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: 
If $z\ne -1$ then $\left(\frac {z}{(z+1)^2}\right)^{-1}=2+z+\frac {1}{z}.$

If $z=a+ib$ with $a,b\in \Bbb R$ then $$|z|=1\implies 1=|z|^2=a^2+b^2\implies 1=(a+ib)(a-ib)\implies \frac {1}{z}=a-ib\implies$$ $$\implies 2+z+\frac {1}{z}=2+2a\in \Bbb R.$$

The fact that $\frac {1}{z}=\overline z=Re(z)-iIm(z)$ when $|z|=1$ is something to calculate ONCE. It's used all the time.
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.