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-

Tinku Tara June 4, 2023 Algebra 0 Comments

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.

$$\mid{z}\mid=\mid{Arg}\:\left(\frac{{a}}{{b}}\pi\right)\mid=\mathrm{1}\wedge{k},\:{n}\in\mathbb{Z}\wedge{b}\neq\mathrm{0}\leqslant{k}<{n}: \\ $$$${x}^{{n}} ={z}\Rightarrow{x}={e}^{\frac{\mathrm{2}{k}+{a}}{{bm}}\pi{i}} \\ $$$$\mathrm{To}\:\mathrm{prove}\:\mathrm{that},\:\mathrm{please}. \\ $$

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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-

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