# Complex Numbers problem

• October 30th 2009, 01:46 PM
sf1903
Complex Numbers problem

How do I solve this? I have tried without any success. Thanks!
• October 30th 2009, 01:53 PM
Plato
Quote:

Originally Posted by sf1903

$1 + i\sqrt 3 = 2\cos \left( {\frac{\pi }
{3}} \right) + i2\sin \left( {\frac{\pi }
{3}} \right)$

$2^9=512$ and $9\left(\frac{\pi}{3}\right)=3\pi$
• October 30th 2009, 01:54 PM
Bruno J.
Hint : $2e^{i\pi/3}=1+\sqrt{3}i$.
• October 30th 2009, 02:48 PM
sf1903
Quote:

Originally Posted by Plato
$1 + i\sqrt 3 = 2\cos \left( {\frac{\pi }
{3}} \right) + i2\sin \left( {\frac{\pi }
{3}} \right)$

$2^9=512$ 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?
• October 30th 2009, 02:50 PM
Bruno J.
Yes! $(2e^{i\pi/3})^9=512e^{3i\pi}=-512$.