Complex Analysis – Page 2

October 5, 2022October 24, 2022

202210051329 Exercise 3.1

(a) Prove that the following are open sets:
i. $\{ z\in\mathbb{C}:|z-1|<|z+\mathrm{i}|\}$ ;
ii. $\mathbb{C}\backslash [0,1]$ .
(b) Prove that the following are not open:
i. $\{ z\in\mathbb{C}:\mathrm{Re}\, z\geqslant 0\}$ ;
ii. $\{ z\in\mathbb{C}:|z|\leqslant 2,\mathrm{Re}\, z>1\}$ .

_{Extracted from H. A. Priestley. (2003). Introduction to Complex Analysis.}

Background.

Definition. (open set)
A set $S\subseteq \mathbb{C}$ is open if, given $z\in S$ , there exists $r>0$ (depending on $z$ ) such that $\mathrm{D}(z;r)\subseteq S$ .

Definition. (open disc)
The open disc centre $a\in\mathbb{C}$ and radius $r>0$ is defined to be

$\mathrm{D}(a;r):=\{ z\in\mathbb{C}: |z-a|<r \}$ .

(a) i.

Warm-up.

$\begin{aligned} |z-1| & = |(x+\mathrm{i}y)-1| \\ & = \sqrt{(x-1)^2+(y)^2} \\ |z+\mathrm{i}| & = |(x+\mathrm{i}y)+\mathrm{i}| \\ & = \sqrt{(x)^2+(y+1)^2} \\ \end{aligned}$

No, just let $r=|z+\mathrm{i}|$ is done.

(a) ii.

Set-up.

In set notation a complex interval number may be represented in the form

$\mathcal{A}=[a,b]+[c,d]\mathrm{i}=\{ x+\mathrm{i}y\, |\, a\leqslant x\leqslant b,c\leqslant y \leqslant d\}$ .

_{Extracted from R. Boche. (1966). Complex Interval Arithmetic with Some Applications.}

Thus $\mathbb{C}\backslash [0,1]$ is equivalent to

$\{z=a+\mathrm{i}b\textrm{ where }a,b\notin [0,1]\}$ ,

as shown in the figure below:

Abortive attempt.

Lemma.

Let $S\subset X$ be a non-empty subset and $U\subset S$ . Then $U$ is open in $S$ if and only if $U=V\cap S$ for some $V\subset X$ which is open in $X$ .

Proof. Necessity. $\forall\, z\in U,\,\exists\, r>0$ s.t. $\mathrm{D}_S(z;r)\subset U$ . Let $\displaystyle{V=\bigcup_{z\in U}\mathrm{D}_X(z;r)}$ . Then $U=V\cap S$ . Note that $V$ is open in $X$ . Sufficiency. $\forall\, z\in U\subset V,\,\exists\, r>0$ s.t. $\mathrm{D}_X(z;r)\subset V$ . Thus $\mathrm{D}_S(z;r)=\mathrm{D}_X(z;r)\cap S\subset V\cap S\subset U$ . $\blacksquare$

Now that $U=\mathbb{C}\backslash [0,1]$ , $S=\mathbb{C}$ , and $X=\mathbb{C}$ … quit this circular reasoning of no use.

Make use of the following

Lemma.

The union of any collection of open sets is open.

Proof.

Let $\displaystyle{S=\bigcup_{\lambda\in I}S_\lambda}$ where $S_\lambda$ is open in $\mathbb{C}$ and $I$ any index set. Then $\forall\, z\in S$ , $z\in S_\lambda$ for some $\lambda$ and so $\exists\, r>0$ s.t. $\mathrm{D}(z;r)\subset S_\lambda\subset S$ . $\blacksquare$

Thereby

$\mathbb{C}\backslash [0,1]=\mathbb{C}(-\infty ,0)\cup\mathbb{C}(1,\infty)$

is open.

QED

(b)

A set $S$ is not open whenever $S^0$ (/ $\textrm{int}\, S$ ), the interior of $S$ , falls short of its whole, i.e., $S\supset S^0\neq S$

For i. and ii. the unbelonging boundaries $\partial S$ ‘s to either are $\{z\in\mathbb{C}:\textrm{Re }z=0\}$ and $\{z\in\mathbb{C}:|z|=2,\mathrm{Re}\, z>1\}$ .

September 29, 2022September 29, 2022

202209291831 Problem 1.1

Prove the parallelogram law

$|z_1+z_2|^2+|z_1-z_2|^2=2[|z_1|^2+|z_2|^2]$

and give a geometric interpretation.

_{Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.}

Draw a picture.

Apply the Law of Cosines,

$\begin{aligned} |z_1+z_2|^2 & =|z_1|^2+|z_2|^2-2(-z_1)\cdot z_2 \\ |z_1-z_2|^2 & =|z_1|^2+|z_2|^2-2z_1\cdot z_2 \\ \end{aligned}$

and that is all I could interpret.

