Find the region of the plane given by the equation $z^2 = i|z|^2$

Question

In order to find the region described by

$$z^2 = i|z|^2$$

I did:

$$z = a+bi \rightarrow |z|^2 = a^2+b^2 \rightarrow i|z|^2 = a^2i+b^2i$$

$$z = a+bi \rightarrow z^2 = a^2 + 2ab -b^2$$

$$z^2 = i|z|^2 \rightarrow a^2 + 2ab -b^2 = a^2i+b^2i \rightarrow (a^2+2ab-b^2)-(a^2+b^2)i = 0 \rightarrow $$

$$a^2+2ab-b^2 = 0$$

which gave me a pair of intersecting lines. Am I right? It's kinda strange that a complex equation gives me a pair of lines

Answer 1

You know that the absolute values of $z^2$ and $i|z^2|$ are the same no matter which complex number $z$ is, since $|i|=1$. And you know $i|z|^2$ is a pure imaginary. So $z^2$ must also be a pure imaginary. If $z= r(\cos\theta+i\sin\theta)$ then $z^2= r^2(\cos(2\theta) + i\sin(2\theta))$. So you need $\cos(2\theta)=0$. That's two straight lines through the origin.

Per Rahul's comment, this is not done yet. The relation $\cos(2\theta)=0$ is also satisfied by some values of $z$ for which $z^2=-i|z^2|$. You have to figure out which parts of those two lines that corresponds to.

Answer 2

It is clear that $z=0$ is in the region. If $z\neq 0$ write $z=re^{it}$ with $r>0$ and $t \in [0,2\pi).$ The condition can be written $$r^2e^{2it}=ir^2,$$ which means $$e^{2it} =i.$$ But $2t \in [0,4\pi).$ So the only possibilities are $2t = \pi/2$ and $2t = 5\pi/2$. Hence $t = \pi/4$ or $t=5\pi/4$. Can you now draw the region ?

Answer 3

Well.. let me transform my comment into an answer.

$z=0$ evidently satisfies. If $z\neq 0$, then: $$ z^2 = i|z|^2 = iz\bar z \\ z = i\bar z \\ Re\{z\} = Im\{z\} $$

All these steps are equivalent. And thus: $z = a + ai$, $\forall a\in\mathbb{R}$ is the solution.