【高校数学C】3-1-2 複素数の極形式｜要点まとめ

絶対値は\(|1+\sqrt{3}i|=\sqrt{1^2+(\sqrt{3})^2}=2\)
偏角は\(\displaystyle \cos\theta=\frac{1}{2},\sin\theta=\frac{\sqrt{3}}{2}\)より、\(\displaystyle \theta=\frac{\pi}{3}\)
よって、
\(\displaystyle 1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)

絶対値は\(|-2-2i|=\sqrt{(-2)^2+(-2)^2}=2\sqrt{2}\)
偏角は\(\displaystyle \cos\theta=-\frac{1}{\sqrt{2}},\sin\theta=-\frac{1}{\sqrt{2}}\)より、\(\displaystyle \theta=\frac{5}{4}\pi\)
よって、
\(\displaystyle -2-2i=2\sqrt{2}\left(\cos\frac{5}{4}\pi+i\sin\frac{5}{4}\pi\right)\)

絶対値は\(|-1|=\sqrt{(-1)^2+0^2}=1\)
偏角は\(\cos\theta=-1,\sin\theta=0\)より、\(\theta=\pi\)
よって、
\(-1=\cos\pi+i\sin\pi\)

絶対値は\(|3i|=\sqrt{0^2+3^2}=3\)
偏角は\(\cos\theta=0,\sin\theta=1\)より、\(\displaystyle \theta=\frac{\pi}{2}\)
よって、
\(\displaystyle 3i=3\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\)

\(\displaystyle \alpha=\sqrt{3}-i=2\left(\cos\frac{11}{6}\pi+i\sin\frac{11}{6}\pi\right)\)
\(\displaystyle \beta=-2+2i=2\sqrt{2}\left(\cos\frac{3}{4}\pi+i\sin\frac{3}{4}\pi\right)\)
\(\displaystyle \alpha\beta=2\left(\cos\frac{11}{6}\pi+i\sin\frac{11}{6}\pi\right)\)
\(\displaystyle \ \ \ ・2\sqrt{2}\left(\cos\frac{3}{4}\pi+i\sin\frac{3}{4}\pi\right)\)
\(\displaystyle =4\sqrt{2}\left(\cos\frac{31}{12}\pi+i\sin\frac{31}{12}\pi\right)\)
\(\displaystyle =4\sqrt{2}\left(\cos\frac{7}{12}\pi+i\sin\frac{7}{12}\pi\right)\)

\(\displaystyle \frac{\alpha}{\beta}=\frac{2\left(\cos\frac{11}{6}\pi+i\sin\frac{11}{6}\pi\right)}{2\sqrt{2}\left(\cos\frac{3}{4}\pi+i\sin\frac{3}{4}\pi\right)}\)
\(\displaystyle =\frac{\sqrt{2}}{2}\left(\cos\frac{13}{12}\pi+i\sin\frac{13}{12}\pi\right)\)