Could you help me with the following please:
Find the conformal mapping of the angle $− \frac{π}{4} <Arg (z) <\frac{π}{2}$ to the right half plane $Im(w)> 0$ such that $w (1 - i) = 2, w (i) = -1, w (0) = 0$.
I have seen similar examples, but take $− \frac{π}{2} <Arg (z) <\frac{π}{2}$, I have tried and cannot find a pattern to be able to solve this exercise. I would greatly appreciate your help.
Regards.
Let’s see whether I can get this right:
You want to increase the angle of the wedge from $135°$ to $180°$, so need a $4/3$ power. That is $f_1(z)=z^{4/3}$, which sends $1-i=2^{1/2}e^{-i\pi/4}$ to $2^{2/3}e^{-i\pi/3}$, $0$ to $0$, and $i=e^{i\pi/2}$ to $e^{2i\pi/3}$.
Needs a rotation of $60°$, so multiply by $e^{i\pi/3}$ to get $f_2(z)=e^{i\pi/3}z$. Unfortunately, $f_2\circ f_1$ sends $0$ to $0$ and $i$ to $-1$ all right, but sends $1-i$ how? Its progress is $1-i\mapsto2^{2/3}e^{-i\pi/3}\mapsto2^{2/3}=\sqrt[3]4$.
So now for $f_3$ all we need is a map of the Upper Half Plane that leaves zero and $-1$ fixed, but moves all other points on the real axis. The general form of such is $$ f_3(z)=\frac z{cz+1+c}\,, $$ for real $c>-1$. (In general, a fractional linear transformation of UHP will be given by $\frac{az+b}{cz+d}$ with real coefficients satisfying $ad-bc>0$.)
Well. we need only find $c$ such that $2^{2/3}$ gets sent to $2$, i.s. such that $$ 2=\frac{2^{2/3}}{2^{2/3}c+1+c}\,, $$ which I solve to get $c=\frac{2^{2/3}-2}{2(2^{2/3}+1)}$, though I may have slipped up in my high-school algebra here. And that does it, setting $w=(f_3\circ f_2\circ f_1)(z)$.