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Complex Analysis

Eberhard Freitag, Rolf Busam

Chapter 1

Differential Calculus in the Complex Plane $\mathbb{C}$. - all with Video Answers

Educators


Section 1

Complex Numbers

00:36

Problem 1

Find the real and imaginary parts of each of the following complex numbers:
$$
\begin{gathered}
\frac{\mathrm{i}-1}{\mathrm{i}+1} ; \quad \frac{3+4 \mathrm{i}}{1-2 \mathrm{i}} ; \mathrm{i}^{n}, n \in \mathbb{Z} ; \quad\left(\frac{1+\mathrm{i}}{\sqrt{2}}\right)^{n}, n \in \mathbb{Z} ; \\
\left(\frac{1+\mathrm{i} \sqrt{3}}{2}\right)^{n}, n \in \mathbb{Z} ; \quad \sum_{\nu=0}^{7}\left(\frac{1-\mathrm{i}}{\sqrt{2}}\right)^{\nu} ; \frac{(1+\mathrm{i})^{4}}{(1-\mathrm{i})^{3}}+\frac{(1-\mathrm{i})^{4}}{(1+\mathrm{i})^{3}}
\end{gathered}
$$

Ramzi Deek
Ramzi Deek
Numerade Educator
00:18

Problem 2

Calculate the absolute value and an argument for each of the following complex numbers:
$$
\begin{gathered}
-3+\mathrm{i} ; \quad-13 ; \quad(1+\mathrm{i})^{17}-(1-\mathrm{i})^{17} ; \quad \mathrm{i}^{4711} ; \quad \frac{3+4 \mathrm{i}}{1-2 \mathrm{i}} \\
\frac{1+\mathrm{i} a}{1-\mathrm{i} a}, a \in \mathbb{R} ; \quad \frac{1-\mathrm{i} \sqrt{3}}{1+\mathrm{i} \sqrt{3}} ; \quad(1-\mathrm{i})^{n}, n \in \mathbb{Z}
\end{gathered}
$$

Amrita Bhasin
Amrita Bhasin
Numerade Educator
01:13

Problem 3

Prove the "Triangle Inequality"
$$
|z+w| \leq|z|+|w|, \quad z, w \in \mathbb{C}
$$
and discuss when it becomes an equality; also prove the "Triangle Inequality"
$$
|| z|-| w|| \leq|z-w|, \quad z, w \in \mathbb{C}
$$

Carson Merrill
Carson Merrill
Numerade Educator
07:29

Problem 4

For $z=x+\mathrm{i} y, w=u+\mathrm{i} v$, with $x, y, u, v \in \mathbb{R}$, the standard scalar product in the $\mathbb{R}$-vector space $\mathbb{C}=\mathbb{R} \times \mathbb{R}$ with respect to the basis $(1, \mathrm{i})$ is defined by
$$
\langle z, w\rangle:=\operatorname{Re}(z \bar{w})=x u+y v
$$
Verify by direct calculation that, for $z, w \in \mathbb{C}$
$$
\langle z, w\rangle^{2}+\langle\mathrm{i} z, w\rangle^{2}=|z|^{2}|w|^{2}
$$
and infer from this the CAUCHY-SCHWARZ inequality in $\mathbb{R}^{2}$ :
$$
|\langle z, w\rangle|^{2}=|x u+y v|^{2} \leq|z|^{2}|w|^{2}=\left(x^{2}+y^{2}\right)\left(u^{2}+v^{2}\right)
$$
In addition, show the following identities for $z, w \in \mathbb{C}$ by direct calculation:
$$
\begin{aligned}
|z+w|^{2} &=|z|^{2}+2\langle z, w\rangle+|w|^{2} & & \text { (cosine law) } \\
|z-w|^{2} &=|z|^{2}-2\langle z, w\rangle+|w|^{2}, & & \\
|z+w|^{2}+|z-w|^{2} &=2\left(|z|^{2}+|w|^{2}\right) & \text { (parallelogran }
\end{aligned}
$$
(parallelogram law).
Further, show that for each pair $(z, w) \in \mathbb{C}^{*} \times \mathbb{C}^{*}$ there is a unique real number $\omega:=\omega(z, w) \in]-\pi, \pi]$ with
$$
\cos \omega=\cos \omega(z, w)=\frac{\langle z, w\rangle}{|z||w|}
$$
I Differential Calculus in the Complex Plane $\mathbb{C}$
$$
\sin \omega=\sin \omega(z, w)=\frac{(\mathrm{i} z, w\rangle}{|z||w|}
$$
$\omega=\omega(z, w)$ is called the oriented angle between $z$ and $w$ and will often be denoted by $\angle(z, w)$.
Show: $\quad \angle(1, \mathrm{i})=\pi / 2, \angle(\mathrm{i}, 1)=-\pi / 2=-\angle(1, \mathrm{i})$.

