# Find the value of Z^69

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• May 3rd 2009, 09:45 PM
zorro
Find the value of Z^69
If $z = \frac{\sqrt{3}+1}{2}$ find the value of $z^{69}$
• May 4th 2009, 12:03 PM
vemrygh
Write the complex number on the form $z=re^{\theta i}$, and then compute $z^{69}=r^{69}e^{69\theta i}.$
• May 4th 2009, 01:30 PM
Soroban
Hello, zorro!

Quote:

$\text{If }z = \frac{\sqrt{3}+1}{2}, \text{ find the value of }z^{69}$

Convert $z$ to polar form: . $z \:=\:\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}$

Then use DeMoivre's Theorem: . $(\cos\theta + i\sin\theta)^n \;=\;\cos(n\theta) + i\sin(n\theta)$

• May 5th 2009, 12:49 AM
Calculus26
Am I missing something here?

I would agree with y'all if it were

[3^(1/2)+i]/2

But as z is not complex??????
• May 5th 2009, 05:16 AM
mr fantastic
Quote:

Originally Posted by Calculus26
Am I missing something here?

I would agree with y'all if it were

[3^(1/2)+i]/2

But as z is not complex??????

Probably a typo. Clarification might be given in a few weeks.
• June 28th 2009, 02:18 AM
zorro
i have tried to solve the problem but dont know what is the right answer .Could u please let me know what is the right answer .........
• June 28th 2009, 03:07 AM
Plato
Quote:

Originally Posted by zorro
i have tried to solve the problem but dont know what is the right answer .Could u please let me know what is the right answer .........

Is the number z complex or real?

Is it $\frac{\sqrt{3}+{\color{red}i}}{2}$ or just $\frac{\sqrt{3}+1}{2}$?
• June 28th 2009, 03:29 AM
aidan
Quote:

Originally Posted by zorro
If $z = \frac{\sqrt{3}+1}{2}$ find the value of $z^{69}$

$z = \frac{\sqrt{3}+1}{2}$ = 1.366025404 approx.

find the value of $z^{69}$
${1.366025404}^{69} = 2221547172.+decimaldigits$
How many decimals places are needed.

or do you need the binominal expansion of this
$\frac{ \left ( \sqrt{3}+1 \right )^{69} } {2^{69}}$
• June 29th 2009, 11:47 PM
zorro
Want to know the right solution
If $z=\frac{\sqrt{3}+1}{2}$ , find the value of $z^{69}$
• June 30th 2009, 12:37 AM
Moo
Okay, let me explain it to you :

In the form you've written, it is not possible to get a nice form for the exact value !
If it were $\frac{\sqrt{3}+i}{2}$, then the problem would be less meaningless.

• June 30th 2009, 12:53 AM
mr fantastic
Quote:

Originally Posted by zorro
If $z=\frac{\sqrt{3}+1}{2}$ , find the value of $z^{69}$

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.
• July 4th 2009, 01:46 PM
aidan

$\frac {479994729504490251817683255296 \times \sqrt{3+1}}{ 590295810358705651712}$
• July 4th 2009, 02:02 PM
Plato
Quote:

Originally Posted by aidan
$\frac {479994729504490251817683255296 \times \sqrt{3+1}}{ 590295810358705651712}$

As you can see your exact solution is a bit off.

Go to Wolfram|Alpha. Type “((sqrt[3]+1)/2)^69” exactly then click the “=” at the right-hand side of the input box.
• July 4th 2009, 08:53 PM
CaptainBlack
Quote:

Originally Posted by aidan

$\frac {479994729504490251817683255296 \times \sqrt{3+1}}{ 590295810358705651712}$

Quote:

Originally Posted by Plato
As you can see your exact solution is a bit off.

Go to Wolfram|Alpha. Type “((sqrt[3]+1)/2)^69” exactly then click the “=” at the right-hand side of the input box.

What is more interesting about aidan's answer (other than the obvious typo: $\sqrt{3+1}$ ) is why is it wrong (at least I think this is of interest).

Cb
• July 4th 2009, 11:54 PM
zorro
Sorry for the typo mistake
Quote:

Originally Posted by Plato
Is the number z complex or real?

Is it $\frac{\sqrt{3}+{\color{red}i}}{2}$ or just $\frac{\sqrt{3}+1}{2}$?

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