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.


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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
$the answers to my former question 60039: z^i =(1/2)−(1/2)i p=0; q=1; a=((√2)/2); α=−(π/4) p=0 ⇒ r=e^(π(−(1/4)+2n)) with n∈Z θ=(1/2)ln 2 z_n =e^(π(−(1/4)+2n)) e^(i((ln 2)/2)) =e^(π(−(1/4)+2n)) (cos ((ln 2)/2) +i sin ((ln 2)/2)) z_0 ≈.428829+.154872i z_(−1) ≈.000800814+.000289214i z_1 ≈229.634+82.9325i z^(1−i) =1+i p=1; q=−1; a=(√2); α=(π/4) −(9/8)−((ln 2)/(4π))≤n<(7/8)−((ln 2)/(4π)) −1.18016≤n<.819841 ⇒ n∈{−1, 0} r=2^(1/4) e^(−π(n+(1/8))) ; r_(−1) =2^(1/4) e^((7π)/8) ; r_0 =2^(1/4) e^(−(π/8)) θ=πn+(π/8)+((ln 2)/4); θ_(−1) =((ln 2)/4)−((7π)/8); θ_0 =((ln 2)/4)+(π/8) z_(−1) ≈−15.6841−9.96443i z_0 ≈.677773+.430602i$
$the answers to my former question 60039 :$ $z^{i} = \frac{1}{2} - \frac{1}{2} i$ $p = 0; q = 1; a = \frac{\sqrt{2}}{2}; α = - \frac{π}{4}$ $p = 0 \Rightarrow r = e^{π (- \frac{1}{4} + 2 n)} with n \in Z$ $θ = \frac{1}{2} \ln 2$ $z_{n} = e^{π (- \frac{1}{4} + 2 n)} e^{i \frac{\ln 2}{2}} = e^{π (- \frac{1}{4} + 2 n)} (\cos \frac{\ln 2}{2} + i \sin \frac{\ln 2}{2})$ $z_{0} \approx . 428829 + . 154872 i$ $z_{- 1} \approx . 000800814 + . 000289214 i$ $z_{1} \approx 229.634 + 82 . 9325 i$ $z^{1 - i} = 1 + i$ $p = 1; q = - 1; a = \sqrt{2}; α = \frac{π}{4}$ $- \frac{9}{8} - \frac{\ln 2}{4 π} ⩽ n < \frac{7}{8} - \frac{\ln 2}{4 π}$ $- 1.18016 ⩽ n < . 819841$ $\Rightarrow n \in {- 1, 0}$ $r = 2^{\frac{1}{4}} e^{- π (n + \frac{1}{8})}; r_{- 1} = 2^{\frac{1}{4}} e^{\frac{7 π}{8}}; r_{0} = 2^{\frac{1}{4}} e^{- \frac{π}{8}}$ $θ = π n + \frac{π}{8} + \frac{\ln 2}{4}; θ_{- 1} = \frac{\ln 2}{4} - \frac{7 π}{8}; θ_{0} = \frac{\ln 2}{4} + \frac{π}{8}$ $z_{- 1} \approx - 15.6841 - 9 . 96443 i$ $z_{0} \approx . 677773 + . 430602 i$
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