6.5 Complex Functions

When one studies a function $f$ from $\mbox{{\bf R}}$ to $\mbox{{\bf R}}$, one often gets information by looking at the graph of $f$, which is a subset of $\mbox{{\bf R}}\times\mbox{{\bf R}}$. If we consider a function $g\colon\mbox{{\bf C}}\to\mbox{{\bf C}}$, the graph of $g$ is a subset of $\mbox{{\bf C}}\times\mbox{{\bf C}}=(\mbox{{\bf R}}\times\mbox{{\bf R}})\times(\mbox{{\bf R}}\times\mbox{{\bf R}})$, and $\mbox{{\bf C}}\times\mbox{{\bf C}}$ is a `` 4-dimensional" object which cannot be visualized. We will now discuss a method to represent functions from $\mbox{{\bf C}}$ to $\mbox{{\bf C}}$ geometrically.

6.31   Example ($f(z)=z^2$.) Let $f\colon\mbox{{\bf C}}\to\mbox{{\bf C}}$ be defined by $f(z)=z^2$. If $z$ is a point in the circle $C(0,r)$, then $z=ru$ where $u$ is a direction, and $f(z)=r^2u^2$ is a point in the circle $C(0,r^2)$ with radius $r^2$. Thus $f$ maps circles of radius $r$ about $0$ into circles of radius $r^2$ about $0$. Let $u_0$ be a direction in $\mbox{{\bf C}}$. If $z$ is on the ray from $0$ passing through $u_0$, then $z=ru_0$ for some $r\in\mbox{${\mbox{{\bf R}}}^{+}$}$ so $f(z)=r^2u_0^2$, which is on the ray from $0$ passing through $u_0^2$. Hence the ray making an angle $\theta$ with the positive real axis gets mapped by $f$ to the ray making an angle $2\theta$ with the positive $x$-axis.

The left part of the figure shows a network formed by semicircles of radius

$$r\in\{.1,.2,.3,\cdots ,.9,1\},$$

and rays making angles

$$\theta\in\left\{0,\pm {\pi \over {16}},\pm {{2\pi}\over {16}},\cdots,\pm {{8\pi}\over {16}}\right\}$$

with the positive $x$-axis. The right part of the figure shows the network formed by circles of radius

$$r^2\in\{.1^2,.2^2,\cdots,.9^2,1\}$$

and rays making angles

$$2\theta\in\left\{ 0,\pm {\pi\over 8},\pm {{2\pi}\over 8},\cdots,\pm{{8\pi}\over 8}\right\}$$

with the positive $x$-axis. $f$ maps each semicircle in the left part of the figure to a circle in the right part, and $f$ maps each ray in the left part to a ray in the right part. Also $f$ maps each curvilinear rectangle on the left to a curvilinear rectangle on the right. Notice that $f(i)=f(-i)$, and in general $f(z)=f(-z)$, so if we know how $f$ maps points in the right half plane, we know how it maps points in the left half plane. The function $f$ maps the right half plane $\{x>0\}$ onto $\mbox{{\bf C}}\setminus\Big( (\mbox{ negative real axis })\cup\{0\}\Big)$.

6.32   Definition (Image of a function.) Let $S,T$ be sets, let $f\colon S\to T$, and let $A$ be a subset of $\mbox{{\rm dom}}(f)$. We define

$$f(A)=\{f(a)\colon a\in A\}$$

and we call $f(A)$ the image of $A$ under $f$. We call $f\left(\mbox{{\rm dom}} (f)\right)$ the image of $f$.

6.33   Example ($f(z)=z^2$, continued) In the figure above 6.5, the right half of the figure is the image of the left half under the function $f$. The figure below 6.5, shows the image of a cat-shaped set under $f$. The cat on the left lies in the first quadrant, so its square lies in the first two quadrants. The tip of the right ear is $${1+i=\sqrt 2\left( {{\sqrt 2}\over 2}+i{{\sqrt 2}\over 2}\right)}$$, with length $\sqrt 2$, and with direction making an angle $${{\pi\over 4}}$$ with the positive real axis. The image of the right ear has length $(\sqrt 2)^2=2$ and makes an angle $${{\pi\over 2}}$$ with the positive $x$-axis. You should examine how the parts of the cat in each curvilinear rectangle on the left part of the figure correspond to their images on the right part.

6.34   Exercise. Let $C$ be the cat shown in the left part of the above figure. Sketch the image of $C$ under each of the functions $g,h,k$ below:

a)

$g(z)=2z$.

b)

$h(z)=iz$.

c)

$k(z)=2iz$.

6.35   Exercise. Let $C$ be the cat shown in the left part of the above figure. Sketch the image of $C$ under $G$, where $G(z)=-z^2$.

6.36   Exercise. Let $z$ be a direction in $\mbox{{\bf C}}$; i.e., let $z\in C(0,1)$. Show that $z^*=z^{-1}$.

6.37   Example. Let $${v(z)={1\over z}}$$ for all $z\in\mbox{{\bf C}}\backslash\{0\}$. If $z$ is in the circle of radius $r$, then $z=ru$ for some direction $u$, and $${ \vert v(z)\vert=\left\vert {1\over {ru}}\right\vert={1\over {\vert r\vert \; \vert u\vert}}={1\over {\vert r\vert}}}$$, so $v$ takes points in the circle of radius $r$ about $0$ to points in the circle of radius $${{1\over r}}$$ about $0$.

