# conformal map from right half disc to upper half plane

I am trying to solve a problem about a conformal map from right half disc to the upper half plane and get stuck on something, I guess small detail. Here`s my question: I want to find a conformal map from right half disk $\{Rez>0, \vert z \vert <1\}$ onto upper half plane that maps $-i$ to $0$ , $+i$ to $\infty$ and $0$ to $-1$. I found the the Mobius transformation that would send send all points to the questions asked which is $\frac {z+i}{z-i}$ and I was trying to determine where the right half disc will go under this map. Can anyone please explain to me how can I find the image of the right half disc under this map? please.

Thank you

However, the circle $|z|=1$ is mapped to a line (because $i\to \infty$) containing the origin (because $-i\to 0$). This line is orthogonal to the real axis (conformality of mapping at $-i$), and is therefore the imaginary axis.
Your mapping carries the semicircle into a quadrant. Which ? Use conformality. The interval from $-i$ to $0$ is mapped to the interval from $0$ to $-1$. Comformality implies that what was to the right walking along first interval will be to the right walking along its image. This implies that the semicircle will be mapped to the second quadrant.
To get the result you wanted, you need to rotate your function by $-\pi/2$ (to map to first quadrant), and then square (to map to upper half plane).
$$-1\frac{(z+i)^2}{(z-i)^2}.$$