# Principal square root of complex numbers

Let $\sqrt.:= r^{1/2}[cos(\theta/2)+isin(\theta/2)], 0 \leq \theta < 2\pi$ define the the particular square root of a complex number.

For what values of z does the equation $\sqrt{z^2} = z$ hold?

I am really sorry, but this question has me stumped and I have no idea how to proceed, hence I couldn't show any working. If someone could please give me a hint.

• The catch is that $z^2 = r^2(\cos (2 \theta) + i \sin (2 \theta)$, but ${1 \over 2} (2 \theta \mod 2 \pi)$ is not necessarily $\theta$ (excuse awful notation). – copper.hat Oct 21 '15 at 16:35
• So for the equality to hold, $\frac{1}{2} (2\theta) mod 2\pi$ should coincide with $\theta$? – getafix Oct 21 '15 at 16:47
• Well, you need to be careful with parentheses, but basically yes. Its all about the angle. – copper.hat Oct 21 '15 at 16:48
• Which is when $\theta$ lies between $0$ and $\pi$? thank you. i got it. – getafix Oct 21 '15 at 16:50
• Again, you need to be a little more precise. You need (using your range of $\theta$ above) $0 \le 2 \theta < 2 \pi$, which gives $0 \le \theta < \pi$. – copper.hat Oct 21 '15 at 16:53

## 1 Answer

The problem arises when $\theta>\pi$. Let $\theta =\pi + \delta$ where $0<\delta <\pi$. Then, we have

$$z^2=r^2e^{i2\delta}$$

on the branch for which arguments are restricted between $0$ and $2\pi$. Then, the square root of $z^2$ is

$$\sqrt{z^2}=re^{i\delta}=re^{i(\theta -\pi)}=-re^{i\theta}\ne z=re^{i\theta}$$

Therefore, the relationship $\sqrt{z^2}=z$ is valid only for $0\le \arg (z) <\pi$.

• Ah thank you! I figured it out with @copper.hat's hint. – getafix Oct 21 '15 at 16:52
• That's good news! Pleased to hear it. – Mark Viola Oct 21 '15 at 16:56
• Thank you so much for the best vote! That means a lot to me! - Mark – Mark Viola Oct 21 '15 at 23:05