10.15 Complex Function and Rules

Revision

$z_0=a+ib\in\mathbb{C}$, then $mult_{z_0}:\mathbb{C}\to \mathbb{C}$ where $x+iy\mapsto (a+ib)(x+iy)$ is $\R$-linear

If we see $\mathbb{C}=\R^2$, are identified $mult_{a+ib}\longleftrightarrow 2\times 2$ matrix $\begin{pmatrix} a&-b\\ b&a \end{pmatrix}$ because $ \begin{pmatrix} a & -b \ b & a \end{pmatrix} \cdot \begin{pmatrix} x \ y \end{pmatrix}=\begin{pmatrix} ax - by \ ay + bx \end{pmatrix}$ and $(a+ib)(x+iy)=(ax-by)+i(ay+bx)$

Moreover, $mult_{e^{i\alpha}}=$ rotation in angle $\alpha$ and $mult_r=$ expansion/contraction by $r$

Definition

$f:\Omega\to \mathbb{C}$ where $\Omega\subset \mathbb{C}$ is an open set, then $\exists z_0\in \Omega=\text{Dom}(f)$

$\exists r:D_{r}(z_{0})\subset \Omega$, then $\left\{z/|z-z_0|<r\right\} \subset \Omega$

Let $|h|<r$ which is small and complex, then $z_0+h$ is still in the domain

Consider $\lim_{h\to 0}\frac{f(z_{0}+h)-f(z_{0})}{h}=?$

Def: $f$ is differentiable in the complex space at $z_0$ if $\exists\lim_{h\to 0}\frac{f(z_{0}+h)-f(z_{0})}{h}=:f'(z_{0})$

Remark

$\lim_{h\to 0}\frac{f(z_{0}+h)-f(z_{0})}{h}=f'(z_{0})\iff \lim_{h\to 0}\frac{|f(z_{0}+h)-f(z_{0})-f'(z_{0})\cdot h|}{|h|}=0$

$f'(z_{0})\cdot h$ is linear in $h$, that is function $mult_{f'(z_0)}:\mathbb{C}\to \mathbb{C}$ is linear with $h$

If $f$ is differentiable (in the $\mathbb{R}^2$ "classical" sense) at $z_0$, then its differential at $z_0$ is given by multiplication by the complex number $f'(z_{0})=a+ib$.

The differential of $f$ is a linear map that takes an increment $h\in\mathbb{C}$ to a new increment in $\mathbb{C}$

The precise statement is $f(z+h)=f(z)+Df(z)[h]+o(|h|),\quad Df(z)[h]=f'(z)\,h.$

Cauchy-Riemann conditions

We know $f(z) = f(x+iy) = u(x,y) + iv(x,y)$ where $u,v: \Omega \subseteq \text{Dom}(f) \longrightarrow \mathbb{R}$ and $z=x+iy$

Example: $f(z) = z^2 = (x+iy)^2$ $=x^2 - y^2 + 2ixy$ where $u(xy) = x^2 - y^2$ and $v(x,y) = 2xy$

$f$ is complex-diff. at $z_{0} \iff$ $\exists \lim\limits_{h=h_1+h_2i \to 0}\frac{u(x+h_{1},y+h_{2}) + iv(x+h_{1},y+h_{2}) - u(x,y) - iv(x,y)}{h_{1}+h_{2}i}= a+ib = f'(z_{0})$

Cauchy-Riemann Conditions: Suppose this limit exists, let $x+iy=z_0$, $h=h_1+ih_2=\Delta z = \Delta x+i\Delta y$

Then $f'(z_0)=\lim_{\Delta z \to0}\frac{f(z+\Delta z) - f(z)}{\Delta z}= \frac{f(z+\Delta x+i\Delta y) - f(x,y)}{\Delta x+i\Delta y}$
Then $\exists \lim_{\substack{h_1\to 0 \\ h_2=0}}(...) = \lim_{\Delta x \to 0}\frac{f(z+\Delta x)-f(z)}{\Delta x}= f'(z_{0})$(A) and $\lim_{\substack{h_1=0 \\ h_2\to 0}} (...) = \lim_{\Delta y \to 0} \frac{f(z+i\Delta y)-f(z)}{i\Delta y} = f'(z_0)$(B)

