Conformal maps from the left half plane to a sector

Question

I'm trying to do a conformal mapping from the left half plane to a sector symmetrical around -X axis. would it be $Z=W^m$ while $m= (\frac{2}{\pi})(\pi-\theta)$. theta is the angle of the sector around the -X axis. thanks Nachum

Answer 1

You are close, but transferring the construction from the right half-plane to the left works a little differently.

For the corresponding construction in the right half-plane, one maps the sector $-\frac{\pi}{2} < \arg w < \frac{\pi}{2}$ (the half-plane) to the sector $-\theta < \arg z < \theta$, and that can be done by multiplying the argument with $\frac{\theta}{\pi/2}$, so

$$z = w^{2\theta/\pi}$$

works then (with the principal branch of the power). We get the desired mapping for the left half-plane by first mapping the left half-plane to the right - $w\mapsto -w$ - and after the mapping to the sector, rotate the sector to obtain the one in the left half-plane - $z\mapsto -z$. Altogether, the map is

$$z = - (-w)^{2\theta/\pi}.$$

You can also combine the two multiplications with $-1$ to a single rotation after raising $w$ to the appropriate power, then you can write the mapping as

$$z = e^{i(\pi-2\theta)} w^{2\theta/\pi}$$

with the right branch of $w^{2\theta/\pi}$. We cannot express the rotation/multiplication with $e^{i(\pi-2\theta)}$ as a power of $w$, a pure power $z = w^\alpha$ does not work, we need the multiplication to rotate the sector obtained from the power to the correct place.