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I have problem solving this equation: $ \left(\frac{1+iz}{1-iz}\right)^4 = \frac12 + i {\sqrt{3}\over 2} $ I know how to solve equations that are like: $ w^4 = \frac12 + i {\sqrt{3}\over 2} $ And I have solved it to: $ w = \cos(-\frac{\pi}{12} + \frac{\pi k}{2})) + i\sin(-\frac{\pi}{12} + \frac{\pi k}{2})) $ But now is: $ w = \frac{1+iz}{1-iz} $ How does one get the complex z? Or am I solving it wrong?

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    @ParthKohli: Yeah, I'm trying to get it to work. :)2012-09-23

2 Answers 2

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$w=\frac{1+\mathrm iz}{1-\mathrm iz}\iff z=\mathrm i\cdot\frac{1-w}{1+w}$ Edit: On the road are the identities $(1-\mathrm iz)\cdot w=1+\mathrm iz$ and $1-w=-\mathrm i\cdot(1+w)\cdot z$.

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    @JulianAssange : The "steps" are made explicit in my answer.2012-09-23
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$ w=\frac{1+iz}{1-iz} $ First, multiply both sides by $1-iz$: $ w(1-iz) = 1+iz $ Expand the left side: $ w-wiz = 1+iz $ Put all terms involving $z$ on one side and those not involving $z$ on the other side: $ w-1=iz+wiz $ Factor $ w-1 = iz(1+w) $ Divide both sides by $i(1+w)$: $ z= \frac{w-1}{i(w+1)} $ Multiply the numerator and denominator by the conjugate, $-i$: $ z = -i\frac{w-1}{w+1} = i\frac{1-w}{1+w}. $