(A) $\lim_{\Delta x \to 0}\frac{u(x+\Delta x, y) + iv(x+\Delta x, y) - u(x, y) - iv(x, y)}{\Delta x}= \lim_{\Delta x \to 0}\frac{u(x+\Delta x, y) - u(x, y)}{\Delta x}+ \lim_{\Delta x \to 0}i \frac{v(x+\Delta x, y) - v(x, y)}{\Delta x}= \frac{\partial u}{\partial x}+ i \frac{\partial v}{\partial x}$

(B) $\lim_{\Delta y \to 0}\frac{u(x, y+\Delta y) + iv(x, y+\Delta y) - u(x, y) - iv(x, y)}{i \Delta y}= \lim_{\Delta y \to 0}-i \left( \frac{u(x, y+\Delta y) - u(x, y)}{\Delta y}\right) + \lim_{\Delta y \to 0}\frac{iv(x, y+\Delta y) - iv(x, y)}{i\Delta y}= \frac{\partial v}{\partial y}- i \frac{\partial u}{\partial y}$

$(A)=(B)\ \Rightarrow\ \frac{\partial u}{\partial x}+i\,\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}-i\,\frac{\partial u}{\partial y}\quad\text{in }\mathbb{C}\Rightarrow \operatorname{Re}(\cdot)\ \text{and}\ \operatorname{Im}(\cdot)\ \text{are the same}$

C–R equations $\begin{cases} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \\\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y} \end{cases}$

Consider $F:\mathbb{R}^{2}\to\mathbb{R}^{2}$, $F(x,y)=(u(x,y),\,v(x,y))$.
Suppose $F$ is differentiable (in the real sense). We will compute the differential of $F$ at some point $(x,y)$.

$DF_{(x,y)}=\begin{pmatrix}\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y}\\[6pt]\dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y}\end{pmatrix}=\begin{pmatrix}a & c\\b & d\end{pmatrix}$

$F(x+h_{1},\,y+h_{2})\approx F(x,y)+DF_{(x,y)} \begin{pmatrix} h_{1} \\ h_{2} \end{pmatrix}=\Big(u(x,y)+u_{x}h_{1}+u_{y}h_{2},\,v(x,y)+v_{x}h_{1}+v_{y}h_{2}\Big )$

By C–R eq. $\begin{cases} u_{x}=v_{y} \\ v_{x}=-u_{y} \end{cases}\implies \begin{pmatrix} u_{x} & u_{y} \\ v_{x} & v_{y} \end{pmatrix}= \begin{pmatrix} a & c \\ b & d \end{pmatrix}\Rightarrow a=d, b=-c$

$\Rightarrow\begin{pmatrix}a & c\\b & d\end{pmatrix}=\begin{pmatrix}a & -b\\b & a\end{pmatrix}\ \leftarrow\ \text{the matrix given by multiplication by a complex number.}$

Example

$f(z)=z^2$, what is $f'(z)$?

Solution
$\lim_{h\to 0}\frac{f(z+h)-f(z)}{h},\quad h\in\mathbb{C}$
For $f(z)=z^{2}$: $\frac{f(z+h)-f(z)}{h}=\frac{(z+h)^{2}-z^{2}}{h}=\frac{2zh+h^{2}}{h}=2z+h$
$\Rightarrow\ f'(z)=\lim_{h\to 0}(2z+h)=2z,\quad \forall z\in\mathbb{C}.$
$f(z)=\overline{z}$, that is $f(x+iy)=x-iy$

Solution

$u(x,y)=x, v(x,y)=-y$, then $Df=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$

By C–R: $v_{x}=1,\ v_{y}=-1\ \Rightarrow\ v_{x}\neq v_{y}\ \Rightarrow\ \text{NOT diff. in the complex sense.}$
$f(z)=|z|^{2}=x^{2}+y^{2}$

Solution

Note that $v(x,y)\equiv 0$, if there exists derivative in complex sense, then it satisfies C-R equation

In general, if $f(x,y)=u(x,y)+iv(x,y)$ and $v(x,y)\equiv 0\ \forall(x,y)\ (\text{Im} f\subset\mathbb{R}\,)$

Then $\frac{\partial v}{\partial y}=0=\frac{\partial v}{\partial x}\ \Rightarrow\ \text{by C–R equation}\ \Rightarrow\ Du=0\ \Rightarrow u \text{ is constant}$

Contradiction
Prove: $f(z)=\frac{1}{z}$ is holomorphic for $z\neq 0$.