September 15, 2021October 24, 2022

202109151516 CYU 7.1

If $a=3+4i$ , what is the product $a^*a$ ?

Background. (Complex conjugate)

The complex conjugate of $a+bi$ is equal to $a-bi$ .

$\begin{aligned} a^*a & = (3-4i)(3+4i) \\ & = (3)(3)+(3)(4i)+(-4i)(3)+(-4i)(4i) \\ & = 9 - 16i^2 \\ & = 9 - 16(-1)\\ & = 9+16 \\ & = 25 \\ \end{aligned}$

February 19, 2021October 24, 2022

202102190301 Homework 1 (Q3)

Find the analytic functions $w(z)=u(x,y)+\mathrm{i}v(x,y)$ if

(a) $u(x,y)=x^3-3xy^2$ ;

(b) $v(x,y)=e^{-y}\sin x$ .

Attempts.

(a)

$\displaystyle{\frac{\partial u}{\partial x}}=3x^2-3y^2$ .

From (first of the Cauchy-Riemann equations)

$\displaystyle{\frac{\partial u}{\partial x}}=\displaystyle{\frac{\partial v}{\partial y}}$ ,

we have

$v=\displaystyle{\int} (3x^2-3y^2)\,\mathrm{d}y+g(x)$

where $g$ is a function independent of $y$ . Then

$v=3x^2y-y^3+g(x)$ .

Now

$\displaystyle{\frac{\partial v}{\partial x}}=6xy+g'(x)$ ;

$\displaystyle{\frac{\partial u}{\partial y}}=-6xy$

From (second of the Cauchy-Riemann equations)

$\displaystyle{\frac{\partial u}{\partial y}}=-\displaystyle{\frac{\partial v}{\partial x}}$ ,

we have

$-6xy=-6xy-g'(x)$

i.e.,

$g'(x)=0$ ;

$g(x)=C$ .

Thus $v(x,y)=3x^2y-y^3+C$ for some constant $C\in\mathbb{R}$ .

Part (b) is left to the reader.

February 19, 2021October 24, 2022

202102190222 Homework 1 (Q2)

Show whether or not the function (a) $f(z)=\mathrm{Re}(z)=x$ ; (b) $f(z)=z^2$ is analytic.

Setup.

Theorem 1. Suppose $f:\Omega \rightarrow \mathbb{C}$ is analytic. Then, writing

$f(z)=f(x+\mathrm{i}y)=u(x,y)+\mathrm{i}v(x,y)$

with $u(x,y)=\mathrm{Re}(f(z))$ and $v(x,y)=\mathrm{Im}(f(z))$ , we have the Cauchy-Riemann equations

$\begin{cases} \displaystyle{\frac{\partial u}{\partial x}} & = \displaystyle{\frac{\partial v}{\partial y}} \\ \displaystyle{\frac{\partial u}{\partial y}} & = \displaystyle{-\frac{\partial v}{\partial x}} \end{cases}$

everywhere on $\Omega$ .

Theorem 2. If $f:\Omega\rightarrow \mathbb{C}$ is of class $\mathcal{C}^1$ , $f=u+\mathrm{i}v$ . Then $(u,v)$ satisfies the Cauchy-Riemann equations if and only if $f$ is analytic.

(a) Now that

$u(x,y)=x$ and $v(x,y)=0$ ,

the partial derivatives are as follows

$\begin{aligned} \frac{\partial u}{\partial x} & = 1 & \qquad \frac{\partial u}{\partial y} = & 0 \\ \frac{\partial v}{\partial y} & = 0 & \qquad -\frac{\partial v}{\partial x} = & 0 \\ \end{aligned}$

Since

$1=\displaystyle{\frac{\partial u}{\partial x}}\neq \displaystyle{\frac{\partial v}{\partial y}}=0$ ,

the Cauchy-Riemann equations are not satisfied, and $f$ is $\textrm{\scriptsize{NOT}}$ analytic.

(b)

$\begin{aligned} f(z) & = z^2 \\ f(x+\mathrm{i}y) & = (x+\mathrm{i}y)^2 \\ & = (x^2-y^2)+\mathrm{i}(2xy) \\ & = u(x,y)+\mathrm{i}v(x,y) \end{aligned}$

where $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$ .

Then,

$\displaystyle{\frac{\partial u}{\partial x}} = 2x$ , $\displaystyle{\frac{\partial u}{\partial y}} = -2y$ , $\displaystyle{\frac{\partial v}{\partial y}} = 2x$ , $\displaystyle{-\frac{\partial v}{\partial x}} = -2y$

and the Cauchy-Riemann equations are satisfied. $f$ is thus analytic.

