Find an analytic function that maps the angle $-\pi/4<\operatorname{arg}(z)<\pi/2$ onto the upper half plane so that $w(1-i)=2$, $w(i)=-1$ , and $w(0)=0$
I'm trying to use this formula
$$\frac{w-w_1}{w-w_3 }\cdot\frac{w_2-w_3}{w_2-w_1 }=\frac{z-z_1}{z-z_3 }\cdot\frac{z_2-z_3}{z_2-z_1 }$$
so I got
$$\frac{w-2}{w-0 }\cdot\frac{-1-0}{-1-2 }=\frac{z-1+i}{z-0 }\cdot \frac{i-0}{i-1+i }$$
$$\frac{w-2}{w }\cdot\frac{1}{3 }=\frac{iz-i-1}{z(2i-1) }$$
$$w-2=3w\cdot\frac{iz-i-1}{z(2i-1) }$$
$$w(1-3\cdot\frac{iz-i-1}{z(2i-1) })=2$$
$$w=\frac{2z(2i-1)}{2iz-z-3iz+3i+3}$$
$$w=\frac{4iz-2z}{-iz-z+3i+3}$$
$$w=\frac{4iz-2z}{(3-z)(1+i)}$$
In order for this to be analytic, $z$ must not be $3$, but the angle $-π/4<\operatorname{arg}(z)<π/2$ also include $3$.
Now I follow the suggestion of GEdgar and tried the power of $z$. I know that $w=z^n$ always map the angle $\pi/n$ onto the upper half plane. In this problem my angle is between $\frac{-\pi}{4}$ and $\frac{\pi}{2}$ so the $arg(w)$ should be between $-n\pi/4$ and $n\pi/2$? and radius of $w$ is $r^n$ for $|z|=r$
Now plug into the function $w=z^n$ I got
$$(1-i)^n=2$$
$$i^n =-1$$
$$0^n=0$$
The last equation work for any $n$, the second equation implies that $n=2$ but the first equation I solve for $n$ and got $n= \ln (\frac{2}{1-i})$ but $n$ is the real number, right?
Actually, I don't have to worry about $n$ because $w=z^n$ is analytic for all $n$, but if this is true then why they have to give me $w(1-i)=2$, $w(i)=-1$ , and $w(0)=0$?
Best Answer
The map $$g(z):=\left(e^{i\pi/4} z\right)^{4/3}$$ maps the open wedge $W:\ -{\pi\over 4}<{\rm Arg}(z)<{\pi\over2}$ onto the upper half plane. The three points $i$, $0$, $1-i$ are lying on the boundary of $W$, but $g$ takes care of them as well. One computes $$g(i)=-1, \quad g(0)=0,\quad g(1-i)=2^{2/3}\ .$$ Not all three points so obtained are at the desired position. We therefore have to set up a Moebius transformation $T$ with $$T:\quad\bigl(-1,0,2^{2/3}\bigr)\mapsto\bigl(-1,0,2\bigr)\ .$$ For this $T$ we can make the following "Ansatz": $$T(z)={az\over d-z}\ .$$ (One has $b=0$ since $T(0)=0$; furthermore we may take $c=-1$ since $T$ cannot be a similarity.) Plugging in the correspondences $-1\mapsto -1$, $2^{2/3}\mapsto 2$ gives $$a=2{1+2^{2/3}\over2-2^{2/3}}\doteq 12.5,\quad d=3{2^{2/3}\over 2-2^{2/3}}\doteq 11.54\ .$$ $T$ maps the real axis (incl. $\infty$) onto the real axis. Since $T'(0)={a\over d}>0$ the upper half-plane is mapped onto the upper half-plane, as desired. The $f$ you are looking for is then given by $f:=T\circ g$.