$\frac{f(z+h)-f(z)}{h}=\frac{\frac{1}{z+h}-\frac{1}{z}}{h}=\frac{z-(z+h)}{(z+h)z\,h} =\frac{-h}{(z+h)z\,h}=-\frac{1}{(z+h)z}$

Fix $z\neq 0$. As $h\to 0$, $-\frac{1}{(z+h)z}\ \longrightarrow\ -\frac{1}{z^{2}}$

Thus $\left(\dfrac{1}{z}\right)'=-\dfrac{1}{z^{2}}$ for $z\neq 0$

Definition

$f$ is called holomorphic at $z_{0}\iff\exists f'(z_0)$ in the complex sense

$f$ is called holomorphic in $\Omega$ if it is holomorphic on $z,\forall z\in \Omega$

Remark on complex limits

$\exists f'(z_0)$ and $g'(z)\implies \exists (f+g)'(z)$
$\exists f'(z),c\in \mathbb{C}\implies \exists(cf)'=cf'$
$\exists\lim_{h\to 0}f_{1}(h)=\ell_{1},\exists\lim_{h\to 0}f_{2}(h)=\ell_{2} \ \Rightarrow\exists\lim_{h\to 0}\big(f_{1}(h)f_{2}(h)\big)=\ell_{1}\ell_{2}$

Idea: $F(x,y,s,t)=(x s - y t,\ x t + y s)$

Proof

Let $f_{1}(h)=x(h)+i\,y(h),\, f_{2}(h)=s(h)+i\,t(h),\,\ell_{1}=a+ib,\ \ell_{2} =c+id.$
The hypothesis $\lim_{h\to 0}f_{1}(h)=\ell_{1},\, \lim_{h\to 0}f_{2}(h)=\ell_{2}$ is equivalent (componentwise in $\mathbb{R}^2$) to $x(h)\to a,\ y(h)\to b,\ s(h)\to c,\ t(h)\to d\quad(h\to 0)$

Define $F:\mathbb{R}^4\to\mathbb{R}^2$ by $F(x,y,s,t)=(x s- y t,\ x t+ y s)$
Then $f_{1}(h)f_{2}(h)=(x s- y t)+i(x t+ y s)\implies(\text{Re},\text{Im})\big(f_{1}( h)f_{2}(h)\big)=F\big(x(h),y(h),s(h),t(h)\big).$

Since the coordinates of $F$ are polynomials, $F$ is continuous on $\mathbb{R}^4$. Hence by the real limit rules (sum/product continuity), $F\big(x(h),y(h),s(h),t(h)\big)\ \longrightarrow F(a,b,c,d)=(ac-bd,\ ad+bc)$ as $h\to 0$
Translating back to $\mathbb{C}$, $\lim_{h\to 0} f_1(h)f_2(h)=(ac-bd)+i(ad+bc)=(a+ib)(c+id)=\ell_1\ell_2.$

Therefore, $\exists\lim_{h\to 0}f_{1}(h)=\ell_{1},\ \exists\lim_{h\to 0}f_{2}(h)=\ell_{2}\ \Rightarrow \exists\lim_{h\to 0}\big(f_{1}(h)f_{2}(h)\big)=\ell_{1}\ell_{2}\quad\text{in }\mathbb{C} .$
$\exists f'$ and $g'\implies \exists (f\cdot g)'=f'g+ fg'$

Example: If $f(z) = z$, $f'(z) = 1$, then $\frac{f(z+h) - f(z)}{h} = \frac{z+h-z}{h} = \frac{h}{h} = 1$
Then $(z \cdot z)' = z'z + zz' = z+z = 2z$. Induction $f(z) = z^m \implies (z^m)' = mz^{m-1}$

Proof

$\frac{f(z+h)g(z+h)-f(z)g(z)}{h}=\frac{f(z+h)g(z+h)-f(z)g(z+h)+f(z)g(z+h)-f(z)g(z)}{h}$ $=\,f(z+h)\,\frac{g(z+h)-g(z)}{h}\;+\;\frac{f(z+h)-f(z)}{h}\,g(z)$

$\begin{cases} f(z+h)\to f(z)\quad(\text{f cont.}) \\ \frac{g(z+h)-g(z)}{h}\to g'(z)\quad(\text{hypothesis}) \\ \frac{f(z+h)-f(z)}{h}\to f'(z) \end{cases}\ \Rightarrow\ (fg)'(z)=f(z)g'(z)+f'(z)g(z)$

Consequence: Polynomials are holomorphic, and the usual derivation rules hold. $\left(\sum_{n=0}^{N} a_n z^{n}\right)'=\sum_{n=1}^{N} n\,a_n\,z^{\,n-1}$

