2020年4月2日木曜日

数学 - 新しい数とその表示ー複素数と複素平面 - 複素平面 - 複素数の極形式 - 絶対値と偏角、商、差、共役複素数

1. 絶対値。

$\begin{array}{l}\left|z+\frac{1}{\stackrel{-}{z}}\right|\\ =\left|\frac{z\stackrel{-}{z}+1}{\stackrel{-}{z}}\right|\\ =\left|\frac{{\left|z\right|}^{2}+1}{\stackrel{-}{z}}\right|\\ =\frac{\left|{\left|z\right|}^{2}+1\right|}{\left|\stackrel{-}{z}\right|}\\ =\frac{{r}^{2}+1}{\left|z\right|}\\ =\frac{{r}^{2}+1}{r}\\ =r+\frac{1}{r}\end{array}$

偏角。

$\begin{array}{l}\mathrm{arg}\left(z+\frac{1}{\stackrel{-}{z}}\right)\\ =\mathrm{arg}\left({\left|z\right|}^{2}+1\right)-\mathrm{arg}\stackrel{-}{z}\\ =\mathrm{arg}\left({r}^{2}+1\right)+\mathrm{arg}z\\ =0+\theta \\ =\theta \end{array}$

2. $\begin{array}{l}r+z\\ =r+r\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)\\ =r\left(\left(1+\mathrm{cos}\theta \right)+i\mathrm{sin}\theta \right)\\ =r\left(1+\mathrm{cos}\left(\frac{\theta }{2}+\frac{\theta }{2}\right)+i\mathrm{sin}\left(\frac{\theta }{2}+\frac{\theta }{2}\right)\right)\\ =r\left(1+{\mathrm{cos}}^{2}\frac{\theta }{2}-{\mathrm{sin}}^{2}\frac{\theta }{2}+2i\mathrm{sin}\frac{\theta }{2}\mathrm{cos}\frac{\theta }{2}\right)\\ =r\left(2{\mathrm{cos}}^{2}\frac{\theta }{2}+2i\mathrm{sin}\frac{\theta }{2}\mathrm{cos}\frac{\theta }{2}\right)\\ =2r\mathrm{cos}\frac{\theta }{2}\left(\mathrm{cos}\frac{\theta }{2}+i\mathrm{sin}\frac{\theta }{2}\right)\end{array}$

よって、 問題の複素数の絶対値、偏角はそれぞれ

$\begin{array}{l}\left|r+z\right|=2r\mathrm{cos}\frac{\theta }{2}\\ \mathrm{arg}\left(r+z\right)=\frac{\theta }{2}\end{array}$

3. $\begin{array}{l}r-z\\ =r-r\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)\\ =r\left(1-\mathrm{cos}\theta -i\mathrm{sin}\theta \right)\\ =r\left(1-{\mathrm{cos}}^{2}\frac{\theta }{2}+{\mathrm{sin}}^{2}\frac{\theta }{2}-2i\mathrm{sin}\frac{\theta }{2}\mathrm{cos}\frac{\theta }{2}\right)\\ =r\left(2{\mathrm{sin}}^{2}\frac{\theta }{2}-2i\mathrm{sin}\frac{\theta }{2}\mathrm{cos}\frac{\theta }{2}\right)\\ =2r\mathrm{sin}\frac{\theta }{2}\left(\mathrm{sin}\frac{\theta }{2}-i\mathrm{cos}\frac{\theta }{2}\right)\\ =2r\mathrm{sin}\frac{\theta }{2}\left(\mathrm{cos}\left(\frac{\theta }{2}-\frac{\pi }{2}\right)+i\mathrm{sin}\left(\frac{\theta }{2}-\frac{\pi }{2}\right)\right)\\ \left|t-z\right|=2\mathrm{sin}\frac{\theta }{2}\\ \mathrm{arg}\left(r-z\right)=\frac{\theta }{2}-\frac{\pi }{2}\end{array}$

4. $\begin{array}{l}\left|\frac{r-z}{r+z}\right|\\ =\frac{2r\mathrm{sin}\frac{\theta }{2}}{2r\mathrm{cos}\frac{\theta }{2}}\\ =\mathrm{tan}\frac{\theta }{2}\\ \mathrm{arg}\left(\frac{r-z}{r+z}\right)\\ =\mathrm{arg}\left(r-z\right)-\mathrm{arg}\left(r+z\right)\\ =\frac{\theta }{2}-\frac{\pi }{2}-\frac{\theta }{2}\\ =-\frac{\pi }{2}\end{array}$