If $z + \frac{1}{z} = r$, where $r$ be a real number and $|r| < 2$. Find $|z|$.

Question

Let $r$ be a real number, $|r| < 2,$ and let $z$ be a complex number such that $$z + \frac{1}{z} = r.$$Find $|z|.$ I am actually stuck on this problem. Any help will be appreciated. Thank you

Answer 1

Multiply the equation by $z$, subtract $rz$, you get $$ z^2 -rz + 1 = 0. $$ Can you take it from here?

Hint: $ax^2 + bx + c = 0$ has solutions $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Edit

There are a bunch of heavy answers now. In my opinion the following solution is much more elementary. Just note that that $|z| = z \bar z$ for any complex number. Then use the quadratic formula to obtain $z = \frac{r - \sqrt{r^2 - 4}}{2}$ and $\bar z = \frac{r + \sqrt{r^2 - 4}}{2}$. Then $$ z \bar z = \left( \frac{r - \sqrt{r^2 - 4}}{2}\right)\left( \frac{r + \sqrt{r^2 - 4}}{2}\right) = \frac{r^2 - r^2 + 4}{4} = 1. $$

Answer 2

Write $\;z=x+iy\,,\,\,x,y\in\Bbb R\;$ . If $\;y=0\;$ then we have

$$\;x+\frac1x=r\implies x^2-rx+1=0$$

but this quadratic's discriminant is negative (why?) so this is impossible.

Thus, $\;y\neq0\;$ , and then

$$r=z+\frac1z=x+iy+\frac{x-iy}{x^2+y^2}=\left(x+\frac x{x^2+y^2}\right)+y\left(1-\frac1{x^2+y^2}\right)i$$

But all the above is real, thus its imaginary part is zero:

$$1-\frac1{x^2+y^2}=0$$

Complete now the argument and the solution.

Answer 3

Observe that $z+\frac{1}{z}=r$ is a real number, that means $$z+\frac{1}{z}=\bar{z}+\frac{1}{\bar{z}}.$$ This means $$z-\bar{z}=\frac{z-\bar{z}}{z\bar{z}}=\frac{z-\bar{z}}{|z|^2}.$$ So either $z=\bar{z}$ or $\color{red}{|z|^2=1}$. For the second case we get $$\color{red}{|z|=1}$$

If $z=\bar{z}$, then $z$ is a real number, call it $z=s$ (s0 $|z|=|s|$). So we need to solve

\begin{align*} s+\frac{1}{s} & =r\\ s^2-rs+1&=0. \end{align*} This gives $$s=\frac{r\pm\sqrt{r^2-4}}{2}.$$ But with $|r|<2$ we cannot have a real solution.

So we can conclude that $\color{blue}{|z|=1}$.

Answer 4

Using the polar coordinates $z = \rho e^{i\phi}$ we have $$ r = z+ \frac 1z = \rho e^{i\phi} + \frac 1\rho e^{-i\phi} = \left( \rho+ \frac 1 \rho \right) \cos \phi + i \left( \rho- \frac 1 \rho \right) \sin \phi \, . $$ The imaginary part must be zero. If $\rho \ne 1$ then necessarily $\sin \phi = 0$, and it follows that $\cos \phi = 1$ and $$ r = \rho+ \frac 1 \rho > 2 $$ is a contradiction. Therefore $\rho = 1$ (and $\cos \phi = r/2$).