Trigonometric functions

We know $e^{x+iy}=e^{x}e^{iy}=e^{x}(\cos y+i\sin y)=e^{x}\cos y+ie^{x}\sin y=u+iv$ where $u=e^{x}\cos y$, $v=e^{x}\sin y$

$f'(z)=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}\iff\lim_{h\to 0}\frac{\lvert f(z+h)-f(z)-f'(z)h\rvert}{\lvert h\rvert}=0$

Show that $f$ is holomorphic at $z$$\iff$$f$ is real differentiable and C-R hold

Let's C–R check $u_{x}=e^{x}\cos y,\ v_{y}=e^{x}\cos y;\ u_{y}=-e^{x}\sin y,\ v_{x}= e^{x}\sin y$
Hence C–R holds.

Thus $e^z:=e^x\cos y+ie^x\sin y$ is holomorphic

$(e^{z})'=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}=\lim_{\Delta x\to 0}\frac{f(z+\Delta x)-f(z)}{\Delta x}=\lim_{\Delta y\to 0}\frac{f(z+i\,\Delta y)-f(z)}{i\,\Delta y}$ $=u_{x}+iv_{x}= \frac{1}{i}(u_{y} +iv_{y})= v_{y} - iu_{y}$

Which means $(e^z)'=u_x+iv_x\text{ or }v_y-iu_y$. But don’t worry, they give the same answer by C–R

Let $e^{x+iy}=e^{x}\cos y+i\,e^{x}\sin y$ with $u=e^{x}\cos y,v=e^{x}\sin y$.

$\big(e^{x+iy}\big)'=v_{y}-iu_{y}= e^{x}\cos y - i\big(-e^{x}\sin y\big)= e^{x} \cos y + i\,e^{x}\sin y$
$= u_{x}+iv_{x}= e^{x}\cos y + i\,e^{x}\sin y\ \Rightarrow\ (e^{z})'=e^{z}$

Remark

$e^{z} = e^{z+2\pi i}$ is not injective.

Hyperbolic functions $\cosh z=\frac{e^{z}+e^{-z}}{2},\ \sinh z=\frac{e^{z}-e^{-z}}{2}$

Recall $e^{ix}=\cos x+i\sin x$ $(x\in\mathbb{R})$.

Hence $\cos x=\frac{e^{ix}+e^{-ix}}{2},\ \sin x=\frac{e^{ix}-e^{-ix}}{2i}$

For $z\in\mathbb{C}$, define $\cos z=\frac{e^{iz}+e^{-iz}}{2},\ \sin z=\frac{e^{iz}-e^{-iz}}{2i}$

$\cos z$ and $\sin z$ are not bounded.

Example $\cos(ix)=\frac{e^{iix}+e^{-iix}}{2}=\cosh x$, then $\cos(iz)=\cosh(z)$

Identity $\cos^{2}z+\sin^{2}z =\left(\frac{e^{iz}+e^{-iz}}{2}\right)^{2}+\left(\frac{e^{iz}-e^{-iz}}{2i} \right)^{2}=\frac{e^{2iz}+e^{-2iz}+2}{4}+\frac{e^{2iz}+e^{-2iz}-2}{-4}=1$

Unity root

We know $e^{2\pi i}=1$, $e^{2\pi i/m}\neq 1$ (for $m>1$) and $\big(e^{2\pi i/m}\big)^{m}=1$

For $k\in\mathbb{Z}$: $\big(e^{(2\pi i/m)\,k}\big)^{m}=e^{2\pi i k}=(e^{2\pi i})^{k}=1$

The $m$-th roots of unity: $\mu_{k}=e^{2\pi i k/m},\ k=0,1,\dots,m-1$ where modulus $=1$, argument $=2\pi k/m$

Example

$m=3$ (cube roots)
$e^{2\pi i/3}=\cos\!\left(\tfrac{2\pi}{3}\right)+i\sin\!\left(\tfrac{2\pi}{3}\right) =-\tfrac{1}{2}+i\,\tfrac{\sqrt{3}}{2}$
$z^{3}=1\ \Longleftrightarrow\ z^{3}-1=0=(z-1)(z^{2}+z+1)$
Solutions:
$z=1,\quad z=\frac{-1\pm i\sqrt{3}}{2}=e^{\pm 2\pi i/3}$
$m=8$ (8th roots)

Graph of complex function

It's hard to draw a graph of complex function usually because we cannot draw a graph in 4 dimension

But we can draw a graph to describe where a line go to