Question

sejam z1 = 16 (cos 135° + i sen 135°) z2 = 2 ( cos 45° + isen 45°) o quonciente de z1 por z2 é o número?

Answer 1

cos(135°) = -cos(45°)

sen(135°) = sen(45°)

Assim:

$Z_1=16(-cos(45^\circ)+i.sen(45^\circ))\\\\Z_1=16(-\frac{\sqrt{2}}{2}+i.\frac{\sqrt{2}}{2})\\\\Z_1=-8\sqrt{2}+i.8\sqrt{2}\\\\\\Z_2=2(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\\\\Z_2 = \sqrt{2}+i\sqrt{2}\\\\\\\frac{Z_1}{Z_2}=\frac{-8\sqrt{2}+i.8\sqrt{2}}{\sqrt{2}+i\sqrt{2}}\\\\\frac{Z_1}{Z_2}=\frac{-8\sqrt{2}+i.8\sqrt{2}}{\sqrt{2}+i\sqrt{2}}.\frac{\sqrt{2}-i\sqrt{2}}{\sqrt{2}-i\sqrt{2}} \\\\\frac{Z_1}{Z_2}=\frac{(-8\sqrt{2}+i.8\sqrt{2})(\sqrt{2}-i\sqrt{2})}{\sqrt{2}^2+\sqrt{2}^2}$

$\frac{Z_1}{Z_2}=\frac{-8*\sqrt{2}^2+i8\sqrt{2}^2+i8\sqrt{2}^2+8\sqrt{2}^2}{2+2}\\\\\frac{Z_1}{Z_2}=\frac{-16+i32+16}{4}\\ \\\frac{Z_1}{Z_2}=\frac{i32}{4}\\\\\frac{Z_1}{Z_2}=8i$