I am confused on dealing with the positive or negative sign when finding the two square roots of a complex number. For example I am solving a similar question to the one here: link
In the answer it is written: $$z^{1/2}=\sqrt{9\left(\cos\frac{\pi}{3}+i\sin \frac{\pi}{3}\right)}$$
Then: $$=\color{red}3\left(\cos\left(2k\pi+\frac{\pi}{3}\right)+i\sin \left(2k\pi+\frac{\pi}{3}\right)\right)^{1/2}$$
Wouldn't $\sqrt{9}$ result in $\pm 3$? Why did he take the positive root only?
I understand that converting a number into polar form and using De Moivre's formula to find the square root would yield this general formula: $$(r(\cos(\theta)+ i \sin(\theta)))^{1/2} = ±\sqrt{r}(\cos(\theta/2) + i \sin(\theta/2))$$
And that is my source of confusion, I can't apperantly see how to deal with the $\pm$ sign. Considering that in the linked answer it was not there from the first place. I would love to be corrected on what I am misunderstanding.
Best Answer
If you read the full answer, it says the square root is
$$3\left(\cos\left(\frac{6k\pi+\pi}{6}\right)+i\sin \left(\frac{6k\pi+\pi}{6}\right)\right)$$ where $k=0, 1.$
The $k=0$ solution is
$$3\left(\cos\left(\frac{\pi}{6}\right)+i\sin \left(\frac{\pi}{6}\right)\right)$$ and the $k=1$ solution is $$3\left(\cos\left(\frac{7\pi}{6}\right)+i\sin \left(\frac{7\pi}{6}\right)\right).$$
Note that one solution is just the negative of the other.