There's another question that I'm -completely- stuck on, I can't even think of how to start solving this one:

Find all complex numbers z for which $\displaystyle z^3 = -4 \bar z$

If someone is able to explain that to me, thanks.

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- Jun 3rd 2008, 03:06 AMWedricComplex numbers
There's another question that I'm -completely- stuck on, I can't even think of how to start solving this one:

Find all complex numbers z for which $\displaystyle z^3 = -4 \bar z$

If someone is able to explain that to me, thanks. - Jun 3rd 2008, 04:26 AMbobak
Hello Wedric

first write z in "mod/arg" form. $\displaystyle z = Re^{i \theta }$ so $\displaystyle z^3 = R^3e^{i 3\theta }$ and $\displaystyle \bar z = Re^{-i \theta}$

$\displaystyle z^3 = -4 \bar z \ \ \Rightarrow \ \ R^3e^{i 3\theta } = -4Re^{-i \theta} $

$\displaystyle R^2e^{i 4\theta } = -4$

$\displaystyle R^2 \cos 4\theta + R^2 i \sin 4 \theta = -4$

Require $\displaystyle \sin 4 \theta = 0$

and $\displaystyle \cos 4\theta = -1$

$\displaystyle R=2$

$\displaystyle 4 \theta = 2 \pi n +\pi $

$\displaystyle \theta = \frac{ \pi }{4} (1+ 2n) \ \ \forall n \in \mathbb{Z}$

$\displaystyle \therefore z = 2e^{i\frac{ \pi }{4} (1+ 2n)}$

Bobak - Jun 3rd 2008, 04:40 AMWedric
Thanks for the reply, I'll mull over that for a while.