# Finding the cube root of a complex number $z$

$\text{Let }z = -2-2i \text{ where }i \text{ is imaginary. Find in Modulus-Argument form the cube roots of }z$

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.

• You need to take the cube root of the radius as well. $\sqrt[3]{2\sqrt{2}}=\sqrt{2}$.
– mjh
Sep 24, 2014 at 5:34
• Thank you! I didn't even notice that, thanks again! @mjh Sep 24, 2014 at 5:45

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 $r^n e^{in\theta}$.