Let $u_0$ be a direction. If $z$ is in the ray from $0$ through $u_0$, then $z=ru_0$ for some $r\in\mbox{${\mbox{{\bf R}}}^{+}$}$, so $${v(z)={1\over r}u_0^{-1}={1\over r}u_0^*}$$. We noted earlier that $u_0^*$ is the reflection of $u_0$ about the real axis, so $v$ maps the ray making angle $\theta$ with the positive real axis into the ray making angle $-\theta$ with the positive real axis. Thus $v$ maps the network of circles and lines in the left half of the figure into the network on the right half.

The circular arcs in the left half of the figure have radii

$$r\in\{.5,.6,.7,\cdots,1.4,1.5\}.$$

Let's see how $v$ maps the vertical line $x=a\;(a\neq 0),\; a\in\mbox{{\bf R}}$. We know that $${v(a)={1\over a}}$$ and $v$ maps points in the upper half plane to points in the lower half plane. Points far from the origin get mapped to points near to the origin. I claim that $v$ maps the line $x=a$ into the circle with center $${{1\over {2a}}}$$ and radius $${ {1\over {2\vert a\vert}}}$$.

Let $L_a=\{z\colon\mbox{\rm Re}(z)=a\}=\{a+iy\colon y\in\mbox{{\bf R}}\}$, so $L_a$ is the set of points in the line $x=a$. Then

$$\begin{eqnarray*} z\in L_a&\mbox{$\Longleftrightarrow$}&z=a+iy \mbox{ for some }... ...ert a-iy\vert}\over {\vert a+iy\vert}}={1\over {\vert 2a\vert}}, \end{eqnarray*}$$

since $\vert w\vert=\vert w^*\vert$ for all $w\in\mbox{{\bf C}}$. Hence,

$$z\in L_a\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\left... ...er z}\in C\left( {1\over {2a}},{1\over {\vert 2a\vert}}\right),$$

and $v$ maps every point in $L_a$ into $${C\left( {1\over {2a}},{1\over {\vert 2a\vert}}\right)}$$. Now I claim that every point in $${C\left( {1\over {2a}},{1\over {\vert 2a\vert}}\right)}$$ (except for $0$) is equal to $v(z)$ for some $z\in L_a$.

Since $w=v\left(v(w)\right)$, it will be sufficient to show that if $${w\in C\left( {1\over {2a}},{1\over {\vert 2a\vert}}\right)\backslash\{0\}}$$, then $v(w)\in L_a$. I want to show

$$\left(\left\vert w-{1\over {2a}}\right\vert={1\over {\vert 2a... ...ce{1ex}$} {1\over w}=a+iy \mbox{ for some } y\in\mbox{{\bf R}}.$$

Well, suppose $${\left\vert w-{1\over {2a}}\right\vert={1\over {\vert 2a\vert}}}$$, and let $${{1\over w}=A+iB}$$ where $A,B\in\mbox{{\bf R}}$. Then $${w={1\over {A+iB}}={{A-iB}\over {A^2+B^2}}}$$, so

$$\begin{eqnarray*} \left\vert w-{1\over {2a}}\right\vert={1\over {\vert 2a\vert}}... ...2)^2}}={A\over {a(A^2+B^2)}} \\ &\mbox{$\Longrightarrow$}&A=a, \end{eqnarray*}$$

so (by definition of $A$)

$$\left\vert w-{1\over {2a}}\right\vert ={1\over {2a}}\mbox{$\h... ...iB \mbox{ where } B\in\mbox{{\bf R}}.\mbox{ $\mid\!\mid\!\mid$}$$

6.38   Exercise. The argument above does not apply to the vertical line $x=0$. Let $L_0=\{iy\colon y\in\mbox{{\bf R}}\}$. Where does the reciprocal function $v$ map $L_0\backslash\{0\}$?

6.39   Entertainment. Let $${v(z)={1\over z}}$$ for all $z\in\mbox{{\bf C}}\backslash\{0\}$. Show that $v$ maps horizontal lines $y=c$ $(c\neq 0)$ into circles that pass through the origin. Sketch the images of the lines

$$x=j,\mbox{ where } j\in\{0,\pm 1,\pm 2,\pm 3\}$$

and the lines

$$y=j,\mbox{ where } j\in\{0,\pm 1,\pm 2,\pm 3\}$$

on one set of axes using a compass. If you've done this correctly, the circles should intersect at right angles.

6.40   Exercise.

a)

Sketch the image of the network of lines and circular arcs shown below under the function $g$, where $g(z)=z^3$ for all $z\in\mbox{{\bf C}}$.

b)

Cube the cat in the picture.

6.41   Note. De Moivre's formula $\left(\cos(\theta)+i\sin\theta\right)^n=\cos(n\theta)+i\sin(n\theta)$, was first stated in this form by Euler in 1749 ([46, pp. 452-454]). Euler named the formula after Abraham De Moivre (1667-1754) who never explicitly stated the formula, but used its consequences several times ([46, pp. 440-450]).

The method for finding $m$th roots of complex numbers:

$$[r(\cos\theta+i\sin\theta)]^{{1\over m}}=r^{{1\over m}}\left[\cos{\theta\over m}+i\sin{\theta\over m}\right]$$

was introduced by Euler in 1749 [46, pp.452-454].

The idea of illustrating functions from the plane to the plane by distorting cat faces is due to Vladimir Arnold (1937-??), and the figures are sometimes called `` Arnold Cats". Usually Arnold cats have black faces and white eyes and noses, as in [3, pp.6-9].