Without (much) calculation show that if $|z| = 1$ then $ \Im\left(\frac{z}{(z+1)^2}\right)=0 $ 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! 
 A: Lets try a more geometric perspective. Look at the following sketch

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 $\rm BD$.
A: Rationalize the denominator and recall that $z\overline z+|z|^2$:
$$\frac z{(z+1)^2}=\frac z{(z+1)^2}\frac{(\overline z+1)^2}{(\overline z+1)^2}=\frac{|z|^2(\overline z+2)+z}{|z+1|^4}=\frac{z+\overline z+2}{|z+1|^4},$$which is clearly real. (Don't know whether that counts as "not much calculation" - it's  better than the mess you'd get starting with $z=x+iy$...)
Now he tells us he needs a "geometric/visual" solution. Fine, but you have to draw the picture yourself:
Draw a little picture with the points $A=0$, $B=z$, $C=z+1$ and $D=1$ labelled. Now $ABC$ is an isoceles triangle, hence $\angle BCA=\angle BAC$. Since the line $y=c$ is parallel to the $x$-axis, $\angle CAD = \angle ACB$. So $\angle BAD=2\angle CAD$. This says that $\arg z=2\arg(z+1)=\arg(z+1)^2$, hence $z/(z+1)^2$ is real.
A: 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.
A: 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.
