Prove Complex Relationship

Question

Question:

Prove that if $z + \frac{1}{z}$ is real then either the magnitude of $z = 1$ or $z$ is real.

I'm struggling with, I found that $2ab$ must be equal to $0$ but I don't see how that help.

Thanks!

Answer 1

setting $z=a+bi$ then we get $z+1/z=a+bi+1/(a+bi)=a+bi+\frac{a-bi}{a^2+b^2}=\frac{a(a^2+b^2+1)+ib(a^2+b^2-1)}{a^2+b^2}$

Answer 2

Here is an easy approach. If $z + \frac{1}{z}$ is real then $z + \frac{1}{z} = \overline{z + \frac{1}{z}}$. Using the fact that $z \overline{z} = |z|^2$ you get

$$z + \frac{1}{z} =\overline{ z + \frac{1}{z}} = \overline{z}+\frac{1}{\overline{z}} = \overline{z} + \frac{z}{|z|^2} \Rightarrow z -\overline{z} = \frac{z- \overline{z}}{|z|^2} \Rightarrow (|z|^2 - 1)(z-\overline{z}) = 0$$

$\mathbb{C}$ is a domain.

Answer 3

Notice that $\frac{1}z=\frac{1}{|z|^2}\bar{z}$. If you split $z=a+bi$, this means your expression is $$a+bi+\frac{a-bi}{|z|^2}.$$ If $b$ is non-zero - that is, $z$ is not real, what must be true of $|z|$ for the imaginary part of the above expression to be $0$?