Show Posts
1
Quiz-3 / Re: Q3 TUT 5101
« on: October 14, 2018, 12:20:54 AM »
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
Suppose $G(z_{1}) = G(z_{2})$ then
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 $D$.
$Arg(z_{1}) + c_{0} +\pi = Arg(z_{2}) + c_{0} +\pi \Rightarrow Arg(z_{1})=Arg(z_{2})$