solving-u-v-w-with-u-v-w-C-finding-all-possible-solutions-I-tested-this-with-several-values-and-found-no-mistake-please-review-and-comment-I-hope-this-will-help-at-least-some-of-you-

Question Number 60175 by MJS last updated on 18/May/19

solving u^{v} = w with u, v, w \in C

finding all possible solutions

I tested this with several values and found

no mistake . please review and comment .

I hope this will help at least some of you .

Commented by MJS last updated on 18/May/19

Commented by MJS last updated on 18/May/19

the answers to my former question 60039 :

z^{i} = \frac{1}{2} - \frac{1}{2} i

p = 0; q = 1; a = \frac{\sqrt{2}}{2}; α = - \frac{π}{4}

p = 0 \Rightarrow r = e^{π (- \frac{1}{4} + 2 n)} with n \in Z

θ = \frac{1}{2} \ln 2

z_{n} = e^{π (- \frac{1}{4} + 2 n)} e^{i \frac{\ln 2}{2}} = e^{π (- \frac{1}{4} + 2 n)} (\cos \frac{\ln 2}{2} + i \sin \frac{\ln 2}{2})

z_{0} \approx . 428829 + . 154872 i

z_{- 1} \approx . 000800814 + . 000289214 i

z_{1} \approx 229.634 + 82 . 9325 i

z^{1 - i} = 1 + i

p = 1; q = - 1; a = \sqrt{2}; α = \frac{π}{4}

- \frac{9}{8} - \frac{\ln 2}{4 π} ⩽ n < \frac{7}{8} - \frac{\ln 2}{4 π}

- 1.18016 ⩽ n < . 819841

\Rightarrow n \in {- 1, 0}

r = 2^{\frac{1}{4}} e^{- π (n + \frac{1}{8})}; r_{- 1} = 2^{\frac{1}{4}} e^{\frac{7 π}{8}}; r_{0} = 2^{\frac{1}{4}} e^{- \frac{π}{8}}

θ = π n + \frac{π}{8} + \frac{\ln 2}{4}; θ_{- 1} = \frac{\ln 2}{4} - \frac{7 π}{8}; θ_{0} = \frac{\ln 2}{4} + \frac{π}{8}

z_{- 1} \approx - 15.6841 - 9 . 96443 i

z_{0} \approx . 677773 + . 430602 i

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