Question

I want to solve w^n = z^n for n when w,z are complex numbers.

Answer 1

More specifically (1+i)^n = (1-i)^n

Answer 2

n = 0? XD

Answer 3

thats one solution actually, I would like the full answer ( you can check yourself that for n = 4 its also true ! )

Answer 4

That's interesting haha

Answer 5

Okay.. then the answer is all integer multiples of 4, right? :D

Answer 6

\[\{n| \quad n = 4k ; \quad k \in \mathbb{Z} \}\]
fancy :D

Answer 7

Right, @giannisl9 ?

Answer 8

I thought of that too, but you have to prove it somehow.
The answer is probable what you wrote

Answer 9

See what I thought of

Answer 10

I already know how to prove it lol.
I was just hoping you did too ^^

Answer 11

Want to show me your train of thoughts ?!

Answer 12

Sure.
\[\Large 1+\color{blue}i = \sqrt{2}\left[\cos\left(\frac\pi4\right)+\color{blue}i\sin\left(\frac\pi4\right)\right]\]
\[\Large 1-\color{blue}i=\sqrt2\left[\cos\left(-\frac\pi4\right)+\color{blue}i\sin\left(-\frac\pi4\right)\right]\]

Answer 13

Ok got your way !
I was hoping for a more straight solution but thats good, representing the complex numbers this way

Answer 14

Thank you

Answer 15

So you can do it from here? :)

Answer 16

Yes :)

Answer 17

Awesome. I always knew you just needed a light nudge ^^`

Answer 18

Thanks for the help !

Answer 19

^.^