Image of complex function $f$ in complex plane

Question

Let $f(z) = \frac{z-i}{z+i}$ be a complex function. We have to look at values in the plane given by $$P = \{ z \ | \ -\text{Im}(z) < \text{Re}(z) < \text{Im}(z)\}$$I need to describe the image of $f$ under $P$. I don't know how to find it. I tried to write the function the following way $$\frac{z-i}{z+i} = \frac{x+(y-1)i}{x+(y+1)i} \cdot \frac{x-(y+1)i}{x-(y+1)i} = \frac{x^2+y^2-1}{x^2+(y+1)^2} - \frac{2x}{x^2+(y+1)^2}i$$ But I don't know how I can find the image of $f$ when I have found this. Can somebody help me?

Answer 1

This is the Cayley transformation. It's well known that Mobius transformations map the Riemann sphere $\hat{\Bbb C}$ bijectively onto itself.

It maps the upper half plane conformally onto the unit disk. The lower half plane goes to the exterior of the disk. $-i\mapsto\infty$.

This particular transformation gives an isomorphism between the Poincaré upper half plane model, and that of the unit disk.

See what happens to the boundary lines of $P$. Mobius transformations map generalized circles to generalized circles. We have $0\mapsto-1,1-i\mapsto1-2i,\infty\mapsto1$. Those three points determine a circle. Then as a test point, $i\mapsto0$. So, it maps the half plane above $y=-x$ to the interior of the circle.

For the other boundary line, $y=x$, $f$ maps it to the circle determined by $-1,1/(1+2i)$ and $1$. Then since $-i\mapsto\infty$, we get the interior of the circle.

So, $f(P)$ is the intersection of the interior of the first circle and the interior of the second.