If $w=T(z)=\sqrt{\frac{1-iz}{z-i}}$, then what is the image of $\{z : |z|<1\}$ under $T$ ?
I rewrote $T$ as $T(z)= e^{\frac{1}{2}log(\frac{1-iz}{z-i})}$, and found image of interior of unit disk under $\frac{1-iz}{z-i}$ but doesn't able to move forward. Any help appreciated.
The image of the unit disk $\mathbb D = \{z\in\mathbb C||z|<1\}$ by $\frac{1-iz}{z-i}$ is the upper half plane $\mathbb H = \{z\in\mathbb C|\Im z>0\}$. It is a Möbius map, and you can check this by directly computing the imaginary part of the forward map and the modulus of the inverse. Taking the the principal branch of the square root, the image of $\mathbb H$ is the upper right quadrant (halve the argument).
Hope this helps.