複素数の累乗 $( \frac{1+\sqrt{2}-i}{1+\sqrt{2}+i} )^{2025}$ の値を求める問題です。ここで、$i$ は虚数単位です。

1. 問題の内容

複素数の累乗 $( \frac{1+\sqrt{2}-i}{1+\sqrt{2}+i} )^{2025}$ の値を求める問題です。ここで、$i$ は虚数単位です。

2. 解き方の手順

まず、複素数 $\frac{1+\sqrt{2}-i}{1+\sqrt{2}+i}$ を簡単にします。分母の共役複素数 $1+\sqrt{2}-i$ を分母と分子にかけます。

$$\frac{1+\sqrt{2}-i}{1+\sqrt{2}+i} = \frac{(1+\sqrt{2}-i)(1+\sqrt{2}-i)}{(1+\sqrt{2}+i)(1+\sqrt{2}-i)} = \frac{(1+\sqrt{2}-i)^2}{(1+\sqrt{2})^2+1}$$

分子を展開します。

$$(1+\sqrt{2}-i)^2 = (1+\sqrt{2})^2 - 2i(1+\sqrt{2}) + i^2 = 1+2\sqrt{2}+2 - 2i(1+\sqrt{2}) - 1 = 2+2\sqrt{2}-2(1+\sqrt{2})i$$

分母を展開します。

$$(1+\sqrt{2})^2+1 = 1+2\sqrt{2}+2+1 = 4+2\sqrt{2} = 2(2+\sqrt{2})$$

よって、

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

$$\frac{1+\sqrt{2}-i}{1+\sqrt{2}+i} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$$

次に、$\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$ を極形式で表します。絶対値は $\sqrt{(\frac{\sqrt{2}}{2})^2+(-\frac{\sqrt{2}}{2})^2} = \sqrt{\frac{1}{2}+\frac{1}{2}} = 1$ です。偏角は $arg(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i) = -\frac{\pi}{4}$ です。

したがって、$\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i = \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) = e^{-i\frac{\pi}{4}}$ となります。

$$(\frac{1+\sqrt{2}-i}{1+\sqrt{2}+i})^{2025} = (e^{-i\frac{\pi}{4}})^{2025} = e^{-i\frac{2025\pi}{4}} = e^{-i(506\pi + \frac{\pi}{4})} = e^{-i(253(2\pi) + \frac{\pi}{4})} = e^{-i\frac{\pi}{4}}$$

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

3. 最終的な答え

$\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$