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April 30th, 2017, 05:07 AM  #1 
Newbie Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0  ideas on how to solve this complex number equation
Hey, Been trying to find the correct way to solve this equation, but each time I am getting the wrong results.. mm.. $\displaystyle Equation: Z(\overline{Z}^2) = Z+i $ Also given Z=1 (module of z = 1) Here is what I did, some definitions: $\displaystyle i = CiS\frac{\pi}{2}\theta = e^{\frac{\pi}{2}\theta} $ $\displaystyle Z = Re^{i\theta} $ $\displaystyle \overline{Z}^2 = R^2e^{2i\theta} $ $\displaystyle Z(\overline{Z}^2) = (Re^{i\theta} \cdot R^2e^{2i\theta}) = R^3e^{i\theta} $ Now from here I plug n' play into the original equation: $\displaystyle R^3e^{i\theta} = Re^{i\theta} +e^{\frac{\pi}{2}\theta} $ My thought process: $\displaystyle Z^3Z=i \Longrightarrow Z(Z^21)=i $ So I will get: 1. Z=i 2. (Z+1)(Z1) = i But, this doesn't lead me to the solution (according to the book) which are: $\displaystyle Z1 = \frac{\sqrt{3}}{2}  \frac{1}{2} \cdot i $ $\displaystyle Z2 = \frac{\sqrt{3}}{2}  \frac{1}{2}\cdot i $ I would love to get some pointers on the direction Thanks, Shany Last edited by shanytc; April 30th, 2017 at 05:11 AM. 
April 30th, 2017, 07:07 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,623 Thanks: 1303 
note $Z=1 \implies r=1$ ... $Z \cdot (\bar{Z})^2 = Z + i$ $e^{i\theta} \cdot e^{2i\theta} = Z + i$ $e^{i\theta} = Z + i$ $\bar{Z} = Z + i$ $abi= (a+bi) + i$ $abi = a + (b+1)i \implies b = b+1 \implies b = \dfrac{1}{2}$ $Z = 1 \implies a^2+b^2 = 1 \implies a = \pm \dfrac{\sqrt{3}}{2}$ $Z = \pm \dfrac{\sqrt{3}}{2}  \dfrac{1}{2}i$ 
April 30th, 2017, 07:46 AM  #3 
Newbie Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0 
Beautiful ! Thank you ... I guess I was overlooking thing that I totally forgot to plug Z=1 ! Thanks! 

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