Q3 TUT 5101

[Victor Ivrii]

Let $D$ be the domain obtained by deleting the ray $\{x\colon x \le 0\}$ from the plane, and let $G(z)$ be a branch of $\log z$ on $D$. Show that $G$ maps $D$ onto a horizontal strip of width of $\pi$,
$$
\{x + iy\colon - \infty < x < \infty,\ c_0 < y < c_0 + 2\pi\},
$$
and that the mapping is one-to-one on $D$.

Draw both domains.$\newcommand{\Arg}{\operatorname{Arg}}$

[Kathy Ngo]

Fix $G(z)$ such that $G(z)=\ln|z|+i(Arg(z) + c_{0} +\pi)$ where $c_{0} \in \mathbb{R}$
let $x=\ln|z|$ and $y=(Arg(z) + c_{0} +\pi)$

Note that

$\forall z\in D, |z|\geq 0 \Rightarrow x=\ln|z|\in(-\infty, \infty)$

and

$\forall z\in D, Arg(z) \in (-\pi, \pi) \Rightarrow y=Arg(z) + c_{0} +\pi \in (c_{0} , c_{0}+2\pi)$

Therefore $G$ maps $D$ onto $\{x+iy: -\infty < x< \infty, c_{0} <y<c_{0}+2\pi\}$.


Suppose $G(z_{1}) = G(z_{2})$ then

$\ln|z_{1}|+i(Arg(z_{1}) + c_{0} +\pi)=\ln|z_{2}|+i(Arg(z_{2}) + c_{0} +\pi)$

therefore, we have

$\ln|z_{1}| = \ln|z_{2}|$
$Arg(z_{1}) + c_{0} +\pi = Arg(z_{2}) + c_{0} +\pi \Rightarrow Arg(z_{1})=Arg(z_{2})$

these two equations imply $z_{1} = z_{2}$, hence the mapping is one-to-one on $D$.

[Victor Ivrii]

Arguably you need to type Arg upright and provide a proper spacing after it. The trouble is that LaTeX (and Mathjax) would not recognize \Arg since it was not defined. But one can use \operatorname{Arg} ; however one could do better then this defining (only once!) \Arg

Code: [Select]

\newcommand{\Arg}{\operatorname{Arg}}(I did it).


Similarly, one can change existing definition

Code: [Select]

\renewcommand{\Re}{\operatorname{Re}}


On the pictures do not use  xxxxxx for "cuts". Just straight line