Help understanding the transform $f(z)=\frac{2z+1}{5z+3}$ 
Consider the Mobius transformation $f(z)=\frac{2z+1}{5z+3}$. Show that this function maps the upper half plane and the lower half planes onto themselves. What can you say about the left and right half planes?

With a little  calculation  I was able to show that if $f(u+iv)=x+iy$, then $x=\frac{(1-3u)(5-2u)-15v^2}{(5-2u)^2+25v^2}$ and $y=\frac{v}{(5u-2)^2+25v^2}$. Based on this I was able to prove the first part. However I can't make much of the second part of the question. But my main concern is was this type of calculation unnecessary? I feel like I'm missing the bigger picture and am unable to see how these fractional transforms work. Any advice?  
 A: Mobius transforms preserve angles and transforms lines/circles into lines/circles (you can say a line is just a circle that goes through $\infty$).
Therefore, the image of the line $i\Bbb R$ is a line or a circle which is :


*

*perpendicular to the image of the line $\Bbb R$, which is $\Bbb R$.

*contains $f(0) = 1/3$

*contains $f(\infty) = 2/5$


So it must be the circle of diameter $[1/3 ; 2/5]$.
Then, you just have to check which side of the line corresponds to which side of the circle.
A: About your main concern: I think this is easier
$$f(z)=\frac{(2z+1)(5\overline z+3)}{|5z+3|^2}=\frac{10|z|^2+11\,\text{Re}\,z+3}{|5z+3|^2}+\color{red}{\frac{\text{Im}\,z}{|5z+3|^2}}i$$
and it follows at once that the upper and lower half planes are mapped onto themselves.
About the left and right half planes:
$$\text{Re}\,z\ge 0\implies10|z|^2+11\,\text{Re}\,z+3>0 $$
so it mapps the right halpf plane to itself. Try now to check about the left plane.
A: Write $w(z)=\frac{2z+1}{5z+3}$ then $z=\frac{1-3w}{5w-2}$. 
The left half plane is $Re(z)\le0$ 
$\Rightarrow$  $z+$$\overline z$$\le0$. On substituting $z$ from above, you get after simplification:
$11(w+$$\overline w$)$-30w$$\overline w-4$$\leq0$  which is an interior of a circle on $w$-plane.
A: If we write $$w=\frac {2z+1}{5z+3}$$ then rearranging, we have $$z=\frac{1-3w}{5w-2}$$
Now let $w=u+iv$ amd this simplifies to $$z=\frac{-2+11u-15u^2-15v^2+iv}{(5u-2)^2+25v^2}$$
Clearly when the imaginary part of $z$ is positive, then so is the imaginary part of $w$
When the real part of $z$ is positive, then $u$ and $v$ satisfy the inequality $$15u^2+15v^2-11u+2<0$$ so points in the right-hand half-plane are mapped to points inside this circle.