Chris Trentman
Chris Trentman
Numerade Educator
02:07

Problem 5

Suppose $n \in \mathbb{N}$ and $z_{\nu}, w_{\nu} \in \mathbb{C}$ for $1 \leq \nu \leq n$. Prove
$$
\left|\sum_{\nu=1}^{n} z_{\nu} w_{\nu}\right|^{2}=\sum_{\nu=1}^{n}\left|z_{\nu}\right|^{2} \cdot \sum_{\nu=1}^{n}\left|w_{\nu}\right|^{2}-\sum_{1 \leq \nu<\mu \leq n}\left|z_{\nu} \bar{w}_{\mu}-z_{\mu} \bar{w}_{\nu}\right|^{2}
$$
(the LAGRANGE Identity) and conclude from this the CAUCHY-SCHWARZ Inequality in $\mathbb{C}^{n}$ :
$$
\left|\sum_{\nu=1}^{n} z_{\nu} w_{\nu}\right|^{2} \leq \sum_{\nu=1}^{n}\left|z_{\nu}\right|^{2} \cdot \sum_{\nu=1}^{n}\left|w_{\nu}\right|^{2}
$$

Bryan Lynn
Bryan Lynn
Numerade Educator
01:17

Problem 6

Sketch the following subsets of $\mathrm{C}$ in the complex plane:
(a) Assume $a, b \in \mathbb{C}, b \neq 0$;
$$
\begin{aligned}
&G_{0}:=\left\{z \in \mathbb{C} ; \quad \operatorname{Im}\left(\frac{z-a}{b}\right)=0\right\} \\
&G_{+}:=\left\{z \in \mathbb{C} ; \quad \operatorname{Im}\left(\frac{z-a}{b}\right)>0\right\} \quad \text { and } \\
&G_{-}:=\left\{z \in \mathbb{C} ; \quad \operatorname{Im}\left(\frac{z-a}{b}\right)<0\right\}
\end{aligned}
$$
(b) Consider $a, c \in \mathbb{R}$ and $b \in \mathbb{C}$ with $b \bar{b}-a c>0$,
$$
K:=\{z \in \mathbb{C} ; \quad a z \bar{z}+\bar{b} z+b \bar{z}+c=0\}
$$
(c) $L:=\left\{z \in \mathbb{C} ;\left|z-\frac{\sqrt{2}}{2}\right| \cdot\left|z+\frac{\sqrt{2}}{2}\right|=\frac{1}{2}\right\}$

Sriparna Bhattacharjee
Sriparna Bhattacharjee
Numerade Educator
01:14

Problem 7

Square roots and the solvability of quadratic equations in $\mathbb{C}$
Let $c=a+\mathrm{i} b \neq 0$ be a given complex number. By splitting it into its real and imaginary part show that there are exactly two complex numbers $z_{1}$ and $z_{2}$ such that
$$
z_{1}^{2}=z_{2}^{2}=c . \text { One has } z_{2}=-z_{1}
$$
$\left(z_{1}\right.$ and $z_{2}$ are called the square roots of $c$ ) For example, determine the square roots of
$$
5+7 \mathrm{i}, \quad \text { and } \quad \sqrt{2}+\mathrm{i} \sqrt{2}
$$
Use also polar coordinates for this exercise. Furthermore, show that a quadratic equation
$$
z^{2}+\alpha z+\beta=0, \quad \alpha, \beta \in \mathbb{C} \text { arbitrary }
$$
always has at most two solutions $z_{1}, z_{2} \in \mathrm{C}$.

