Question

Тема: Комплексные числа в тригонометрической форме.
Найти модуль комплексного числа:
z=-2+3i

Answer 1

$z=Re+i*Im\\\\ r=|z|=\sqrt{(Re)^2+(Im)^2}\\\\ cos(\phi)=\frac{Re}{r}=\frac{Re}{\sqrt{(Re)^2+(Im)^2}}\\\\ sin(\phi)=\frac{Im}{r}=\frac{Im}{\sqrt{(Re)^2+(Im)^2}}\\\\ z=r*[cos(\phi)+i*sin(\phi)]$

$z=-2+{3}*i\\\\ Re=-2\ \ \ Im={3}\\\\ r=|z|=\sqrt{(Re)^2+(Im)^2}=\sqrt{(-2)^2+({3})^2}=\sqrt{13}\\\\ cos(\phi)=\frac{Re}{r}=\frac{-2}{\sqrt{13}}=-\frac{2}{13}\\\\ sin(\phi)=\frac{Im}{r}=\frac{{3}}{\sqrt{13}}=\frac{3}{\sqrt{13}}\\\\ \phi=arctg(\frac{Im}{Re})=arctg(-\frac{3}{2})=\pi-arctg(\frac{3}{2})\\\\ z=r*[cos(\phi)+i*sin(\phi)]\\\\ z=\sqrt{13}*[cos(\pi-arctg(\frac{3}{2}))+i*sin(\pi-arctg(\frac{3}{2}))]$

$z=|z|*e^{i\phi}=\sqrt{13}*e^{i*[\pi-arctg(\frac{3}{2})]}$

Ответ: $\sqrt{13}*[cos(\pi-arctg(\frac{3}{2}))+i*sin(\pi-arctg(\frac{3}{2}))]=\sqrt{13}*e^{i*[\pi-arctg(\frac{3}{2})]}$