First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.5

Solution

Given $z=-2-2i$ .

|z|=\sqrt{(-2)^2+(-2)^2}=\sqrt{8}=2\sqrt{2}

\tan\theta=\frac{y}{x}=\frac{-2}{-2}=1\Rightarrow \theta=45^\circ

Since $x<0$ and $y<0$ , $z$ lies in Quadrant III, so:

\theta=45^\circ+180^\circ=225^\circ=\frac{5\pi}{4}

\boxed{|z|=2\sqrt{2},\quad \operatorname{Arg}(z)=\frac{5\pi}{4}\ (\text{or }225^\circ)}