# Complex roots for a 5th root

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 $$e^{9\pi i/5}=e^{-\pi i/5}$$.
• But this range of values is in quadrant 3 right? Don't the arguments have the be within quadrant $\text{IV}$ and quadrant $\text{I}$? Jan 31 at 10:41
• 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? Jan 31 at 10:46
• 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? Jan 31 at 10:53
• 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) : consistency with corresponding Real Analysis convention  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. Jan 31 at 10:58
• 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 $x^2 = 4.$ Jan 31 at 10:59