Find the value of Z^69

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• Jul 5th 2009, 12:07 AM
malaygoel
Quote:

Originally Posted by zorro
It is
$\frac{\sqrt{3}+i}{2}$
Now please can any one give me the right solution to this question?

The solution has been given above...see the first two replies of this thread.
• Jul 5th 2009, 12:16 AM
Prove It
Quote:

Originally Posted by zorro
It is
$\frac{\sqrt{3}+i}{2}$
Now please can any one give me the right solution to this question?

$z = \frac{\sqrt{3}}{2} + \frac{1}{2}i$.

Changing to polars gives:

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

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

$= \sqrt{1}$

$= 1$

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

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

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

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

Now, using DeMoivre's Theorem...

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

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

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

$=-i$.
• Jul 5th 2009, 12:40 AM
CaptainBlack
Quote:

Originally Posted by zorro
It is
$\frac{\sqrt{3}+i}{2}$
Now please can any one give me the right solution to this question?

See Soroban's post.

We are more interested in you acquiring a fishing net than giving you this fish.

CB
• Jul 5th 2009, 12:54 AM
zorro
Quote:

Originally Posted by mr fantastic
There is no simple exact answer to the question you have posted. This has been said several times now.

So either the question has a mistake in it or an approximate answer is required. If the former, then the likely question and its answer has been given in this thread. If the latter, the answer has also been given.

I don't see how any further progress can be made here unless the OP is prepared to acknowledge and address the above issues.

Sorry fr the mistake the question is:

If $z = \frac{\sqrt{3}+i}{2}$ ; find the value $z^{69}$
• Jul 5th 2009, 01:05 AM
Prove It
Quote:

Originally Posted by zorro
Sorry fr the mistake the question is:

If $z = \frac{\sqrt{3}+i}{2}$ ; find the value $z^{69}$

Take the time to read the responses you've gotten. Two people have told you a method of how to solve it, and I've posted the solution for you.
• Jul 5th 2009, 01:18 AM
mr fantastic
Quote:

Originally Posted by zorro
It is
$\frac{\sqrt{3}+i}{2}$
Now please can any one give me the right solution to this question?

Posts #2 and #3 tell you what to do. What part of their instruction don't you understand?
• Dec 14th 2009, 12:25 PM
zorro
I didnt get the middle portion
Quote:

Originally Posted by mr fantastic
Posts #2 and #3 tell you what to do. What part of their instruction don't you understand?

How is $cos(69. \frac{\pi}{6}) + i sin(69. \frac{ \pi}{6})$ = $cos \frac{3 \pi}{2} + i sin \frac{3 \pi}{2}$

$\frac{69\pi}{6} = \frac{23\pi}{2}$.
This angle is the same as $\frac{3\pi}{2}$, just having gone around the unit circle five times.