1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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). $\endgroup$ – copper.hat Oct 21 '15 at 16:35
  • $\begingroup$ So for the equality to hold, $\frac{1}{2} (2\theta) mod 2\pi$ should coincide with $\theta$? $\endgroup$ – getafix Oct 21 '15 at 16:47
  • 1
    $\begingroup$ Well, you need to be careful with parentheses, but basically yes. Its all about the angle. $\endgroup$ – copper.hat Oct 21 '15 at 16:48
  • $\begingroup$ Which is when $\theta$ lies between $0$ and $\pi$? thank you. i got it. $\endgroup$ – getafix Oct 21 '15 at 16:50
  • 1
    $\begingroup$ 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 $0 \le \theta < \pi$. $\endgroup$ – copper.hat Oct 21 '15 at 16:53
1
$\begingroup$

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 $0\le \arg (z) <\pi$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.