Let $z_0$ be any complex number, $t$ any real number and $α∈(0,2π)$.
Find a conformal mapping from the sector $\{z:0≤arg(z−z_0)≤α\}$ onto the upper half plane, $\{w:Im(w)≥0\}$ such that $z_0$ is mapped to $t$.
EDIT:
My mapping is $$w = f(z) = (z-z_0)^{\large\frac{\pi}{\alpha}} + t$$
which easily shows that $z_0$ maps to $t$, as desired. But I am not totally sure of whether my mapping is onto the UHP, as needed. Here are my thoughts:
Suppose first that $z_1 = f(z) = (z-z_0)^{\large\frac{\pi}{\alpha}} + t$
for some arbitrary $z_1$ from the UHP.
Using Tanuj's suggestion (please see his answer below), and working backwards, then, looking at the arbitrary complex number $z_1$ from the UHP, and translating it by some fixed real number $t$, has the effect of translating the real part of $z_1$ -- so that $(z_1-t)$ lives in either quadrant I or quadrant II in the UHP. So it must have argument at most equal to $\pi$. And so $(z_1-t)^{\frac{\alpha}{\pi}}$ must have argument $\le \alpha$, and so we are done. I.e., we have shown that any arbitrary $z_1$ from the UHP is an image point of the mapping $(z-z_0)^{\frac{\pi}{\alpha}} + t$.
What do you think?
Thanks,
Choose a point $z_1$ in the upper half plane.
$$ z_1 = (z-z_0)^{\pi \over \alpha} + t $$
$$ \implies (z - z_0) = (z_1 -t)^{\alpha \over \pi} $$ So, all we need to check is if $\arg((z_1 -t)^{\alpha \over \pi}) \leq \alpha$.
Check that $\arg(z_1 -t) \leq \pi$ and we are done.