Let n be a positive integer. Describe the complex solutions of the equation $(z − 1)^n = z^n$ in standard form.
Thanks a lot!
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3Why? What have you tried on the problem so far; what are you stuck on, what do you know about complex numbers? – Steven Stadnicki Sep 19 '18 at 01:59
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i tried z^n= cos(nθ)+isin(nθ), but don't know how can i deal with (z-1)^n – user372376 Sep 19 '18 at 02:21
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2Hint: $0$ is clearly not a solution, so the equation is equivalent to $$\left(\frac{z - 1}{z}\right)^n = 1.$$ – Theo Bendit Sep 19 '18 at 02:21
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3Also, I am presuming you recall the idea of the $n^{th}$ roots of unity (1)? – eSurfsnake Sep 19 '18 at 02:28
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1related https://math.stackexchange.com/questions/2905854 – Nosrati Sep 19 '18 at 02:51
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1oh! Thanks! so now i got 1-1/z=cos(2kπ/n)+i*sin(2kπ/n) for k=0,1,2....n-1 and then solve for z? – user372376 Sep 19 '18 at 02:57
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1You should only get $n-1$ solutions. Do you see why? Which root of unity do you have to discard? – Steve D Sep 19 '18 at 04:19