# Conformal mapping of the unit circle minus a smaller circle

Map the region inside the circle $|z| = 1$ and outside the circle $|z-1/2| = 1/2$ conformally onto the unit disk.

I was thinking of using some scaling and shifting to get from the unit circle to the circle $|z-1/2| = 1/2$. but I'm not sure how that might be useful since it doesn't help with getting the region we want. The other idea I had was to work backwards starting with the unit circle to get to the region we want.

Source: Spring 1992

## 1 Answer

Hint: What happens to that region under the map $z\to (1+z)/(1-z)?$

• I think this mapping takes the unit circle to the right half plane? but I'm not sure how it affects the inner circle. – iYOA Jun 4 '18 at 23:28
• Yes on the right half plane. As for the other circle, a linear fractional transformation takes a circle through a pole to what? – zhw. Jul 12 '18 at 5:20
• oh I see, it becomes a vertical line through x=1. Is there a nice way to show that though? I just used trig, but I feel like I'm missing some complex analysis technique. So now we're left with the right half plane minus the strip from 0 to 1 right? – iYOA Jul 13 '18 at 5:30
• $\mathbb R$ goes to $\mathbb R.$ Big circle therefore goes to a line $\perp$ to $\mathbb R$ and so does the small circle. $-1$ is on the big circle and goes to $0$, $0$ is on small circle and goes to $1$ Our region therefore goes to $\{x+iy: 0<x<1\}.$ – zhw. Jul 13 '18 at 15:29
• ah yes, I inverted it. So to finish it off, you would multiply it by $i$ to rotate it, then use $z \to e^z$ get from the horizontal strip to the unit disk? – iYOA Jul 14 '18 at 3:45