$\text{Let }z = -2-2i \text{ where }i \text{ is imaginary. Find in Modulus-Argument form the cube roots of }z$
So far I've done this $$r = \sqrt8 = 2 \sqrt2 \\ \alpha = \frac{-\pi + \frac{\pi}{4}}{3} = \frac{-\pi}{4} $$
Which then leads me to believe that the first answer in Mod-Arg form is $$ z_0 = 2\sqrt2 (\cos\frac{-\pi}{4} + i\sin\frac{-\pi}{4}) $$
Then I increases the value of K by $1$ (Using De Moivre's Theorem for finding nth roots)
$$ r = 2\sqrt2 \\ \alpha = \frac{(-\pi + \frac{\pi}{4})+2\pi(1)}{3} = \frac{5\pi}{12} \\ \text{Thus } z_1 = 2\sqrt2(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}) $$
and finally, where $K = n-1 =2$
$$ r = 2\sqrt2 \\ \alpha = \frac{(-\pi + \frac{\pi}{4})+2\pi(2)}{3} = \frac{13\pi}{12} \\ $$
But $\frac{13\pi}{12} > \pi$, so $\frac{13\pi}{12} -2\pi = \frac{-11\pi}{12}$, so $$ z_2 = 2\sqrt2(\cos\frac{-11\pi}{12} + i\sin\frac{-11\pi}{12}) $$
The actual answers have exactly the same arguments, except the modulus is only $\sqrt2$, why so? Did I miss something in my working out? Thanks in advance.
Seems like you forgot to take the cube root of $2\sqrt{2}$, after which, you get the answer.
If you look take a complex number $re^{i\theta}$ to the $n$th power, you get $r^n e^{in\theta}$.