightarrow x=0$ and $y=0.$ Also if $z_{1}\cdot z_{2}=0,$ then either $z_{1}=0$ or $z_{2}=0.$ \vspace{1pt} We now deal with dividing one complex number by another. Suppose $z_{2}\neq 0,$ then \vspace{1pt} \begin{equation*} \dfrac{z_{1}}{z_{2}}=\dfrac{x_{1}+iy_{1}}{x_{2}+iy_{2}} \end{equation*} \vspace{1pt} However, this is not an expression in the form $a+bi,$ so we do some algebraic manipulation to get it into this form. \vspace{1pt} \begin{equation*} \dfrac{z_{1}}{z_{2}}=\left( \dfrac{x_{1}+iy_{1}}{x_{2}+iy_{2}}\right) \left( \dfrac{x_{2}-iy_{2}}{x_{2}-iy_{2}}\right) =\dfrac{x_{1}x_{2}+y_{1}y_{2}}{% x_{2}^{2}+y_{2}^{2}}+i\left( \dfrac{x_{2}y_{1}-x_{1}y_{2}}{% x_{2}^{2}+y_{2}^{2}}\right) \end{equation*} \vspace{1pt} \paragraph{Example:} \begin{eqnarray*} \dfrac{1+4i}{-2+2i} &=&\left( \dfrac{1+4i}{-2+2i}\right) \left( \dfrac{-2-2i% }{-2-2i}\right) \\ &=&\dfrac{-2-2i-8i-8i^{2}}{\left( -2\right) ^{2}-\left( 2i\right) ^{2}} \\ &=&\dfrac{-2-10i+8}{4+4} \\ &=&\dfrac{6-10i}{8} \\ &=&\allowbreak \frac{3}{4}-\frac{5}{4}i \end{eqnarray*} \vspace{1pt} If $z=x+iy,$ then the number $\bar{z}=x-iy$ is called the \emph{complex conjugate} of $z.$ Note that $z\cdot \bar{z}=x^{2}+y^{2}.$ \ The absolute value of a complex number, which is a \emph{real} number, is defined to be \vspace{1pt} \begin{equation*} \left| z\right| =\sqrt{x^{2}+y^{2}}=\sqrt{z\bar{z}} \end{equation*} The diagram below shows how one may graph $z$ and $\bar{z}.$ \begin{center} \FRAME{dtbpF}{3.1185in}{2.0081in}{0pt}{}{}{6-1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 3.1185in;height 2.0081in;depth 0pt;original-width 4.3439in;original-height 2.7812in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/6-1.bmp';file-properties "XNPEU";}} \vspace{1pt} \end{center} The fact that $z\bar{z}=x^{2}+y^{2}=\left| z\right| ^{2}$ leads to two fundamental properties of complex numbers. The triangle inequality \vspace{1pt} \begin{equation*} \left| \text{ }z_{1}+z_{2}\right| \leq \left| \text{ }z_{1}\right| +\left| \text{ }z_{2}\right| \end{equation*} and the fact that $\left| \text{ }z_{1}-z_{2}\right| =$ the distance between $z_{2}$ and $z_{2}.$ \vspace{1pt} \paragraph{Example:} Verify the triangle inequality for $z_{1}=3-5i$ and $z_{2}=-2+i.$ \begin{equation*} 3-5i+\left( -2+i\right) =\allowbreak 1-4i \end{equation*}% so $\left| 1-4i\right| =\allowbreak \sqrt{17}.$ Also, $\left| 3-5i\right| =\allowbreak \sqrt{34}$ and $\left| -2+i\right| =\allowbreak \sqrt{5}.$ Clearly $\sqrt{17}\thickapprox \allowbreak 4.\,\allowbreak 123\,1\leq \sqrt{% 34}+\sqrt{5}\thickapprox \allowbreak 8.\,\allowbreak 067$ \vspace{1pt} \end{document} %%%%%%%%%%%%%%%%%%%% End /document/lec_4_10_00.tex %%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%% Start /document/graphics/6-1.bmp %%%%%%%%%%%%%%%%%% BudPzF@@@@@@@XCA@@@J@@@@aF@@@lP@@@P@@`@@@@@@@p`mA@@qN@@@D{@@@@@@@@@@@@@@@@@ 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