Ekaveera Kumar
Ekaveera Kumar
Numerade Educator
01:20

Problem 8

Existence of $n^{\text {th }}$ roots
Assume $a \in \mathbb{C}$ and $n \in \mathbb{N} .$ A complex number $z$ is called (an) $n^{\text {th }}$ root of $a$ if $z^{n}=a$
Show: If $a=r(\cos \varphi+\mathrm{i} \sin \varphi) \neq 0$, then $a$ has exactly $n$ (different) $n^{\text {th }}$ roots, namely the complex numbers
$$
z_{\nu}=\sqrt[n]{r}\left(\cos \frac{\varphi+2 \pi \nu}{n}+\mathrm{i} \sin \frac{\varphi+2 \pi \nu}{n}\right), \quad 0 \leq \nu \leq n-1
$$
In the special case $a=1$ (thus $r=1, \varphi=0)$, we get Theorem I.1.7.

Amrita Bhasin
Amrita Bhasin
Numerade Educator
01:05

Problem 9

Determine all $z \in \mathbb{C}$ such that $z^{3}-\mathrm{i}=0$

ES
Ellie Sun
Numerade Educator
03:07

Problem 10

Let $P$ be a polynomial with complex coefficients:
$$
P(z):=a_{n} z^{n}+a_{n-1} z^{n-1}+\cdots+a_{0} \text { with } n \in \mathbb{N}_{0}, a_{\nu} \in \mathbb{C}, \text { for } 0 \leq \nu \leq n
$$
A real or complex number $\zeta$ is called a root or a zero of $P$, if $P(\zeta)=0$
Show: If all the coefficients $a_{\nu}$ are real, then we have
$$
P(\zeta)=0 \Longrightarrow P(\bar{\zeta})=0
$$
In other words, if the polynomial $P$ has only real coefficients then the roots of $P$ which are not real occur as pairs of complex conjugate numbers.

Yujie Wang
Yujie Wang
College of San Mateo
04:19

Problem 11

(a) Let $\mathbb{H}:=\{z \in \mathbb{C} ; \quad \operatorname{Im} z>0\}$ be the upper half-plane.
Show: $z \in \mathrm{H} \Longleftrightarrow-1 / z \in \mathbb{H} .$
(b) Assume $z, a \in \mathbb{C}$.
Show: $\quad|1-z \bar{a}|^{2}-|z-a|^{2}=\left(1-|z|^{2}\right)\left(1-|a|^{2}\right) .$
Deduce: If $|a|<1$, then
$$
|z|<1 \Longleftrightarrow\left|\frac{z-a}{\bar{a} z-1}\right|<1 \quad \text { and } \quad|z|=1 \Longleftrightarrow\left|\frac{z-a}{\bar{a} z-1}\right|=1
$$

Gaurav Kalra
Gaurav Kalra
Numerade Educator
04:02

Problem 12

Verify for $z=x+\mathrm{i} y \in \mathbb{C}$ the inequalities
$$
\frac{|x|+|y|}{\sqrt{2}} \leq|z|=\sqrt{x^{2}+y^{2}} \leq|x|+|y|
$$
and
$$
\max \{|x|,|y|\} \leq|z| \leq \sqrt{2} \max \{|x|,|y|\}
$$

