Geometrically interpreting $\Im ({z}/{(z + 1)^2} )$ Here is the question from Visual Complex Analysis by Needham.

Show geometrically that if |z| = 1 then 
$\Im\left(\frac{z}{(z + 1)^2}\right) = 0$
What other points apart from the unit circle satisfy this equation?

I know that z is a point on the circle, and (z+1) is a point on the unit circle translated by a unit vector, but I do not know what squaring a complex number or taking its reciprocal corresponds to in geometry.
I also have no idea how to proceed with the second half of the question ("What other points apart from the unit circle satisfy this equation?")
 A: Hopefully, the following qualifies as a (partially) geometric argument...
Quite obviously, all real $\,z \in \Bbb R \setminus \{ -1 \}\,$ satisfy the given condition that $\,\frac{z}{(z+1)^2} \in \Bbb R\,$.
Otherwise, $\,\frac{z}{(z+1)^2} \in \Bbb R$ $\iff \frac{(z+1)^2}{z} \in \Bbb R$ $\iff z+\frac{1}{z}+2 \in \Bbb R$ $\iff z + \frac{1}{z}\in \Bbb R\,$. Then the problem reduces to showing that the latter happens iff $\,\frac{1}{z} = \bar z$ $\iff |z| = 1\,$. This could be argued geometrically by first noting that the locus of points $\,w\,$ such that $z + w \in \Bbb R$ is the horizontal line $\,w \in \{x - \operatorname{Im}(z) \mid x \in \Bbb R\}\,$. However, in this case $\,w = \frac{1}{z} = \frac{1}{|z|^2} \cdot \bar z\,$, so $\,w\,$ must also lie on the oblique ray $\,w \in \{ \lambda \bar z \mid \lambda \in \Bbb R^+\}\,$. The intersection of the two is the single point $\,w = \bar z\,$, which completes the proof.
Incidentally, the above also answers the second part of the question.
A: You forgot $z\ne-1$


*

*Non-geometrically


$$\Im\left(\frac{z}{(z + 1)^2}\right) = \frac{\left(\frac{z}{(z + 1)^2}\right) - \overline{\left(\frac{z}{(z + 1)^2}\right)}}{2i}$$
Thus, $$\Im\left(\frac{z}{(z + 1)^2}\right) = 0 \iff \left(\frac{z}{(z + 1)^2}\right) = \overline{\left(\frac{z}{(z + 1)^2}\right)} = \left(\frac{\overline{z}}{\overline{(z + 1)^2}}\right) = \left(\frac{\overline{z}}{(\overline{z} + 1)^2}\right) \tag{*}$$
For $|z|=1$, we have that $\overline{z} = \frac 1 z$. Therefore, the last term in $(*)$ becomes
$$\left(\frac{\overline{z}}{(\overline{z} + 1)^2}\right) = \left(\frac{\frac 1 z}{(\frac 1 z + 1)^2}\right) = \left(\frac{z}{(z + 1)^2}\right)$$
Therefore, $\Im\left(\frac{z}{(z + 1)^2}\right) = 0$ for $|z| = 1$ but $z \ne -1$.


*Geometrically


(see below)


*Other points


Just solve $(*)$ with $z=re^{i \phi}$ and $\overline{z}=re^{-\phi i}$ to get $r=1$ or $e^{2 i \phi} = 1$. Similarly, if we solve with $z=x+iy$, we'll get $y=0$ (the real line) or $x^2+y^2=1$, except of course for $z \ne -1$.
So, imaginary part of that fraction is zero if $z$ is on the real line of $\mathbb C$ or on the unit circle, except of course for $z \ne -1$
A: Here's a very nice purely geometric solution that some students of mine came up with.
$\mathrm{Im}\left(\frac{z}{(z+1)^2}\right) = 0$ if and only if $\arg(z) = \arg((z+1)^2) + k\pi$, with $k = 0$ or $k=1$. Additionally, for $w \in \mathbb C$, we have $\arg(w^2) = 2\arg(w)$. We will show $\arg(z) = \arg((z+1)^2)$, or more specifically, that
\begin{equation}
\arg(z) = 2\arg(z+1).
\end{equation}
Consider the parallelogram whose vertices are $0$, $1$, $z$, and $z+1$. Let $\theta_0 = |\arg(z)|$ be the angle at the vertices $0$ and $z+1$, and let $\theta_1$ be the angle at the vertices $1$ and $z$. Since $|z| = 1$, the segment from $0$ to $z$ and the segment from $z$ to $z+1$ both have length $1$, so all sides of this parallelogram are equal. Bisect the parallelogram with a line segment from $0$ to $z+1$; we now have an isosceles triangle with vertices $0$, $1$, and $z+1$ whose angles are:

*

*$\theta_1$ at $1$;

*$\theta_2 := |\arg(z+1)|$ at $0$; and

*$\theta_0 - \theta_2$ at $z+1$.

Since the legs of the triangle from $0$ to $1$ and from $1$ to $z+1$ come from the parallelogram, they are of equal length, and respectively are across from the angles $\theta_0 - \theta_2$ and $\theta_2$. So $\theta_0 - \theta_2 = \theta_2$, so $|\arg(z)| = 2|\arg(z+1)|$. Since $\mathrm{Im}(z) = \mathrm{Im}(z+1)$, $\arg(z)$ and $\arg(z+1)$ have the same sign, so we have proven the above equation.
