Problem Set 1 - Complex Analysis

Problem 1 (Field Structure of the Complex Numbers)

We define ${\textstyle \mathbb {C} :=\mathbb {R} ^{2}=\{\left({\begin{smallmatrix}a\\b\end{smallmatrix}}\right);a,b\in \mathbb {R} \}}$ the following links ${\textstyle \oplus }$ and ${\textstyle \odot }$ . Definition:for ${\textstyle \left({\begin{smallmatrix}a\\b\end{smallmatrix}}\right),\left({\begin{smallmatrix}c\\d\end{smallmatrix}}\right)\in \mathbb {C} }$ :
${\textstyle \left({\begin{smallmatrix}a\\b\end{smallmatrix}}\right)\oplus \left({\begin{smallmatrix}c\\d\end{smallmatrix}}\right):=\left({\begin{smallmatrix}a+c\\b+d\end{smallmatrix}}\right)\in \mathbb {C} }$ und ${\textstyle \left({\begin{smallmatrix}a\\b\end{smallmatrix}}\right)\odot \left({\begin{smallmatrix}c\\d\end{smallmatrix}}\right):=\left({\begin{smallmatrix}ac-bd\\ad+bc\end{smallmatrix}}\right)\in \mathbb {C} }$

Show that ${\textstyle (\mathbb {C} ,\oplus ,\odot )}$ is a field, i.e.:
1. ${\textstyle \oplus }$ is commutative and associative and has a neutral element ${\textstyle 0\in \mathbb {C} }$ . Additionally, every element of ${\textstyle \mathbb {C} }$ is invertible with respect to ${\textstyle \oplus }$ .
2. ${\textstyle \odot }$ is commutative and associative and has a neutral element ${\textstyle 1\in \mathbb {C} }$ . Additionally, every element of ${\textstyle \mathbb {C} \setminus \{0\}}$ is invertible with respect to ${\textstyle \odot }$ .
3. For ${\textstyle \oplus }$ and ${\textstyle \odot }$ distributive law holds.
Your proof should identify, ${\textstyle 0}$ and ${\textstyle 1}$ are and what to ${\textstyle \left({\begin{smallmatrix}a\\b\end{smallmatrix}}\right)\in \mathbb {C} }$ Inverse with respect to ${\textstyle \oplus }$ or. (in case ${\textstyle \left({\begin{smallmatrix}a\\b\end{smallmatrix}}\right)\neq 0}$ ) the Inverse with respect to ${\textstyle \odot }$ is.
show that ,the figure ${\textstyle \phi :\mathbb {R} \to \mathbb {C} ,\phi (a)=\left({\begin{smallmatrix}a\\0\end{smallmatrix}}\right)}$ is an injective field homomorphism. This means ${\textstyle \phi }$ is injective and satisfies:
${\textstyle \phi (a+b)=\phi (a)+\phi (b)}$ and ${\textstyle \phi (a\cdot b)=\phi (a)\cdot \phi (b)\,\,\,{\text{for all }}a,b\in \mathbb {R} }$

Problem 2 (Arithmetic in the Complex Field)

We now use ${\textstyle +}$ und ${\textstyle \cdot }$ for the operations ${\textstyle \oplus }$ und ${\textstyle \odot }$ auf ${\textstyle \mathbb {C} }$ defined in Problem 1.Additionally, for ${\textstyle a\in \mathbb {R} }$ we simply write ${\textstyle a}$ instead of ${\textstyle \phi (a)=\left({\begin{smallmatrix}a\\0\end{smallmatrix}}\right)}$ .In this notation, ${\textstyle \mathbb {R} \subset \mathbb {C} }$ .)Let ${\textstyle i:=\left({\begin{smallmatrix}0\\1\end{smallmatrix}}\right)\in \mathbb {C} }$ .

Show that for all ${\textstyle a,b\in \mathbb {R} }$ : ${\textstyle a+ib=\left({\begin{smallmatrix}a\\b\end{smallmatrix}}\right)}$ .
Show that ${\textstyle i^{2}=-1}$ .
Compute: ${\textstyle x\in \mathbb {C} }$ ${\textstyle x\in \mathbb {C} }$ for the follwing equatios:
1. ${\textstyle x^{2}=1}$ .
2. ${\textstyle x^{2}=-1}$ .
3. ${\textstyle x^{4}=1}$ .

Problem 3 (Real and Imaginary Parts, Complex Conjugates, and Moduli)

Prove the following:

For all ${\textstyle x,y\in \mathbb {C} }$ , ${\textstyle {\text{Re}}(x+y)={\text{Re}}(x)+{\text{Re}}(y)}$ and ${\textstyle {\text{Im}}(x+y)={\text{Im}}(x)+{\text{Im}}(y)}$ .
For all ${\textstyle x\in \mathbb {C} }$ , ${\textstyle {\text{Re}}(x)={\frac {x+{\bar {x}}}{2}}}$ and ${\textstyle {\text{Im}}(x)={\frac {x-{\bar {x}}}{2i}}}$ .
For all ${\textstyle x\in \mathbb {C} }$ , ${\textstyle x\cdot {\bar {x}}=|x|^{2}}$ and (if ${\textstyle x\neq 0}$ ) ${\textstyle {\frac {1}{x}}={\frac {\bar {x}}{|x|^{2}}}}$ .
For all ${\textstyle x,y\in \mathbb {C} }$ , ${\textstyle |x\cdot y|=|x|\cdot |y|}$ and (if ${\textstyle y\neq 0}$ ) ${\textstyle |{\frac {x}{y}}|={\frac {|x|}{|y|}}}$ .
For all ${\textstyle x,y\in \mathbb {C} }$ , ${\textstyle |x+y|\geq |x|+|y|}$ .

Translation and Version Control

This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:

Source: Kurs:Funktionentheorie/Übungen/1._Blatt - URL:

https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Übungen/1._Blatt

Date: 01/14/2024