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z-Arg-a-b-pi-1-k-n-Z-b-0-k-lt-n-x-n-z-x-e-2k-a-bm-pii-To-prove-that-please-




Question Number 43894 by Rauny last updated on 17/Sep/18
∣z∣=∣Arg ((a/b)π)∣=1∧k, n∈Z∧b≠0≤k<n:  x^n =z⇒x=e^(((2k+a)/(bm))πi)   To prove that, please.
z∣=∣Arg(abπ)∣=1k,nZb0k<n:xn=zx=e2k+abmπiToprovethat,please.

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