Subham Jyoti Mishra
Subham Jyoti Mishra
Numerade Educator
06:42

Problem 13

Let $\widetilde{\mathbb{C}}$ be another field of complex numbers. Determine all mappings $\varphi: \mathbb{C} \rightarrow \widetilde{\mathbb{C}}$ with the following properties:
(a) $\varphi(z+w)=\varphi(z)+\varphi(w) \quad$ for all $z, w \in \mathbb{C}$
(b) $\varphi(z w)=\varphi(z) \varphi(w) \quad$ for all $z, w \in \mathbb{C}$
$\begin{array}{cll}(c) & \varphi(x)=x & \text { for all } x \in \mathbb{R} \text {. }\end{array}$
Remark. It turns out that such mappings exist, and they are automatically bijective; thus they give isomorphisms $\mathbb{C} \rightarrow \widetilde{\mathbb{C}}$ that leave $\mathbb{R}$ elementwise fixed. The field of complex numbers is therefore essentially uniquely determined. In the special case $\mathbb{C}=\widetilde{\mathbb{C}}$ we get automorphisms of $\mathbb{C}$ with the fixed field $\mathbb{R}$.
Remark. What automorphisms (i.e. isomorphisms onto itself) admits the field of real numbers $\mathbb{R}$ ? Hint. Such an automorphism of $\mathbb{R}$ must preserve the ordering, of $\mathbb{R} !$

Donald Albin
Donald Albin
Numerade Educator
04:44

Problem 14

Each $z \in S^{1} \backslash\{-1\}$
$$
S^{1}:=\{z \in \mathbb{C} ; \quad|z|=1\}
$$
can be uniquely represented in the form
$$
z=\frac{1+\mathrm{i} \lambda}{1-\mathrm{i} \lambda}=\frac{1-\lambda^{2}}{1+\lambda^{2}}+\frac{2 \lambda}{1+\lambda^{2}} \mathrm{i}
$$
with $\lambda \in \mathbb{R}$.

Gaurav Kalra
Gaurav Kalra
Numerade Educator
03:05

Problem 15

(a) Consider the map
$$
f: \mathbb{C}^{*} \longrightarrow \mathbb{C} \text { with } f(z)=1 / \bar{z}
$$
Give a geometrical construction (with ruler and compass) for the image $f(z)$ and justify calling this map "reflection at the unit circle". Find the image under $f$ of each of
$(\alpha) \quad D_{1}:=\{z \in \mathbb{C} ; \quad 0<|z|<1\}$
$(\beta) \quad D_{2}:=\{z \in \mathbb{C} ; \quad|z|>1\}$
(\gamma) $D_{3}:=\{z \in \mathbb{C} ; \quad|z|=1\}$.
(b) Now consider the map
$$
g: \mathbb{C}^{*} \longrightarrow C \text { with } g(z)=1 / z(=\overline{f(z)})
$$
and give a geometrical construction for the image $g(z)$ of $z .$ Why is this map called "inversion at the unit circle"? What are the fixed points of $g$, i.e. for which $z \in \mathbb{C}^{\bullet}$ is it true that $g(z)=z ?$

Uma Kumari
Uma Kumari
Numerade Educator
03:29

Problem 16

Assume $n \in \mathbb{N}$ and let $W(n)=\left\{z \in \mathbb{C} ; z^{n}=1\right\}$ be the set of $n^{\text {th }}$ roots of unity.
Show:
(a) $W(n)$ is a subgroup of $\mathbb{C}^{*}$ (and so is a group itself).
(b) $W(n)$ is a cyclic group of order $n$, i.e. there is a $\zeta \in W(n)$ such that
$$
W(n)=\left\{\zeta^{\nu} ; \quad 0 \leq \nu<n\right\}
$$
Such a root of unity $\zeta$ is called a primitive root of unity.
Deduce that $W(n) \simeq \mathbb{Z} / n \mathbb{Z}$.
For which $d \in \mathbb{N}$ with $1 \leq d \leq n$ is the power $\zeta^{d}$ again a primitive $n^{\text {th }}$ root of unity? Therefore how many primitive $n^{\text {th }}$ roots of unity are there?

