If $\displaystyle z = \frac{\sqrt{3}+1}{2}$ find the value of $\displaystyle z^{69}$
Hello, zorro!
$\displaystyle \text{If }z = \frac{\sqrt{3}+1}{2}, \text{ find the value of }z^{69}$
Convert $\displaystyle z$ to polar form: .$\displaystyle z \:=\:\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}$
Then use DeMoivre's Theorem: .$\displaystyle (\cos\theta + i\sin\theta)^n \;=\;\cos(n\theta) + i\sin(n\theta)$
$\displaystyle z = \frac{\sqrt{3}+1}{2}$ = 1.366025404 approx.
find the value of $\displaystyle z^{69}$
$\displaystyle {1.366025404}^{69} = 2221547172.+decimaldigits$
How many decimals places are needed.
or do you need the binominal expansion of this
$\displaystyle \frac{ \left ( \sqrt{3}+1 \right )^{69} } {2^{69}}$
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.
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.