Determine the image of $S$ under $f(z)=\frac{z^2+2}{z^2+1}.$

Question

Let $S$ be the region $\{z:0<|z|<\sqrt{2}, \ 0 < \text{arg}(z) < \pi/4\}$. Determine the image of $S$ under the transformation

$$f(z)=\frac{z^2+2}{z^2+1}.$$

I'm facing some difficulties on problems of this nature that includes Möbius transformations. First I can see that under $z\mapsto z^2$ doubles the argument and squares the modulus. So now we have that our $S$ is transformed to $$S_1=\{z:0<|z|<2, \ 0 < \text{arg}(z) < \pi/2\}$$

We can now let $M(z)=\frac{z+2}{z+1}.$ But here we need to parametrize the boundary of $S_1$. Thus we have that

$$\gamma_1=2it, \quad t\in[0,1]$$ $$\gamma_2=2t, \quad t\in[0,1]$$ $$\gamma_3=2e^{it}, \quad t\in[0,\pi/2]$$

So,

$$M(\gamma_1)=\frac{2it+2}{2it+1}=\frac{4t^2+2}{4t^2+1}-i\frac{2t}{4t^2+1},$$

$$M(\gamma_2)=\frac{2t+2}{2t+1}$$

$$M(\gamma_3)=\frac{2e^{it}+2}{2e^{it}+1}$$

substituting $t=0$ and $t=1$ in $M(\gamma_1)$ we get the two points $2$ and $\frac{6}{5}-i\frac{2}{5}$. This means that $\gamma_1$ is mapped to a circle or line going through those points.

Doing the same for $M(\gamma_2)$ we get that the two points are $2$ and $\frac{4}{3}$.

However, plotting these points dont really make sense. Also, I'm not sure how to handle $M(\gamma_3)$, it gets very messy. This leads me to assume that the method I'm using is very inefficient.

Any way around this?

Answer 1

I think this might help:

$$f(z)=\frac{2+z^2}{1+z^2}=1+\frac{1}{1+z^2}$$

If $z=x+iy$:

$$\frac{1}{1+z^2}=\frac{1}{(x^2-y^2+1)+2ixy}=\frac{(x^2-y^2+1)-2ixy}{(x^2-y^2+1)^2+4x^2y^2}$$

So:

$$f(z)=\frac{(x^2-y^2+1)^2+4x^2y^2+(x^2-y^2+1)-2ixy}{(x^2-y^2+1)^2+4x^2y^2}$$

Answer 2

It helps to consider not only those segments or arcs, but the whole line or circle that they are part of.

$M$ maps $(0, 2, \infty)$ on the real line to $(2, \frac 43, 1)$, therefore the segment $\gamma_2$ from $0$ to $2$ is mapped to the segment from $2$ to $\frac 43$ on the real line.

$(0, 2i, \infty)$ on the imaginary line are mapped to $(2, \frac 65 - \frac 25 i, 1)$, therefore the segment from $0$ to $2i$ is mapped to an arc on the circle $C$ through $(2, \frac 65 - \frac 25 i, 1)$. This would already be sufficient to determine $C$. It becomes even easier if we use that analytic mappings preserve angles, so that $C$ must intersect the real line at right angles. It follows that $C$ is the circle with center $\frac 32$ and radius $\frac 12$, and the segment $\gamma_2$ from $0$ to $2i$ is mapped to the arc from $2$ to $\frac 65 - \frac 25 i$ on that circle.

In a similar manner you can conclude that the arc $\gamma_3$ from $2$ to $2i$ is mapped to the arc from $\frac 43$ to $\frac 65 - \frac 25 i$ on the circle with center $\frac 23$ and radius $\frac 23$.