1. ## Riemann Mapping (Complex)

Use conformal maps or combinations of conformal maps such as linear fractional transformations, powers, roots, sin z, log z, etc., to find a one-to-one analytic function mapping the given region D onto the upper half-plane U.

(1) $D = \{z: |Arg z| < \alpha\}, \alpha \leq \pi$

(2) $D = \{z= x + iy: x,y > 0\}$

(3) $D = \{z= x+ iy: |y-1| < 2\}$

(4) $D = \{z: |z-z_0| < r_0\}$

I'll try to do (1)

$-\alpha < Arg z < \alpha$

$z=re^{i\gamma}$
$z^{\alpha}=r^{\alpha}e^{i\theta\alpha}$

$-\alpha\gamma < \theta\gamma <\alpha\gamma$

$2\alpha\gamma = \pi$

$\gamma = \frac{\pi}{2\alpha}$

Answer: $iz^{\frac{\pi}{2\alpha}}$

If anyone could tell me if I'm on the right track and provide me with some help on the other problems, I would really appreciate it, thanks!

Use conformal maps or combinations of conformal maps such as linear fractional transformations, powers, roots, sin z, log z, etc., to find a one-to-one analytic function mapping the given region D onto the upper half-plane U.

(1) $D = \{z: |Arg z| < \alpha\}, 0< \alpha \leq \pi$
Try to visual this set. Use several values to $\alpha$ to motive you. If $\alpha = \pi$ then $|\arg z|$ is the cut-plane at the non-positive axis (cut along $(-\infty,0]$). The square root map $z\to z^{1/2}$ (where the branch of the square root is choose along the same cut i.e. principle branch) will map this set to $|\arg z| < \tfrac{\pi}{2}$ i.e. the right half-plane. But you want the upper-half plane to rotate your picture one-quater turn i.e. $z\to iz$ will do the trick. This gives the conformal mapping $z\to iz^{1/2}$. Thus, in general if we want to shrink (or expand) $|\arg z| < \alpha$ to $|\arg z| < \tfrac{\pi}{2}$ we should use $z\mapsto z^{\pi / 2\alpha}$ as you correctly said. But this only gives us the right half plane, therefore, we need $z\mapsto iz^{\pi /2\alpha}$. Good job .

(2) $D = \{z= x + iy: x,y > 0\}$
Again visualize this set. This is the first quadrant. You can also think of this set as $\{ r e^{i\theta} | r>0, 0<\theta<\tfrac{\pi}{2} \}$. Consider the mapping $z\mapsto z^2$. Under this mapping $re^{i\theta} \mapsto r^2 e^{2i\theta}$. Therefore, $0 < 2\theta < \pi$ while $r^2 > 0$. This is the upper half-plane. That is therefore the conformal mapping that will do the trick.
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Again visualize this set. This is the horizontal infinite strip between $-1$ and $3$. Let us raise this stip up to the positive side by the mapping $z\mapsto z+1$. Now we will compress this strip to $0 < y < \pi$ the reason for this will soon be apparent. We do this by a dilation mapping in this case $z\mapsto \tfrac{\pi}{4}z$. Note the following mapping $z\mapsto e^z$ and look what happens with the points. The point $x+iy$, $-\infty < x < \infty, 0 < y < \pi$ becomes $e^{x+iy} = e^x e^{iy}$. And what is that? That is the upper half plane!
Therefore, the required mapping is $z\mapsto e^{ \pi (z+1)/4}$.