# Problem Set 1 - Complex Analysis
## Problem 1 (Field Structure of the Complex Numbers)

We define C:=R$^2$={(ab);a, b∈R}{\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 (ab), (cd)∈C{\textstyle \left({\begin{smallmatrix}a\\b\end{smallmatrix}}\right), \left({\begin{smallmatrix}c\\d\end{smallmatrix}}\right)\in \mathbb {C} }:
(ab)⊕(cd):=(a+cb+d)∈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 (ab)⊙(cd):=(ac−bdad+bc)∈C{\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 (C, ⊕, ⊙){\textstyle (\mathbb {C} , \oplus , \odot )} is a field, i.e.: Your proof should identify, 0{\textstyle 0} and 1{\textstyle 1} are and what to (ab)∈C{\textstyle \left({\begin{smallmatrix}a\\b\end{smallmatrix}}\right)\in \mathbb {C} } Inverse with respect to ⊕{\textstyle \oplus } or. (in case (ab)≠0{\textstyle \left({\begin{smallmatrix}a\\b\end{smallmatrix}}\right)\neq 0}) the Inverse with respect to ⊙{\textstyle \odot } is.
** ⊕{\textstyle \oplus } is commutative and associative and has a neutral element 0∈C{\textstyle 0\in \mathbb {C} }. Additionally, every element of C{\textstyle \mathbb {C} } is invertible with respect to ⊕{\textstyle \oplus }.
** ⊙{\textstyle \odot } is commutative and associative and has a neutral element 1∈C{\textstyle 1\in \mathbb {C} }. Additionally, every element of C∖{0}{\textstyle \mathbb {C} \setminus \{0\}} is invertible with respect to ⊙{\textstyle \odot }.
** For ⊕{\textstyle \oplus } and ⊙{\textstyle \odot } distributive law holds.
* show that ,the figure ϕ:R→C, ϕ(a)=(a0){\textstyle \phi :\mathbb {R} \to \mathbb {C} , \phi (a)=\left({\begin{smallmatrix}a\\0\end{smallmatrix}}\right)} is an injective field homomorphism. This means ϕ{\textstyle \phi }is injective and satisfies: ϕ(a+b)=ϕ(a)+ϕ(b){\textstyle \phi (a+b)=\phi (a)+\phi (b)} and ϕ(a⋅b)=ϕ(a)⋅ϕ(b)for all a, b∈R{\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 C{\textstyle \mathbb {C} } defined in Problem 1.Additionally, for a∈R{\textstyle a\in \mathbb {R} } we simply write a{\textstyle a} instead of ϕ(a)=(a0){\textstyle \phi (a)=\left({\begin{smallmatrix}a\\0\end{smallmatrix}}\right)}.In this notation, R⊂C{\textstyle \mathbb {R} \subset \mathbb {C} }.)Let i:=(01)∈C{\textstyle i:=\left({\begin{smallmatrix}0\\1\end{smallmatrix}}\right)\in \mathbb {C} }.
* Show that for all a, b∈R{\textstyle a, b\in \mathbb {R} }: a+ib=(ab){\textstyle a+ib=\left({\begin{smallmatrix}a\\b\end{smallmatrix}}\right)}.
* Show that i$^2$=−1{\textstyle i^{2}=-1}.
* Compute: x∈C{\textstyle x\in \mathbb {C} } for the follwing equatios:
** x$^2$=1{\textstyle x^{2}=1}.
** x$^2$=−1{\textstyle x^{2}=-1}.
** x$^4$=1{\textstyle x^{4}=1}.

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

Prove the following:
* For all x, y∈C{\textstyle x, y\in \mathbb {C} }, Re(x+y)=Re(x)+Re(y){\textstyle {\text{Re}}(x+y)={\text{Re}}(x)+{\text{Re}}(y)} and Im(x+y)=Im(x)+Im(y){\textstyle {\text{Im}}(x+y)={\text{Im}}(x)+{\text{Im}}(y)}.
* For all x∈C{\textstyle x\in \mathbb {C} }, Re(x)=(x+x$^¯$)/2 {\textstyle {\text{Re}}(x)={\frac {x+{\bar {x}}}{2}}} and Im(x)=x−x$^¯$/2i {\textstyle {\text{Im}}(x)={\frac {x-{\bar {x}}}{2i}}}.
* For all x∈C{\textstyle x\in \mathbb {C} }, x⋅x$^¯$=|x|$^2${\textstyle x\cdot {\bar {x}}=|x|^{2}} and (if x≠0{\textstyle x\neq 0}) 1/x=x$^¯$/|x|$^2$ {\textstyle {\frac {1}{x}}={\frac {\bar {x}}{|x|^{2}}}}.
* For all x, y∈C{\textstyle x, y\in \mathbb {C} }, |x⋅y|=|x|⋅|y|{\textstyle |x\cdot y|=|x|\cdot |y|} and (if y≠0{\textstyle y\neq 0}) | x/y |=|x|/|y| {\textstyle |{\frac {x}{y}}|={\frac {|x|}{|y|}}}.
* For all x, y∈C{\textstyle x, y\in \mathbb {C} }, |x+y|≥|x|+|y|{\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