a) Преобразуем z2 в алгебраическую форму:
б) Преобразуем z1 в тригонометрическую форму:
Делаем деление и получаем ответ:
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Answers & Comments
a) Преобразуем z2 в алгебраическую форму:![z_2 = 6e^{2\pi i} = 6(\cos 2\pi + i\sin 2\pi) = 6(1+i\cdot 0) = 6+0i = 6\\ ></p><p>Делаем деление и получаем ответ: <img src=](https://tex.z-dn.net/?f=z_2%20%3D%206e%5E%7B2%5Cpi%20i%7D%20%3D%206%28%5Ccos%202%5Cpi%20%2B%20i%5Csin%202%5Cpi%29%20%3D%206%281%2Bi%5Ccdot%200%29%20%3D%206%2B0i%20%3D%206%5C%5C)
б) Преобразуем z1 в тригонометрическую форму:
Делаем деление и получаем ответ:![\frac{z_1}{z_2} = \frac{|z_1|}{|z_2|} (\cos (\varphi_1-\varphi_2) + i\sin (\varphi_1-\varphi_2)) = \frac{2}{\sqrt{2}} (\cos (-\frac{\pi}{3}+\frac{\pi}{12})+i\sin (\frac{\pi}{3}+\frac{\pi}{12})) = \sqrt 2 (\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4})) \frac{z_1}{z_2} = \frac{|z_1|}{|z_2|} (\cos (\varphi_1-\varphi_2) + i\sin (\varphi_1-\varphi_2)) = \frac{2}{\sqrt{2}} (\cos (-\frac{\pi}{3}+\frac{\pi}{12})+i\sin (\frac{\pi}{3}+\frac{\pi}{12})) = \sqrt 2 (\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}))](https://tex.z-dn.net/?f=%5Cfrac%7Bz_1%7D%7Bz_2%7D%20%3D%20%5Cfrac%7B%7Cz_1%7C%7D%7B%7Cz_2%7C%7D%20%28%5Ccos%20%28%5Cvarphi_1-%5Cvarphi_2%29%20%2B%20i%5Csin%20%28%5Cvarphi_1-%5Cvarphi_2%29%29%20%3D%20%5Cfrac%7B2%7D%7B%5Csqrt%7B2%7D%7D%20%28%5Ccos%20%28-%5Cfrac%7B%5Cpi%7D%7B3%7D%2B%5Cfrac%7B%5Cpi%7D%7B12%7D%29%2Bi%5Csin%20%28%5Cfrac%7B%5Cpi%7D%7B3%7D%2B%5Cfrac%7B%5Cpi%7D%7B12%7D%29%29%20%3D%20%5Csqrt%202%20%28%5Ccos%28-%5Cfrac%7B%5Cpi%7D%7B4%7D%29%2Bi%5Csin%28-%5Cfrac%7B%5Cpi%7D%7B4%7D%29%29)