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?

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