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For $n\in\mathbb{N}$, determine real and imaginary part of $\left(\frac{i+1}{|i+1|}\right)^n$.

Attempt: $$\left(\frac{i+1}{|i+1|}\right)^n = \frac{(i+1)^n}{|i+1|^n} = (i+1)^n~\cdot~|i+1|^{-n}=\left(\sum_{k=0}^{n}\binom{n}{k}i^{n-k}\right)~\cdot~\left(\sum_{k=0}^{n}\left|\binom{n}{k}i^{n-k}\right|\right)^{-1}$$

Up until this point, this seems useful for me. However, I can't find any way to determine whether the exponent of $i$ is odd or even. I also tried to write down the binomial coefficient of $\binom{n}{k}$, with no success. Thanks for any help!

Righter
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losdelkelle
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1 Answers1

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Note that $$|a+bi| = \sqrt{a^2+b^2} \implies |1+i| =\sqrt 2$$ and therefore $$z = \left(\frac{1+i}{|1+i|}\right)^n = \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i\right)^n = \left(\cos \frac\pi4 + i\sin\frac\pi4\right)^n$$ Then, Moivre's formula implies $$\Im z= \sin\frac{\pi n}{4}$$

VIVID
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    You can also complete solution by writing $cos(\pi /4) + isin(\pi / 4) = e^{i \pi / 4}$ and therefore, $z = e^{i n \pi / 4}$. So, $Re(z) = cos(n\pi / 4)$ and $Im(z) = sin(n \pi / 4)$ – Snowball May 07 '21 at 08:09
  • @Snowball Yes, right. – VIVID May 07 '21 at 08:11