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 August 31st, 2012, 02:59 AM #1 Member   Joined: Feb 2012 Posts: 39 Thanks: 0 Complex equation I'm perplexed by the following problem: "Find the number of solutions for n>2 and n€N of the following equation: $z^{n-1}=i\overline{z}$ So far I have worked out: Z=0 is always a solution. I can rewrite the original equation like this: $z^{n}=i|z|^{2}$ Furthermore, I can see the following: $|z|^{n-1}=|\overline{z}|$ so, $|z|=1$. Now, I have to solve $z^{n}=i$. I have no clue how to go about, so any help would be greatly appreciated.
 August 31st, 2012, 03:35 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Complex equation $z^n= r^3 e^{n i \theta} = i = e^{i \frac{\pi}{2}}$ Implies $r= 1$ and $n \theta= \frac{\pi}{2} + 2 k \pi$ for k = 0,1,2,...,n $\theta= \frac{\pi}{2n} + \frac{2 k \pi}{n}$ for k = 0,1,2,...,n $z= e^{i \theta}$ and we are done. EDIT : @skipjack : Yes, sorry. I have edited that.
 September 2nd, 2012, 01:13 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,471 Thanks: 2039 Dividing 2k? by n gives 2k?/n, not k?/n.

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