Question

dau 5 stele
va rooooooog!

Answer 1

Forma trigonometrica: $re^{i\theta}$.

$\theta = \tan^{-1}(\frac{b}{a}) \ \ \ \ \ r = \sqrt{a^2 + b^2}$

$z_{1} = (1+i\sqrt{3}) \\a = 1 \ \ b = \sqrt{3} => \theta = \tan^{-1}(\frac{\sqrt{3}}{1}) = \frac{\pi}{3}\\ r = \sqrt{1^2 + \sqrt{3}^2} =\sqrt{4} = 2 => z = 2e^{i\frac{\pi}{3}}$

$z_{2} = \sqrt{3} + i \ \ \ \theta = \frac{\pi}{6} \ \ \ \ r = 2 => z_{2} = 2e^{i\frac{\pi}{6}}$

$z_{1} \cdot z_{2} = 2e^{i\frac{\pi}{3}} \cdot 2e^{i\frac{\pi}{6}} = 4e^{\frac{5\pi}{6}} = 4(-\frac{\sqrt{3}}{2} + \frac{1}{2}i) = -2\sqrt{3} + 2i$

$z_{3} = 2 + 2i \ \ \ \theta = \tan^{-1}(\frac{2}{2}) = \frac{\pi}{4}\\r = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} => z_{3} = 2\sqrt{2}e^{i\frac{\pi}{4}}$

$z_{4} = 1 - i\sqrt{3} \ \ \ \ \theta = \tan^{-1}(-\frac{1}{\sqrt{3}}) = \frac{11\pi}{6} \ \ \ r = \sqrt{1^2 + \sqrt{3}^2} = \sqrt{4} = 2 => z_{4} = 2e^{i\frac{11\pi}{6}}$

$z_{3} \cdot z_{4} = 2\sqrt{2}e^{i\frac{\pi}{4}} \cdot 2e^{i\frac{11\pi}{6}} = 4\sqrt{2}e^{i\frac{\pi}{12}}$