How to understand this equation in a complex plane

Question

the equation: $$z=z^*u \quad\forall u=e^{i\theta}$$ is said to be line passing through O and in the direction of $u$

I'm stuck. how to understand that?

Answer 1

Let $z=x+iy$. Then, $z=z^*u $ becomes

$$x+iy = (x-iy)e^{i\theta}=(x-iy)(\cos\theta+i\sin\theta) =(\cos\theta+y\sin\theta)+i(\sin\theta-y\cos\theta)$$

which leads to

$$x=x\cos\theta+y\sin\theta,\>\>\>\>\>y=x\sin\theta-y\cos\theta$$

Rearrange the equalities to get,

$$\sin \frac\theta2 \left(y\cos\frac\theta2-x\sin\frac\theta2 \right)=0,\>\>\>\>\> \cos\frac\theta2 \left(y\cos\frac\theta2-x\sin\frac\theta2 \right)=0$$

Since $\sin \frac\theta2$ and $\cos\frac\theta2$ cannot be zero at the same time,

$$y\cos\frac\theta2-x\sin\frac\theta2=0$$

Thus, $z$ represents a line passing through O and with its tangent angle $\theta/2$.