Complex number equality problem

[v4valour]

Homework Statement

The problem states that you need to solve the following equation (without a calculator) : z^5 = z̅

Homework Equations

z=a+bi and z̅=a-bi

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.

Thank you in advance.

 

[Samy_A]

(is z̅ * z^5 = z^6 ??)

No, or more correctly, not necessarily.

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

 

[Ray Vickson]

Homework Statement

The problem states that you need to solve the following equation (without a calculator) : z^5 = z̅

Homework Equations

z=a+bi and z̅=a-bi

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.

Thank you in advance.

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

 

[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

 

[v4valour]

The polar form results in : (re^itheta)⁶=re^-itheta
e^6itheta=e^{-i theta}
e^7itheta=0
cos(7theta) + isin(7theta)=0

 

[Samy_A]

The polar form results in : (re^itheta)⁶=re^-itheta

The 6 should be 5.

e^6itheta=e^{-i theta}
e^7itheta=0

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.

 

[Ray Vickson]

The polar form results in : (re^itheta)⁶=re^-itheta
e^6itheta=e^{-i theta}
e^7itheta=0
cos(7theta) + isin(7theta)=0

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