I have to evaluate,

$\displaystyle (1-\sqrt3i)^5$

so far I have that I found,

$\displaystyle r(\cos\theta + i\sin\theta)$

where r = 2

and that tan (theta) = $\displaystyle -\sqrt3$

but where do i go now?

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- Mar 24th 2009, 09:12 PMdriverfan2008De Moivre Theorem Help
I have to evaluate,

$\displaystyle (1-\sqrt3i)^5$

so far I have that I found,

$\displaystyle r(\cos\theta + i\sin\theta)$

where r = 2

and that tan (theta) = $\displaystyle -\sqrt3$

but where do i go now? - Mar 24th 2009, 09:23 PMChris L T521
Solving for $\displaystyle \theta$, you should have $\displaystyle \theta=-\frac{\pi}{3}$

Therefore, the complex number in polar form is the same as $\displaystyle 2\left[\cos\left(-\tfrac{\pi}{3}\right)+i\sin\left(-\tfrac{\pi}{3}\right)\right]=2e^{i\left(-\frac{\pi}{3}\right)}$

Thus, $\displaystyle \left(1-\sqrt{3}i\right)^5=\left(2e^{i\left(-\frac{\pi}{3}\right)}\right)^5=32e^{i\left(-\frac{5\pi}{3}\right)}=32e^{i\frac{\pi}{3}}$ $\displaystyle =32\left[\cos\left(\tfrac{\pi}{3}\right)+i\sin\left(\tfrac{ \pi}{3}\right)\right]=\boxed{16+16\sqrt{3}i}$ - Mar 24th 2009, 09:27 PMdriverfan2008
How did you go from,

$\displaystyle =32\left[\cos\left(\tfrac{\pi}{3}\right)+i\sin\left(\tfrac{ \pi}{3}\right)\right]$

to

$\displaystyle =\boxed{16+16\sqrt{3}i}$ - Mar 24th 2009, 09:51 PMProve It