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 $w(0)=0$ 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 $w(0)=0$?
 A: Disclaimer. At the risk of giving a wrong answer - it will become apparent
at the bottom line why - nevertheless here comes. Something is wrong, either in the
question or in my answer. My question is, of course: where is the error?
The following is a related issue @ Mathematics Stack Exchange:

Applications of conformal mapping
And the main keyword is found @ Wikipedia as:


Schwarz–Christoffel mapping

Using that theory we have for $-\pi/4<\operatorname{arg}(z)<\pi/2$ , at $z=0$ :
$$
\frac{dz}{dw} = K'(w-0)^{(\pi/2+\pi/4)/\pi-1} = K' w^{-1/4}
\quad \Longrightarrow \quad z = K w^{3/4} + C 
$$
With $K = 4/3\cdot K'$. Substitute $z=0$ and $w=0$ herein and find that $C=0$.
Next step:
$$
i = K(-1)^{3/4} = K e^{i\,3\pi/4} = K/\sqrt{2}(-1+i)
\quad \Longrightarrow \quad
K = \frac{1-i}{\sqrt{2}}
$$
It follows that:
$$
w(z) = \left(\frac{\sqrt{2}\,z}{1-i}\right)^{4/3}
= \left(e^{i\pi/4} z\right)^{4/3} = g(z)
$$
Consequently:
$$
w(1-i) = \left(\sqrt{2}\right)^{4/3} = 2^{2/3} \ne 2
$$
This is not consistent with the OP's question and it explains the above disclaimer.

Update. It seems that my wrong answer has served as an Ansatz
for the correct answer by Christian Blatter: $\,w(z) = T(g(z))$ ;
so it has served a purpose and I'm not going to delete it.
A contour plot of both $\,g(z)\,$ - on the left - and $\,T(g(z))\,$ - on the
right - shows the differences.The viewports are $\,-0.1 < x < 1.9$ ,
$-1 < y < +1\,$ with $\,z = x+i\,y$ ; isolines of the Imaginary parts
are in $\color{red}{red}$ and isolines of the Real parts are in
$\color{green}{green}$ . Contour levels are between $-4$ and $+4$
with steps of $4/30$.

A: 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 $f:=T\circ g$.
