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April 30th, 2017, 05:07 AM   #1
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From: Israel

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Question 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.
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April 30th, 2017, 07:07 AM   #2
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note $|Z|=1 \implies r=1$ ...

$Z \cdot (\bar{Z})^2 = Z + i$

$e^{i\theta} \cdot e^{-2i\theta} = Z + i$

$e^{-i\theta} = Z + i$

$\bar{Z} = Z + i$

$a-bi= (a+bi) + i$

$a-bi = a + (b+1)i \implies -b = b+1 \implies b = -\dfrac{1}{2}$

$|Z| = 1 \implies a^2+b^2 = 1 \implies a = \pm \dfrac{\sqrt{3}}{2}$

$Z = \pm \dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i$
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April 30th, 2017, 07:46 AM   #3
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From: Israel

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Beautiful ! Thank you ... I guess I was overlooking thing that I totally forgot to plug |Z|=1 !

Thanks!
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