De Moivre Theorem Help

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March 24th 2009, 09:12 PM
driverfan2008

De Moivre Theorem Help

I have to evaluate,
$(1-\sqrt3i)^5$

so far I have that I found,
$r(\cos\theta + i\sin\theta)$
where r = 2

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

but where do i go now?
March 24th 2009, 09:23 PM
Chris L T521

Quote:

Originally Posted by driverfan2008

I have to evaluate,
$(1-\sqrt3i)^5$

so far I have that I found,
$r(\cos\theta + i\sin\theta)$
where r = 2

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

but where do i go now?

Solving for $\theta$ , you should have $\theta=-\frac{\pi}{3}$

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

Thus, $\left(1-\sqrt{3}i\right)^5=\left(2e^{i\left(-\frac{\pi}{3}\right)}\right)^5=32e^{i\left(-\frac{5\pi}{3}\right)}=32e^{i\frac{\pi}{3}}$ $=32\left[\cos\left(\tfrac{\pi}{3}\right)+i\sin\left(\tfrac{ \pi}{3}\right)\right]=\boxed{16+16\sqrt{3}i}$
March 24th 2009, 09:27 PM
driverfan2008

How did you go from,

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

to

$=\boxed{16+16\sqrt{3}i}$
March 24th 2009, 09:51 PM
Prove It

Quote:

Originally Posted by driverfan2008

How did you go from,

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

to

$=\boxed{16+16\sqrt{3}i}$

What's $\cos{\frac{\pi}{3}}$ equal?

What's $\sin{\frac{\pi}{3}}$ equal?

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