Write $z=5-5i$ in the polar form with $z=Arg(z)$
My work
Let $z=5-5i\in \mathbb{C}$ then
$r=\sqrt{5^2+5^2}=5\sqrt{2}$
$\theta=tan^{-1}(\frac{-5}{5})=\frac{-\pi}{4}$
By definition, $-\pi<Arg(z)\leq \pi$ then
I have problem trying to find that angle. Can someone help me?
$\endgroup$ 31 Answer
$\begingroup$Actually, you already finished your work. You found the the module is $5\sqrt{2}$, the argument is $-\frac{\pi}{4}$, so the number is $z=5\sqrt{2}e^{-i\frac{\pi}{4}}=5\sqrt{2}(\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}))=5\sqrt{2}(\cos(\frac{\pi}{4})-i\sin(\frac{\pi}{4}))$.
$\endgroup$ 5
