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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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Please add appropriate MathJax code so your question is understandable. – Parth Kohli Sep 23 '12 at 14:49
@ParthKohli: Yeah, I'm trying to get it to work. :) – Curtain Sep 23 '12 at 14:50
Thanks for the edit. – Curtain Sep 23 '12 at 14:52
up vote 4 down vote accepted

$$w=\frac{1+\mathrm iz}{1-\mathrm iz}\iff z=\mathrm i\cdot\frac{1-w}{1+w}$$ 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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What are the steps you are taking in order to get z? I've struggled around a bit with the conjugate, but that gets me to z squared. – Curtain Sep 23 '12 at 14:55
See Edit. (No conjugate at all in the picture.) – Did Sep 23 '12 at 14:58
Thanks! That solved it! – Curtain Sep 23 '12 at 15:08
@JulianAssange : The "steps" are made explicit in my answer. – Michael Hardy Sep 23 '12 at 18:06

$$ w=\frac{1+iz}{1-iz} $$ 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, $-i$: $$ z = -i\frac{w-1}{w+1} = i\frac{1-w}{1+w}. $$

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