第11章 11.23

$2 + i \Rightarrow \theta_1$、$3 + i \Rightarrow \theta_2$ とおくと

$\cos \theta_1 = \frac{2}{\sqrt{5}}$ $\sin \theta_1 = \frac{1}{\sqrt{5}}$ $\cos \theta_2 = \frac{3}{\sqrt{10}}$ $\sin \theta_2 = \frac{1}{\sqrt{10}}$

$\cos(\theta_1 + \theta_2) = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 = \frac{2}{\sqrt{5}} \cdot \frac{3}{\sqrt{10}} - \frac{1}{\sqrt{5}} \cdot \frac{1}{\sqrt{10}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$

よって $\theta_1 + \theta_2 = \frac{\pi}{4}$