My Math Forum ideas on how to solve this complex number equation

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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: $\displaystyle Z^3-Z=i \Longrightarrow Z(Z^2-1)=i$ So I will get: 1. Z=i 2. (Z+1)(Z-1) = 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: 2,579 Thanks: 1275 note $|Z|=1 \implies r=1$ ... $Z \cdot (\bar{Z})^2 = Z + i$ $e^{i\theta} \cdot e^{-2i\theta} = Z + i$ $e^{-i\theta} = Z + i$ $\bar{Z} = Z + i$ $a-bi= (a+bi) + i$ $a-bi = a + (b+1)i \implies -b = b+1 \implies b = -\dfrac{1}{2}$ $|Z| = 1 \implies a^2+b^2 = 1 \implies a = \pm \dfrac{\sqrt{3}}{2}$ $Z = \pm \dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i$ Thanks from shanytc
 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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