solve for z∈C: (a+bi)^z =b+ai


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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

	$t h a n k y o u$
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