# Thread: Find the value of Z^69

1. Originally Posted by zorro
It is
$\displaystyle \frac{\sqrt{3}+i}{2}$
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.

2. Originally Posted by zorro
It is
$\displaystyle \frac{\sqrt{3}+i}{2}$
Now please can any one give me the right solution to this question?
$\displaystyle z = \frac{\sqrt{3}}{2} + \frac{1}{2}i$.

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...

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

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

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

$\displaystyle =-i$.

3. Originally Posted by zorro
It is
$\displaystyle \frac{\sqrt{3}+i}{2}$
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

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

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

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

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.

6. Originally Posted by zorro
It is
$\displaystyle \frac{\sqrt{3}+i}{2}$
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?

7. ## I didnt get the middle portion

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

8. $\displaystyle \frac{69\pi}{6} = \frac{23\pi}{2}$.

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

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