if $Re(z\overline w)=|z||w|$ then $w=tz$ where $t\in \mathbb R$

Question

I was given the following exercise in complex numbers:

Assume that for some $z,w \in \mathbb C$: $Re(z\overline w)=|z||w|$

Show that this implies $w=tz$ where $t$ is some positive real.

What I did: we know that

$$Re(z\overline w)=\frac{z\overline w+\overline z w}{2}=|z||w|$$

Then $$Re(z\overline w)^2=(\frac{z\overline w+\overline z w}{2})^2=\frac{(z\overline w+\overline z w)(z\overline w+\overline z w)}{4}=\frac{z^2\overline w^2+2z\overline z w\overline w+\overline z^2w^2}{4}=|z|^2|w|^2=z \overline z w\overline w$$

Then overall we see that $z^2\overline w^2+\overline z^2w^2=2z\overline z w\overline w$

I don't see how this implies that $w=tz$. I don't see the connection.

What am I missing?

Answer 1

You have $$\frac{\overline zw+z\overline w}{2}=|z||w|$$ Dividing both sides by $z\overline z=|z|^2$, we get: $$\frac12\left(\frac wz+\overline{\frac wz}\right)=\frac{|w|}{|z|}$$ or: $$\operatorname{Re}\left(\frac wz\right)=\left|\frac wz\right|$$ If $\operatorname{Re}(x)=|x|$, then we know that $x$ is real. Thus, $\frac wz$ is real.

Answer 2

Did you try writing $z$ and $w$ in the polar format? Writing

$$z=r_1e^{i\theta_1}\\ w=r_2e^{i\theta_2}$$

Makes it much easier to calculate what $z\overline{w}$ is.