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Question Number 60175 by MJS last updated on 18/May/19
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.
$$\mathrm{solving}\:{u}^{{v}} ={w}\:\mathrm{with}\:{u},\:{v},\:{w}\:\in\mathbb{C} \\ $$$$\mathrm{finding}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solutions} \\ $$$$\mathrm{I}\:\mathrm{tested}\:\mathrm{this}\:\mathrm{with}\:\mathrm{several}\:\mathrm{values}\:\mathrm{and}\:\mathrm{found} \\ $$$$\mathrm{no}\:\mathrm{mistake}.\:\mathrm{please}\:\mathrm{review}\:\mathrm{and}\:\mathrm{comment}. \\ $$$$\mathrm{I}\:\mathrm{hope}\:\mathrm{this}\:\mathrm{will}\:\mathrm{help}\:\mathrm{at}\:\mathrm{least}\:\mathrm{some}\:\mathrm{of}\:\mathrm{you}. \\ $$
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
$$\mathrm{the}\:\mathrm{answers}\:\mathrm{to}\:\mathrm{my}\:\mathrm{former}\:\mathrm{question}\:\mathrm{60039}: \\ $$$${z}^{\mathrm{i}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$$${p}=\mathrm{0};\:{q}=\mathrm{1};\:{a}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}};\:\alpha=−\frac{\pi}{\mathrm{4}} \\ $$$${p}=\mathrm{0}\:\Rightarrow\:{r}=\mathrm{e}^{\pi\left(−\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{n}\right)} \:\mathrm{with}\:{n}\in\mathbb{Z} \\ $$$$\theta=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$${z}_{{n}} =\mathrm{e}^{\pi\left(−\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{n}\right)} \mathrm{e}^{\mathrm{i}\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} =\mathrm{e}^{\pi\left(−\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{n}\right)} \left(\mathrm{cos}\:\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\:+\mathrm{i}\:\mathrm{sin}\:\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right) \\ $$$${z}_{\mathrm{0}} \approx.\mathrm{428829}+.\mathrm{154872i} \\ $$$${z}_{−\mathrm{1}} \approx.\mathrm{000800814}+.\mathrm{000289214i} \\ $$$${z}_{\mathrm{1}} \approx\mathrm{229}.\mathrm{634}+\mathrm{82}.\mathrm{9325i} \\ $$$$ \\ $$$${z}^{\mathrm{1}−\mathrm{i}} =\mathrm{1}+\mathrm{i} \\ $$$${p}=\mathrm{1};\:{q}=−\mathrm{1};\:{a}=\sqrt{\mathrm{2}};\:\alpha=\frac{\pi}{\mathrm{4}} \\ $$$$−\frac{\mathrm{9}}{\mathrm{8}}−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\pi}\leqslant{n}<\frac{\mathrm{7}}{\mathrm{8}}−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\pi} \\ $$$$−\mathrm{1}.\mathrm{18016}\leqslant{n}<.\mathrm{819841} \\ $$$$\Rightarrow\:{n}\in\left\{−\mathrm{1},\:\mathrm{0}\right\} \\ $$$${r}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{e}^{−\pi\left({n}+\frac{\mathrm{1}}{\mathrm{8}}\right)} ;\:{r}_{−\mathrm{1}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{e}^{\frac{\mathrm{7}\pi}{\mathrm{8}}} ;\:{r}_{\mathrm{0}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{e}^{−\frac{\pi}{\mathrm{8}}} \\ $$$$\theta=\pi{n}+\frac{\pi}{\mathrm{8}}+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}};\:\theta_{−\mathrm{1}} =\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}−\frac{\mathrm{7}\pi}{\mathrm{8}};\:\theta_{\mathrm{0}} =\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}+\frac{\pi}{\mathrm{8}} \\ $$$${z}_{−\mathrm{1}} \approx−\mathrm{15}.\mathrm{6841}−\mathrm{9}.\mathrm{96443i} \\ $$$${z}_{\mathrm{0}} \approx.\mathrm{677773}+.\mathrm{430602i} \\ $$

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