The solution has been given above...see the first two replies of this thread.

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- Jul 5th 2009, 12:07 AMmalaygoel
- Jul 5th 2009, 12:16 AMProve It
$\displaystyle z = \frac{\sqrt{3}}{2} + \frac{1}{2}i$.

Changing to polars gives:

$\displaystyle |z| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2}$

$\displaystyle = \sqrt{\frac{3}{4} + \frac{1}{4}}$

$\displaystyle = \sqrt{1}$

$\displaystyle = 1$

$\displaystyle \theta = \arctan{\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}$

$\displaystyle = \arctan{\frac{1}{\sqrt{3}}}$

$\displaystyle = \frac{\pi}{6}$.

So $\displaystyle z = \frac{\sqrt{3}}{2} + \frac{1}{2}i = 1\,\textrm{cis}\,{\frac{\pi}{6}}$.

Now, using DeMoivre's Theorem...

$\displaystyle z^{69} = 1^{69}\,\textrm{cis}\,{\frac{69\pi}{6}}$

$\displaystyle = \textrm{cis}\,{\frac{3\pi}{2}}$

$\displaystyle = \cos{\frac{3\pi}{2}} + i\sin{\frac{3\pi}{2}}$

$\displaystyle =-i$. - Jul 5th 2009, 12:40 AMCaptainBlack
- Jul 5th 2009, 12:54 AMzorro
- Jul 5th 2009, 01:05 AMProve It
- Jul 5th 2009, 01:18 AMmr fantastic
- Dec 14th 2009, 12:25 PMzorroI didnt get the middle portion
- Dec 14th 2009, 02:01 PMProve It
$\displaystyle \frac{69\pi}{6} = \frac{23\pi}{2}$.

This angle is the same as $\displaystyle \frac{3\pi}{2}$, just having gone around the unit circle five times.