# 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?")

• "... I do not know what squaring a complex number or taking its reciprocal corresponds to in geometry". Hint: $z=re^{i\theta}\mapsto r^2e^{2i\theta}$
– asd
Sep 2, 2018 at 2:37
• 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 $\text{Im}(w)\equiv 0\pmod{\pi}$. Sep 2, 2018 at 4:10
• 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 $|z|=1$. Sep 2, 2018 at 4:11

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 $$$$\arg(z) = 2\arg(z+1).$$$$ Consider the parallelogram whose vertices are $$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 $z \ne -1$