Question: The title pretty much says it. 

> 
> Given that $z$ lies on the unit circle i.e. $|z| = 1$
>  show that 
> 
> 
> $$ \Im\left(\frac{z}{(z+1)^2}\right)=0 $$
> 
> 
> with little to no calculation involved( a geometric/visual solution beeing optimal).
> 
> 
> 

Here is my (best guess at a) solution:

$ \Im\left(\dfrac{z}{(z+1)^2}\right)=0 $ implies that the angle between $z$ and $(z+1)^2$ is a multiple of pi. Hence to show that the result is true we must show/argue that $z$ and $z+1$ lie on the same line.

This is what I have so far, but I don't see how to continue. Any help would be appreciated!

Answer: 
Lets try a more geometric perspective. Look at the following sketch

[![enter image description here](https://i.stack.imgur.com/e1tgr.png)](https://i.stack.imgur.com/e1tgr.png)

Here the point $\rm A$ represents $z$ and the point $\rm C$ represents $z+1$. Our goal is to show that $\angle \rm ABD=2\angle \rm ABC=2\angle \rm CBD$, which then proves that $z$ and $\left(z+1\right)^{2}$ both have the same angle and so $\frac{z}{\left(z+1\right)^{2}}$ is real. But that's easy because the triangle $\Delta\rm ABC$ is isosceles (which is true given ${\rm AB}=\left|z\right|=1$) so $\angle \rm ABC=\angle \rm ACB$, and also $\angle \rm ACB=\angle \rm CBD$ because $\rm AC$ is parallel to $\rm BD$.
Another "easy-ish" way is to make the parametrization $z = e^{i\theta}$, then

$$ \frac{z}{(1+z)^2} = \frac{1}{e^{i\theta} + 2 + e^{-i\theta}} = \frac{1}{2 + \cos \theta} $$

which is a real number. 

To me that's very little calculation, but I can see your book wants a geometric solution without explicitly saying it.