$n$ が自然数のとき、$(\frac{1+i}{\sqrt{2}})^n - (\frac{1-i}{\sqrt{2}})^n$ の値を求める。

1. 問題の内容

$n$ が自然数のとき、$(\frac{1+i}{\sqrt{2}})^n - (\frac{1-i}{\sqrt{2}})^n$ の値を求める。

2. 解き方の手順

まず、$1+i$ と $1-i$ を極形式で表す。

$1+i = \sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))$

$1-i = \sqrt{2}(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}))$

したがって、

$\frac{1+i}{\sqrt{2}} = \cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})$

$\frac{1-i}{\sqrt{2}} = \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4})$

ド・モアブルの定理より、

$(\frac{1+i}{\sqrt{2}})^n = (\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))^n = \cos(\frac{n\pi}{4}) + i\sin(\frac{n\pi}{4})$

$(\frac{1-i}{\sqrt{2}})^n = (\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}))^n = \cos(-\frac{n\pi}{4}) + i\sin(-\frac{n\pi}{4}) = \cos(\frac{n\pi}{4}) - i\sin(\frac{n\pi}{4})$

$(\frac{1+i}{\sqrt{2}})^n - (\frac{1-i}{\sqrt{2}})^n = (\cos(\frac{n\pi}{4}) + i\sin(\frac{n\pi}{4})) - (\cos(\frac{n\pi}{4}) - i\sin(\frac{n\pi}{4}))$

$= 2i\sin(\frac{n\pi}{4})$

3. 最終的な答え

$2i\sin(\frac{n\pi}{4})$