0
$\begingroup$

I'm asked to find all 5th roots of the complex number $z=-243$.

For this question. I got the complex roots as $3e^{\frac{i \pi}{5}}$,$3e^{\frac{3i \pi}{5}}$,$3e^{\pi}$,$3e^{\frac{7 i \pi}{5}}$,$3e^{\frac{9 i \pi}{5}}$.

However, wolfram alpha is giving me negative angles for the last two roots. Why is that? Is it wrong to use $\frac{7 \pi}{5}$ and $\frac{9 \pi}{5}$ for my angles?

Wolfram alpha says $3e^{\frac{-i \pi}{5}}$,$3e^{-\frac{3i \pi}{5}}$ as the last two values instead of $3e^{\frac{7 i \pi}{5}}$,$3e^{\frac{9 i \pi}{5}}$.

I get that the range of $\arctan(x)$ is between $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$ and I notice that subtracting $2 \pi$ from $\frac{7 \pi}{5}$ and $\frac{9 \pi}{5}$ I get the angles from wolfram alpha but I'm not sure how that works.

Aren't the angles for $\arctan(x)$ suppose to fall between $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$ or quadrant $\text{IV}$ and $\text{I}$? The angles I get aren't falling in that range should would I not subtract $\pi$ and not $2\pi$?

$\endgroup$
1
$\begingroup$

Both answers are correct, since $e^{7\pi i/5}=e^{-3\pi i/5}$ and $e^{9\pi i/5}=e^{-\pi i/5}$.

$\endgroup$
7
  • $\begingroup$ But this range of values is in quadrant 3 right? Don't the arguments have the be within quadrant $\text{IV}$ and quadrant $\text{I}$? $\endgroup$ Jan 31 at 10:41
  • $\begingroup$ Why is that? Let's see a simpler example: what are the square roots of $i$? They are$$\sqrt2e^{\pi i/4}\quad\text{and}\quad\sqrt2e^{5\pi i/4}.$$But $\frac{5\pi}4$ belongs to the third quadrant. Should I somehow reject this root because of that. Are you claiming that there are no complex numbers with an argument from the second or the third quadrant? $\endgroup$ Jan 31 at 10:46
  • $\begingroup$ Yes that's exactly what I am claiming. As far as I know the argument of a complex number $a+bi$ is $\arctan\left(\frac{b}{a}\right)$ is it not? Isn't the range of $\arctan(x)$ from $\left( -\frac{\pi}{2},\frac{\pi}{2} \right)$? So shouldn't the arguments all lie in that range only? $\endgroup$ Jan 31 at 10:53
  • $\begingroup$ Start with the arbitrary convention that the principal argument (Arg) of $z$ is in $(-\pi,\pi]$. Then, add the 2nd arbitrary convention that if $(\theta) = $ Arg$(z)$, then the principal argument of $z^{(1/n)} ~:~n\in\mathbb{Z^+}$ is $(\theta)/n.$ These two arbitrary conventions have two purposes (only) :[1] consistency with corresponding Real Analysis convention [2] Allow brief unambiguous discussion (e.g. presentation of a Complex Analysis problem). It is never intended that these conventions would imply that other values of $\theta$ are outlawed. $\endgroup$ Jan 31 at 10:58
  • $\begingroup$ For example: the Real Analysis convention is that $\sqrt{4}$ is $+2$. However, this convention is not intended to outlaw $(-2)$ as a legitimate root of $x^2 = 4.$ $\endgroup$ Jan 31 at 10:59

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.