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 $2\pi$?


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

  • $\begingroup$ 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

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