Complex Numbers problem

**sf1903** · October 30th 2009, 01:46 PM

Calculate: (1+3i)^9

How do I solve this? I have tried without any success. Thanks!

**Plato** · October 30th 2009, 01:53 PM

Originally Posted by sf1903

Calculate: (1+3i)^9

$1 + i\sqrt 3 = 2\cos \left( {\frac{\pi }<br /> {3}} \right) + i2\sin \left( {\frac{\pi }<br /> {3}} \right)$
$2^9=512$ and $9\left(\frac{\pi}{3}\right)=3\pi$

**Bruno J.** · October 30th 2009, 01:54 PM

Hint : $2e^{i\pi/3}=1+\sqrt{3}i$ .

**sf1903** · October 30th 2009, 02:48 PM

Originally Posted by Plato

$1 + i\sqrt 3 = 2\cos \left( {\frac{\pi }<br /> {3}} \right) + i2\sin \left( {\frac{\pi }<br /> {3}} \right)$
$2^9=512$ and $9\left(\frac{\pi}{3}\right)=3\pi$

If im correct the answer should be -512

512*Cos(3π)+i*512*Sin(3π) = -512 + 0 = -512

Or am i out in the blue?

**Bruno J.** · October 30th 2009, 02:50 PM

Yes! $(2e^{i\pi/3})^9=512e^{3i\pi}=-512$ .

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