February 19, 2021October 24, 2022

202102190159 Homework 1 (Q1)

(a) Find the reciprocal of $x+\mathrm{i}y$ , working entirely in the Cartesian representation.

(b) Repeat part (a), working in polar form but expressing the final result in Cartesian form.

Solution.

(a)

$\begin{aligned} \frac{1}{x+\mathrm{i}y} & = \bigg( \frac{1}{x+\mathrm{i}y}\bigg) \bigg( \frac{x-\mathrm{i}y}{x-\mathrm{i}y} \bigg) \\ & = \frac{x-\mathrm{i}y}{x^2-(\mathrm{i}y)^2} \\ & = \frac{x-\mathrm{i}y}{x^2+y^2} \\ & = \bigg( \frac{x}{x^2+y^2} \bigg) - \mathrm{i}\bigg( \frac{y}{x^2+y^2} \bigg) \end{aligned}$

(b)

Let

$x+\mathrm{i}y\stackrel{\mathrm{def}}{=}z=re^{\mathrm{i}\varphi}$

where $r=\sqrt{x^2+y^2}$ and $\varphi = \arctan\bigg( \displaystyle{\frac{y}{x}} \bigg)$ .

Then, the reciprocal of $z=x+\mathrm{i}y$ is

$\begin{aligned} \frac{1}{z} & = \frac{1}{re^{\mathrm{i}\varphi}} \\ & = \frac{1}{r}\, e^{\mathrm{i}(-\varphi )} \\ & = \frac{1}{r}\, \mathrm{cis}(-\varphi )\\ & = \frac{1}{r}\big(\cos (-\varphi)+\mathrm{i}\sin (-\varphi )\big)\\ & = \frac{1}{r}(\cos\varphi -\mathrm{i}\sin\varphi ) \\ & = \frac{1}{\sqrt{x^2+y^2}}\bigg[ \frac{x}{\sqrt{x^2+y^2}} - \mathrm{i}\Big( \frac{y}{\sqrt{x^2+y^2}} \Big) \bigg] \\ & = \bigg( \frac{x}{x^2+y^2} \bigg) - \mathrm{i}\bigg( \frac{y}{x^2+y^2} \bigg) \end{aligned}$

August 18, 2020March 15, 2023

202008181129 Homework 1 (Q1)

Solve for $z\in \mathbb{C}$ in the following equations:

(a) $z^4+z^3+z^2+z+1=0$ ,

(b) $3z^3+29z^2+497z-169=0$ .

Attempts.

(a) Take notice that $z\neq 1$ . Try and see having both sides of the equation multiplied by $(1-z)$ ,

$\begin{aligned} 0 & = (1-z)(z^4+z^3+z^2+z+1) \\ 0 & = 1-z^5 \\ z^5 & = 1 = 1(1+0\,\mathrm{i}) \\ z^5 & = e^{\mathrm{i}(2n\pi)}\qquad \qquad \textrm{where }n=0,1,2,3,4. \\ \end{aligned}$

$\therefore z=e^{\mathrm{i}(\frac{2n\pi}{5})}$ .

( $n=0$ is rejected for $z\neq 1=e^{\mathrm{i}(\frac{2(0)\pi}{5})}$ .)

In polar expression $z=e^{\mathrm{i\frac{2\pi}{5}}},\, e^{\mathrm{i\frac{4\pi}{5}}},\, e^{\mathrm{i\frac{6\pi}{5}}},\, e^{\mathrm{i\frac{8\pi}{5}}}$ .

In trigonometric expression

$z=\mathrm{cis}(\frac{2n\pi}{5})=\mathrm{cis}(\frac{2\pi}{5}),\, \mathrm{cis}(\frac{4\pi}{5}),\, \mathrm{cis}(\frac{6\pi}{5}),\, \mathrm{cis}(\frac{8\pi}{5})$

I.e.,

$\begin{aligned} z^1 & = \cos 72^\circ +\mathrm{i}\sin 72^\circ = \mathrm{cis\,} 72^\circ \\ z^2 & = \cos 144^\circ +\mathrm{i}\sin 144^\circ = \mathrm{cis\,} 144^\circ \\ z^3 & = \cos 216^\circ +\mathrm{i}\sin 216^\circ = \mathrm{cis\,} 216^\circ \\ z^4 & = \cos 288^\circ +\mathrm{i}\sin 288^\circ = \mathrm{cis\,} 288^\circ \\ \end{aligned}$

(b)Let $f(z)=3z^3+29z^2+497z-169=0$ . Then $(3z-1)$ is a factor, because $\frac{1}{3}$ is a zero (i.e., $f(\frac{1}{3}) = \frac{1}{9}+\frac{29}{9}+\frac{497}{3}-169=0$ ).