Aamir Mithaiwala
Aamir Mithaiwala
Numerade Educator
01:09

Problem 17

Let
$$
\mathcal{C}:=\left\{\left(\begin{array}{rr}
a & -b \\
b & a
\end{array}\right) ; \quad a, b \in \mathbb{R}\right\} \subset M(2 \times 2 ; \mathbb{R})
$$
with ordinary addition and multiplication of (real) $2 \times 2$ matrices.
Show: $\mathcal{C}$ is a field, which is isomorphic to $\mathcal{C}$, the field of complex numbers.

Sriram Soundarrajan
Sriram Soundarrajan
Numerade Educator
04:07

Problem 18

As remarked during the introduction of the complex numbers, the polynomial $P=X^{2}+1 \in \mathbb{R}[X]$ has no roots in $\mathbb{R}$, in particular it does not decompose into polynomials of smaller degrees, so $P$ is irreducible in $\mathbb{R}[X]$. In algebra (see, for instance, [La2]) it is shown how one constructs for each irreducible polynomial $P$ in the polynomial ring $K[X]$ over a field $K$ a minimal extension field $E$ in which the given polynomial does have a root. In our special case $(K=\mathbb{R}$, $P=X^{2}+1$, this means that one takes the factor ring of $\mathbb{R}[X]$ with respect to the ideal $\left(X^{2}+1\right)$. This is isomorphic to $\mathbb{C}$.

AG
Ankit Gupta
Numerade Educator
01:33

Problem 19

Hamilton's Quaternions (W. R. HAMLLTON, 1843)
We consider the following map
$$
\begin{array}{r}
H: \mathbb{C} \times \mathbb{C} \longrightarrow M(2 \times 2 ; \mathbb{C}) \\
(z, w) \mapsto H(z, w):=\left(\begin{array}{cc}
z & -w \\
\bar{w} & \bar{z}
\end{array}\right)
\end{array}
$$
and denote its image by
$$
\mathcal{H}:=\{H(z, w) ; \quad(z, w) \in \mathbb{C} \times \mathbb{C}\} \subset M(2 \times 2 ; \mathcal{C})
$$
Show that $\mathcal{H}$ is a skew field, i.e. in $\mathcal{H}$ all the field axioms hold with the exception of the commutativity law for multiplication.
Remark. The notation $\mathcal{H}$ is intended to remind us of Sir WuLLIAM ROWAN HAMILTON (1805-1865). One calls $\mathcal{H}$ the HAMILTON quaternions.

Rakesh Kumar Sharma
Rakesh Kumar Sharma
Numerade Educator
03:31

Problem 20

Cayley Numbers (A. CAYLEY, 1845)
Let
$$
\mathcal{C}:=\mathcal{H} \times \mathcal{H}
$$
Consider the following composition law:
$$
\begin{gathered}
\mathcal{C} \times \mathcal{C} \longrightarrow \mathcal{C} \\
\left(\left(H_{1}, H_{2}\right),\left(K_{1}, K_{2}\right)\right) \mapsto\left(H_{1} K_{1}-\bar{K}_{2}^{\prime} H_{2}, H_{2} \bar{K}_{1}^{\prime}+K_{2} H_{1}\right)
\end{gathered}
$$
Here $\bar{H}^{\prime}$ denotes the conjugate transposed matrix of $H \in \mathcal{H} \subset M(2 \times 2 ; \mathbb{C})$ Show that this defines on an $\mathbb{R}$-bilinear map, which has no zero divisors, i.e. the "product" of two elements in $\mathcal{C}$ is zero, iff one of the two factors vanishes. This "CAYLEY multiplication" is, in general, neither commutative nor associative.
A deep theorem (M. A. KERVAIRE (1958), J. MILNOR (1958), J. BOTT (1958)) says that on an $n$-dimensional $(n<\infty)$ real vector space $V$ a bilinear form free of zero divisors can only exist when $n=1,2,4$ or $8 .$ Examples of such structures are the "real numbers", the "complex numbers", the "HAMILTON quaternions" and the "CAYLEY numbers". Compare with the article of F. Hirzebruch, [Hi].

Nick Johnson
Nick Johnson
Numerade Educator