Найти все значения корня 4-ой степени из z=cos(pi/2)+isin(pi/4)
Нету ли описки в угле синуса или косинуса? ----------------------------------------------------------------
извините, sin(pi/2) тоже
![z=e^{i(\pi/2+2\pi k)} \\
\sqrt[4]{z}=e^{\frac{1}{4}i(\pi/2+2\pi k)}=e^{i(\frac{\pi}{8}+\frac{\pi k}{2})}=\cos(\frac{\pi}{8}+\frac{\pi}{2}k)+i\sin(\frac{\pi}{8}+\frac{\pi}{2}k), k=0,...,3](https://tex.z-dn.net/?f=z%3De%5E%7Bi%28%5Cpi%2F2%2B2%5Cpi+k%29%7D+%5C%5C+%0A+%5Csqrt%5B4%5D%7Bz%7D%3De%5E%7B%5Cfrac%7B1%7D%7B4%7Di%28%5Cpi%2F2%2B2%5Cpi+k%29%7D%3De%5E%7Bi%28%5Cfrac%7B%5Cpi%7D%7B8%7D%2B%5Cfrac%7B%5Cpi+k%7D%7B2%7D%29%7D%3D%5Ccos%28%5Cfrac%7B%5Cpi%7D%7B8%7D%2B%5Cfrac%7B%5Cpi%7D%7B2%7Dk%29%2Bi%5Csin%28%5Cfrac%7B%5Cpi%7D%7B8%7D%2B%5Cfrac%7B%5Cpi%7D%7B2%7Dk%29%2C+k%3D0%2C...%2C3 "z=e^{i(\pi/2+2\pi k)} \\
\sqrt[4]{z}=e^{\frac{1}{4}i(\pi/2+2\pi k)}=e^{i(\frac{\pi}{8}+\frac{\pi k}{2})}=\cos(\frac{\pi}{8}+\frac{\pi}{2}k)+i\sin(\frac{\pi}{8}+\frac{\pi}{2}k), k=0,...,3")