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.

  • 1
    $\begingroup$ $\sqrt[n]{x}$ always means the principal $n$th root of $x,$ so $\sqrt{9}=3$ and I disagree with the linked answer's usage of that symbol to mean all the roots of its input. $\quad$ And to summarise Golden Ratio's answer: putting a $\pm$ front of one of the roots (as in your suggested formula) is equivalent to adding a multiple of $\frac{2\pi}2=\pi$ to its argument to create the other root (as in the linked answer). $\endgroup$
    – ryang
    Mar 14, 2023 at 11:33

2 Answers 2


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.

  • $\begingroup$ Can you please clarify (and why) whether the answer is 3e^(-5πi/6) or 3e^(7πi/6) or whether they are both correct? Rajpoot writes the answer as 3e^(-5πi/6) but multiplying this with 3e^(πi/6) gives 9e^(-2πi/3) rather than the intended answer. $\endgroup$
    – sevenn0
    Mar 14, 2023 at 10:13
  • $\begingroup$ @sevenn0 They are the same since $e^{7\pi i /6}=e^{-5\pi i /6}\times e^{2\pi i}=e^{-5\pi i /6}\times 1=e^{-5\pi i /6}$ $\endgroup$ Mar 14, 2023 at 10:16
  • $\begingroup$ Thanks for the response. The answer of the original question is required in -π < θ < π, so I assume the correct answer would be to state $e^{-5\pi i /6}$ ? $\endgroup$
    – sevenn0
    Mar 14, 2023 at 10:19
  • 1
    $\begingroup$ @sevenn0 That's correct. Of course, without that convention, $e^{-5 \pi i/6 +2\pi i n}$ for any integer $n$ would work. $\endgroup$ Mar 14, 2023 at 10:22
  • 1
    $\begingroup$ Ah, thanks for clarifying that too. I am still new at the topic so that cleared it for me. $\endgroup$
    – sevenn0
    Mar 14, 2023 at 10:31

In general, assigned a complex number:

$$ z := x + \text{i}\,y $$

with $x,y \in \mathbb{R}$, calculate modulus $\rho \in [0,\infty)$ and phase $\theta \in (-\pi,\pi]$ as follows:

$$ \rho \equiv \sqrt{x^2+y^2}, \quad \quad \quad \theta \equiv \text{atan2}(y,x) $$

the nth roots of $z$ are trivially equal to:

$$ \left(\sqrt[n]{z}\right)_k = \sqrt[n]{\rho}\left(\cos\left(\frac{\theta+2\,k\,\pi}{n}\right)+\text{i}\sin\left(\frac{\theta+2\,k\,\pi}{n}\right)\right) \quad \text{with} \; k=1,2,\dots,n $$

where the signs are taken into account by the sine and cosine functions.

In this case, being:

$$ z := \frac{9\sqrt{3}+9\,\text{i}}{\sqrt{3}-\text{i}} = \frac{9\sqrt{3}+9\,\text{i}}{\sqrt{3}-\text{i}}\,\frac{\sqrt{3}+\text{i}}{\sqrt{3}+\text{i}} = \frac{9}{2} + \frac{9\sqrt{3}}{2}\,\text{i} $$

it follows that:

$$ \rho \equiv \sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{9\sqrt{3}}{2}\right)^2} = 9, \quad \quad \quad \theta \equiv \text{atan2}\left(\frac{9\sqrt{3}}{2},\frac{9}{2}\right) = \frac{\pi}{3} $$

so the 2-th roots of $z$ are:

$$ \left(\sqrt[2]{z}\right)_k = \sqrt[2]{\rho}\left(\cos\left(\frac{\theta+2\,k\,\pi}{2}\right)+\text{i}\sin\left(\frac{\theta+2\,k\,\pi}{2}\right)\right) \quad \text{with} \; k=1,2 $$


$$ \left(\sqrt[2]{z}\right)_1 = -\frac{3\sqrt{3}}{2} - \frac{3}{2}\,\text{i}\,, \quad \quad \quad \left(\sqrt[2]{z}\right)_2 = \frac{3\sqrt{3}}{2} + \frac{3}{2}\,\text{i}\,. $$

  • $\begingroup$ Multiplying the final two answers gives a negative result (while the original equation is positive). Could you please clarify why (or if it is wrong)? $\endgroup$
    – sevenn0
    Mar 14, 2023 at 10:16

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