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?")

  • 1
    $\begingroup$ "... I do not know what squaring a complex number or taking its reciprocal corresponds to in geometry". Hint: $z=re^{i\theta}\mapsto r^2e^{2i\theta}$ $\endgroup$
    – asd
    Sep 2, 2018 at 2:37
  • $\begingroup$ This is not a geometric argument, so it is placed as a comment. If $z\neq 0$, then $z=\exp(w)$ for some $w\in\mathbb{C}$. Then, $$f(z)=\frac{z}{(z+1)^2}=\frac{1}{2+2\,\cosh(w)}=\frac{1}{4}\,\text{sech}^2\left(\frac{w}{2}\right)\,.$$ Thus, for all $z\in\mathbb{C}\setminus\{-1\}$, the imaginary part of $f(z)$ is $0$ if and only if $z=0$, $\text{Re}(w)=0$, or $\text{Im}(w)\equiv 0\pmod{\pi}$. $\endgroup$ Sep 2, 2018 at 4:10
  • $\begingroup$ In the case where $\text{Re}(w)=0$, we have precisely that $|z|=1$. In the case where $\text{Im}(w)\equiv 0\pmod{\pi}$, we get $z=\pm r$, where $r>0$, or equivalently, $z\in\mathbb{R}\setminus\{0\}$. That is, for all $z\in\mathbb{C}\setminus\{-1\}$, $\dfrac{z}{(z+1)^2}\in\mathbb{R}$ if and only if $z\in\mathbb{R}$ or $|z|=1$. $\endgroup$ Sep 2, 2018 at 4:11

3 Answers 3


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.


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.


You forgot $z\ne-1$

  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 $z \ne -1$.

  1. Geometrically

(see below)

  1. 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 $z \ne -1$


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