MAT334-2018F > Quiz-3

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: ---\newcommand{\Arg}{\operatorname{Arg}}
--- End code ---
(I did it).

Similarly, one can change existing definition

--- Code: ---\renewcommand{\Re}{\operatorname{Re}}
--- End code ---

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


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