Hello i know that $(i+1)^{17}=256+256i$ is. But how do you get to that result?
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3The easiest systematic way IMO is to write $1+i$ in complex exponential form and then use Euler's formula. – projectilemotion Jun 20 '22 at 15:59
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To raise to a large power use repeated squaring, viz. $z^{17}=(((z^2)^2)^2)^2z$. – J.G. Jun 20 '22 at 16:04
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Use the binomial expasion: $$(1+i)^{17}=\sum_{k=0}^{17}\binom{17}{k}1^{17-k}i^{k}=\sum_{k=0}^{17}\binom{17}{k}i^k$$
Just use the circularity of the powers of $i$
Best way is using the polar form of the complex numbers: $$1+i=\sqrt2e^{i\pi/4}\implies (1+i)^{17}=2^{17/2}e^{17\pi i/4}=256\sqrt2(\cos(17\pi/4)+i\sin(17\pi/4))=256\sqrt2(\cos(pi/4)+i\sin(\pi/4))=256(1+i)$$
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The binomial sum that you simplified to takes a lot of effort to solve. It's not optimal at all. That being said, your other method is perfectly fine! – DatBoi Jun 20 '22 at 16:49
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Oh yes, i only pointed it because i always like to give different solutions for the same problem. But yes it's to complicated compared with the second one hahaha – Guillermo García Sáez Jun 20 '22 at 16:51
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$$i+1=\sqrt{2}\left(\frac 1{\sqrt{2}}+\frac i{\sqrt2}\right)=e^{\pi i/4}\sqrt{2}$$
Then
$$\begin{align*}(i+1)^{17} & =256e^{17\pi i/4}\sqrt{2}\\ & =256\sqrt{2}\left(\cos\frac {\pi}4+i\sin\frac {\pi}4\right)\\ & =256+256i\end{align*}$$
Frank W
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