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

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 $\frac 23$.

