Quick Question regarding De Moivre's Theorem

Question

If $z^2 = 2 - 2i$ find z using the theorem of De Moivre

For this question, i first expressed it in polar form which is $$2\sqrt{2}\left(\cos{\frac{7\pi}4} + i\sin{\frac{7\pi}4}\right)$$

Now because their is 2 roots I would just do $\frac{(315 + 360K)}2$ which is $157.5 + 180K$ now subbing that into the formula we would get (assuming $k = 0$ and $1$ respectively)

$$\frac{2^3}4\left(\cos{\frac{7\pi}8} + i\sin{\frac{7\pi}8}\right)$$ $$\frac{2^3}4\left(\cos{\frac{15\pi}8} + i\sin{\frac{15\pi}8}\right)$$

But the real answer is $\frac{2^3}4 \left(\cos{\frac{7\pi}8} + i\sin{\frac{7\pi}8}\right)$ or $-\frac{2^3}4 \left(\cos{\frac{7\pi}8} + i\sin{\frac{7\pi}8}\right)$

Am think I'm either missing something or doing something wrong here. Any help would be greatly appreciated.

Answer 1

Your answer is equivalent. Note $$\frac{15\pi}{8} = \pi + \frac{7\pi}{8},$$ so $$\cos \frac{15\pi}{8} = -\cos \frac{7\pi}{8},$$ and $$\sin \frac{15\pi}{8} = -\sin \frac{7\pi}{8},$$ as a result of the trigonometric identities $$\cos(\theta+\pi) = -\cos\theta, \quad \sin(\theta+\pi) = -\sin \theta.$$

Answer 2

Hint: it may help to draw on a picture where $\cos \frac{15\pi}{8} + i \sin \frac{15\pi}{8}$ and $-\big(\cos \frac{7\pi}{8} + i \sin \frac{7\pi}{8}\big)$ are.