# Algebra in the complex domain **URL:** \ **Category:** Factual Questions\ **Created:** [September 24, 2003, 12:19pm UTC](https://boards.straightdope.com/t/algebra-in-the-complex-domain/203523 "2003-09-24T12:19:32Z")\ **Posts on this page:** 2\ **Page:** 2
**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\ **Post date:** [December 1, 2003, 11:04pm UTC](https://boards.straightdope.com/t/algebra-in-the-complex-domain/203523/21 "2003-12-01T23:04:02Z")
Saying that x[sup]y[/sup] = 1 for complex numbers is really saying that e[sup]y ln(x)[/sup] = 1. If we write x = r e[sup]i theta[/sup] and y = a + bi, this means that exp((a + bi) ( ln r + i theta + 2 pi i n) ) = 1 exp((a ln r - b theta - 2 pi b n) + i (b ln r + a theta + 2 pi a n) ) = 1 This equation will hold if a ln r - b theta - 2 pi b n = 0 and b ln r + a theta + 2 pi a n = 2 pi m for **some** integer values of m and n. In your case, 5[sup] 2 pi i / ln 5[/sup], you have r = 5, theta = 0, a = 0, b = 2 pi / ln 5; this works for n = 0, m = 1. In summary: your math is correct, and it’s not true that x[sup]y[/sup] = 1 implies that |x| = 1, since you can play fast & loose with the values of m & n your equation is true for. I hope this helps; I have a feeling I didn’t explain this all that well, so other attempts are probably welcome. ---
**Author:** ![MC\_Master\_of\_Ceremonies](https://avatars.discourse-cdn.com/v4/letter/m/bb73d2/32.png) [@MC\_Master\_of\_Ceremonies](https://boards.straightdope.com/u/MC_Master_of_Ceremonies)\ **Post date:** [December 1, 2003, 11:17pm UTC](https://boards.straightdope.com/t/algebra-in-the-complex-domain/203523/22 "2003-12-01T23:17:10Z")
No your explantion was what I was looking for, I thought that the assumption if x[sup]y[/sup] = 1 then |x| = 1 was proably incorrect as it was a rather vague assumption made whilst looking at the cyclotomic equation. [Previous page](https://boards.straightdope.com/t/algebra-in-the-complex-domain/203523.md?page=1)