Is this exercise in complex correctly done?

Question

Find all $z \in \mathbb C$ such:

$$|z|=1, Re(z^4)=-Im(z^4) $$

So what I thougt was:

First, let $z$ be $$z = |z|e^{ix+2k\pi}, |z|=1$$

then,

$$z^4 = e^{i4x+8k\pi}$$

given

$$Re(z^4)=-Im(z^4)$$

this only happens if $4x+8k\pi = \pi/4$ or $4x+8k\pi = \pi5/4$, then

$$x= \pi/16-2k\pi$$

or

$$x= \pi5/16-2k\pi$$

score 2 · Answer 1 · answered Feb 19 '14 at 16:00

2

yes but it would be more correct if you provide general solution equate angle to $$\pi/4+(2n+1)\pi/2$$ instead of $$\pi/4$$ and $$5\pi/4$$

answered Feb 19 '14 at 16:00

Raja Sekar

score 1 · Accepted Answer · edited Feb 19 '14 at 16:05

1

Just to make sure that I'm following we have:

$z^4=e^{i\left(4\theta+8k\pi\right)}=\cos(4\theta+8k\pi)+i\sin(4\theta+8k\pi)$

We want: $Re(z^4)=-Im(z^4) \Rightarrow \cos(4\theta+8k\pi)= -\sin(4\theta+8k\pi)$

Hint: $\cos(\theta)=-\sin(\theta)$, when $\theta=\frac{3\pi}{4}+\pi$

edited Feb 19 '14 at 16:05

user129714

answered Feb 19 '14 at 15:59

Mr.Fry

1

Since OP was equally lax with the notation, I want to point out what I am sure everyone knows-- that it should be $z^4 = e^{i(4 \theta + 8 k \pi)}$ – mlg4080 Feb 19 '14 at 16:01
Oh my God your right, I apologize for that. – Mr.Fry Feb 19 '14 at 16:13

2 Answers2