Complex number and polar coordinates True or false: the polar coordinates of $-1-i$ are $-\sqrt{2}\operatorname{cis}\frac{\pi}{4}$
In my opinion it's true:
$\tan\theta=\frac{-1}{-1}=1\Rightarrow \theta=\frac{5\pi}{4}$, $r=\sqrt{(-1)^2+(-1)^2}=\sqrt{2}$
Therefore we get: $z=\sqrt{2}\operatorname{cis}\frac{5\pi}{4}=\sqrt{2}(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4})=\sqrt{2}(-\cos\frac{\pi}{4}-i\sin\frac{\pi}{4})=-\sqrt{2}\operatorname{cis}\frac{\pi}{4}$
that means it's true. But it was written in the algebra book that because of $-2$, it's not polar coordinate. Is the proof right?
 A: 
Notice:
$$z=\Re[z]+\Im[z]i=|z|e^{\varphi i}=|z|\cos(\varphi)+|z|\sin(\varphi)i$$
Where:

*

*$\Re[z]$ is the real part of the number $z$

*$\Im[z]$ is the imaginary part of the number $z$

*$|z|$ is the abolsute value of $z$ and defined as $|z|:=\sqrt{\Re^2[z]+\Im^2[z]}$

*$\varphi$ is the argument of $z$ and defined as $\varphi:=\arg[z]$

So, when we have the number $s=-1-i$, we get that:
$$s=-1-i=|-1-i|e^{\arg[-1-i]i}=|-1-i|\cos(\arg[-1-i])+|-1-i|\sin(\arg[-1-i])i$$
Now, when we solve the abolsute value, we get that:
$$|s|=|-1-i|=\sqrt{\Re^2[-1-i]+\Im^2[-1-i]}=\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}$$
Now, when we solve the argument, we get that:
$$\varphi=\arg[-1-i]=\pi+\arctan\left(\frac{\Im[-1-i]}{\Re[-1-i]}\right)=\pi+\arctan\left(\frac{-1}{-1}\right)=\pi+\arctan(1)=\frac{5\pi}{4}$$
So, when $\color{red}{k\in\mathbb{Z}}$
(this 'changes' the argument, not the absolute value because that has to be $|z|\in\mathbb{R}^+_0$):
$$-1-i=\sqrt{2}e^{\left(2\pi k+\frac{5\pi}{4}\right)i}=\sqrt{2}\cos\left(2\pi k+\frac{5\pi}{4}\right)+\sqrt{2}\sin\left(2\pi k+\frac{5\pi}{4}\right)i$$
A: No, it's false. The $r$ part must be positive and is computed by
$$
r=\sqrt{(-1-i)(-1+i)}=\sqrt{1+1}=\sqrt{2}
$$
This already answers the true/false question. If you want to find the $\theta$ part, you need to find $\theta$ such that
$$
\cos\theta=-\frac{1}{\sqrt{2}},\quad\sin\theta=-\frac{1}{\sqrt{2}}
$$
and this is clearly
$$
\theta=\pi+\frac{\pi}{4}=\frac{5\pi}{4}
$$
