# Complex number equality problem

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1. Nov 18, 2015

### v4valour

1. The problem statement, all variables and given/known data
The problem states that you need to solve the following equation (without a calculator) : z^5 = z̅

2. Relevant equations
z=a+bi and z̅=a-bi

3. The attempt at a solution
So far I've tried multiplying both sides by z̅: z̅ * z^5 = |z̅|^2
(is z̅ * z^5 = z^6 ??) |z|^6 = |z̅|^2

I then divided both sides by |z|^2, but this has lead to an incorrect answer. The answer given is z^6=1 ,
which then leads to a root of unity equation.

2. Nov 18, 2015

### Samy_A

No, or more correctly, not necessarily.

What happens when you multiply both sides by z?
(Note that $z\bar z = |z|²$)

3. Nov 18, 2015

### Ray Vickson

Have you tried the polar representation $z = r e^{i \theta}$?

4. Nov 19, 2015

### v4valour

Hi
I have tried r5( cos(5 theta) + i sin(5 theta) ) = r( cos(theta) +i sin(theta) ) but this lead to:
cos(5 theta) = cos (- theta) and i sin(5 theta) = i sin(-theta)
which is very difficult to solve without a calculator. Using the polar form results in similarly difficult equations.
Thanks

5. Nov 19, 2015

### v4valour

The polar form results in : (reitheta)6=re-itheta
e6itheta=e-i theta
e7itheta=0
cos(7theta) + isin(7theta)=0

6. Nov 19, 2015

### Samy_A

The 6 should be 5.
This is wrong
If $e^{ix}=e^{iy}$, then $e^{i(x-y)}=1$

Actually, wether you use my hint, or Ray's, you should easily get to the root of unity equation you expect.

Last edited: Nov 19, 2015
7. Nov 19, 2015

### Ray Vickson

The polar form give conditions on both $r$ and $\theta$, but you have said nothing about $r$.