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Question Number 110246 by Her_Majesty last updated on 28/Aug/20

s o l v e f o r z \in C : {(a + b i)}^{z} = b + a i

Answered by mr W last updated on 28/Aug/20

θ = \tan^{- 1} \frac{b}{a}

z \ln (a + b i) = \ln (b + a i)

z (\ln \sqrt{a^{2} + b^{2}} + i θ) = \ln \sqrt{a^{2} + b^{2}} + i (\frac{π}{2} - θ)

z = \frac{\ln \sqrt{a^{2} + b^{2}} + i (\frac{π}{2} - θ)}{\ln \sqrt{a^{2} + b^{2}} + i θ}

z = \frac{[\ln \sqrt{a^{2} + b^{2}} + i (\frac{π}{2} - θ)] [\ln \sqrt{a^{2} + b^{2}} + i θ]}{\ln^{2} \sqrt{a^{2} + b^{2}} + θ^{2}}

z = \frac{\ln^{2} \sqrt{a^{2} + b^{2}} + (\frac{π}{2} - θ) θ + \frac{π}{2} \ln \sqrt{a^{2} + b^{2}} i}{\ln^{2} \sqrt{a^{2} + b^{2}} + θ^{2}}

Commented by Her_Majesty last updated on 28/Aug/20

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