【高校数学C】3-1-2 複素数の極形式｜問題集

絶対値は\(|1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}\)
偏角は\(\displaystyle \cos\theta=\frac{1}{\sqrt{2}},\sin\theta=-\frac{1}{\sqrt{2}}\)より、\(\displaystyle \theta=\frac{7}{4}\pi\)
よって、
\(\displaystyle 1-i=\sqrt{2}\left(\cos\frac{7}{4}\pi+i\sin\frac{7}{4}\pi\right)\)

絶対値は\(|-\sqrt{3}+i|=\sqrt{(-\sqrt{3})^2+1^2}=2\)
偏角は\(\displaystyle \cos\theta=-\frac{\sqrt{3}}{2},\sin\theta=\frac{1}{2}\)より、\(\displaystyle \theta=\frac{5}{6}\pi\)
よって、
\(\displaystyle -\sqrt{3}+i=2\left(\cos\frac{5}{6}\pi+i\sin\frac{5}{6}\pi\right)\)

絶対値は\(|2|=\sqrt{2^2+0^2}=2\)
偏角は\(\cos\theta=1,\sin\theta=0\)より、\(\theta=0\)
よって、
\(2=2(\cos0+i\sin0)\)

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

\(z+1\)
\(\displaystyle =\frac{1}{2}+i\frac{\sqrt{3}}{2}+1\)
\(\displaystyle =\frac{3}{2}+i\frac{\sqrt{3}}{2}\)
\(\displaystyle =\sqrt{3}\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)\)
\(\displaystyle =\sqrt{3}\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)\)

\(\displaystyle \alpha=2+2i=2\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\)
\(\displaystyle \beta=\sqrt{3}+i=2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)\)
\(\displaystyle \alpha\beta=2\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)・2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)\)
\(\displaystyle =4\sqrt{2}\left\{\cos\left(\frac{\pi}{4}+\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{4}+\frac{\pi}{6}\pi\right)\right\}\)
\(\displaystyle =4\sqrt{2}\left(\cos\frac{5}{12}\pi+i\sin\frac{5}{12}\pi\right)\)

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

\(\displaystyle =\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)z\)
よって、
原点を中心に\(\displaystyle \frac{\pi}{4}\)回転

\(\displaystyle =\left\{\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)\right\}z\)
よって、
原点を中心に\(\displaystyle -\frac{\pi}{2}\)回転

\(\displaystyle \left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)(4-2i)\)
\(\displaystyle =\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)(4-2i)\)
\(=(1+2\sqrt{3})+i(2-\sqrt{3})\)

\(\displaystyle \left(\cos\frac{2}{3}\pi+i\sin\frac{2}{3}\pi\right)(4-2i)\)
\(\displaystyle =\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)(4-2i)\)
\(=(\sqrt{3}-2)+i(2\sqrt{3}+1)\)

\(\displaystyle \left\{\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)\right\}(4-2i)\)
\(=-i(4-2i)\)
\(=-2-4i\)

\(\displaystyle \left\{\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right\}(4-2i)\)
\(\displaystyle =\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)(4-2i)\)
\(=(2-\sqrt{3})-i(1+2\sqrt{3})\)

\(\displaystyle \gamma-\alpha=\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)(\beta-\alpha)\)
\(\displaystyle \gamma-\alpha=\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)(4+2i)\)
\(\displaystyle \gamma=(2\sqrt{3}-1)+i(\sqrt{3}+2)+1+i\)
\(=(2\sqrt{3})+i(\sqrt{3}+3)\)