Negative modulus in complex numbers?

Question

This is an introducory task from an exam.

If $z = -2(\cos{5} - i\sin{5})$, then what are:

$Re(z), Im(z), arg(z)$ and $ |z|$?

First of all, how is it possible that the modulus is negative $|z|=-2$? Or is the modulus actually $|z|= 2$ and the minus is kind of in front of everything, and that's why the sign inside of the brackets is changed as well? That would make some sense.

I assume $arg(z) = 5$. How do I calculate $Re(z) $ and $Im(z)$? Something like this should do the job?

$$arg(z) = \frac{Re(z)}{|z|}$$

$$5 = \frac{Re(z)}{2}$$

$$10 = Re(z)$$

And analogically with $Im(z):$

$$arg(z) = \frac{Im(z)}{|z|}$$

$$5 = \frac{Im(z)}{2} \Rightarrow Im(z) = Re(z) = 10$$

I'm sure I'm confusing something here because, probably somewhere wrong $\pm$ signs.

Help's appreciated.

And finally: is there some good calculator for complex numbers? Let's say I have a polar form and I want to find out the $Re(z), Im(z)$ and such. Wolframalpha seems like doesn't work fine for that.

Answer 1

Currently, the number is not in polar form, as it should be in the form $r(\cos(\theta) + i \sin(\theta))$, where $r \ge 0$. Note the $+$ sign, and the non-negative number $r$ out the front. Every complex number, including the one given, has a polar form (in fact, infinitely many), and from this you can read off the modulus and argument. But, since this is not in polar form, you need to do some extra work.

First, try absorbing the minus sign into the brackets:

$$2(-\cos 5 + i \sin 5).$$ Then, recalling that $\sin(\pi - x) = \sin(x)$ and $\cos(\pi - x) = -\cos(x)$, we get $$2(\cos(\pi - 5) + i \sin(\pi - 5)).$$ This is now in polar form. The modulus is $2$, and one of the infinitely many arguments is $\pi - 5$.

Answer 2

If $z=a+bi$, with $a,b\in\mathbb R$, then $\lvert z\rvert=\sqrt{a^2+b^2}$; in particular, $\lvert z\rvert\geqslant0$ for any $z\in\mathbb C$.

Actually, $\bigl\lvert-2(\cos5-i\sin5)\bigr\rvert=2$, $\operatorname{Re}\bigl(-2(\cos5-i\sin5)\bigr)=-2\cos5$, and $\operatorname{Im}\bigl(-2(\cos5-i\sin5)\bigr)=2\sin5$.