Find all complex numbers satisfying the equation

Question

Find all complex numbers $z$ satisfying the equation $$\left|z+\frac{1}{z}\right|=2.$$

Answer 1

Write that $z+\frac{1}{z} = 2e^{i\theta},\ \theta \in \mathbb{R}$

then multiply both sides by $z$, and solve the 2nd degree equation in $z$: $$z^2-2e^{i\theta}z+1=0$$

Answer 2

The solution set for $z$ is the union of the two circles of radius $\sqrt2$ centred respectively at $\def\ii{\mathbf i}\ii$ and at $-\ii$.

First I'll show that there are precisely $4$ solutions on every line through the origin, with the exception of the real line on which $\pm1$ are the only solutions. Write the square of the equation as $(z+\frac1z)\overline{(z+\frac1z)}=4$ which gives $$ 4=|z|^2 + \frac z{\overline z}+\frac {\overline z}z+\frac1{|z|^2} =|z|^2 +\frac1{|z|^2} + 2\cos(2\arg(z)), $$ which on any half-line starting at the origin (for which $\arg z$ is constant) gives a quadratic equation in $r=|z|^2$ with two positive solutions $r=m\pm\sqrt{m^2-1}$ where $m=2-\cos(2\arg(z))\geq1$, which coincide if and only if $\arg z\equiv0\pmod\pi$.

For the remainder I only have the rather unsatisfactory solution of showing that the points of those two circles are effectively solutions of the original equation. This is easily verified for the purely imaginary values $(\pm1\pm\sqrt2)\ii$. For the others fix $\lambda\in\Bbb R$, and consider the points on the circle centred at$~\ii$ that are of the form $a(1+\lambda \ii)$. Application of the equation for the circle gives $$ a=\frac{\lambda\pm\sqrt{2\lambda^2+1}}{\lambda^2+1} $$ for which one easily verifies that $$ \frac1{a(1+\lambda \ii)} = (\frac{\lambda\mp\sqrt{2\lambda^2+1}}{\lambda^2+1})(\lambda\ii-1). $$ Now the fact that one has $\left|a(1+\lambda \ii)+\frac1{a(1+\lambda \ii)}\right|=2$ is a simple calculation. Indeed the real part of the number $a(1+\lambda \ii)+\frac1{a(1+\lambda \ii)}$ is $2\frac{\sqrt{2\lambda^2+1}}{\lambda^2+1}$ while its imaginary part is $2\frac{\lambda^2}{\lambda^2+1}$, and one can use the fact that $\bigl(\sqrt{2\lambda^2+1}\bigr)^2+(\lambda^2)^2=(\lambda^2+1)^2$. The other circle results from applying $z\mapsto -z$ to the first one.

Answer 3

Clearly, $z\ne0$

$$\implies|z^2+1|=2|z|$$

Let $z=re^{iy}$ where $r(\ge0),y$ are real

$$2r^2=(r^2\cos2y+1)^2+(r^2\sin2y)^2=r^4+2r^2\cos2y+1$$

$$(r^2)^2-4r^2\sin^2y+1=0$$

$$\implies r^2=\frac{4\sin^2y\pm\sqrt{(4\sin^2y)^2-4}}2=2\sin^2y\pm\sqrt{4\sin^4y-1}$$

As $4\sin^4y-1<4\sin^4y, r^2=2\sin^2y+\sqrt{4\sin^4y-1}$

As $r^2$ is real, we need $4\sin^4y-1\ge0\iff2\sin^2y\ge1\iff\cos2y=1-2\sin^2y\le0$

$\cos2y\le0\iff2m\pi+\dfrac\pi2\le2y\le2m\pi+\dfrac{3\pi}2$ where $m$ is any integer