how do I calculate the phi (/theta/angle/degree/no idea how it translates correctly) when calculating the roots of a complex number? I have seen the "formula" (principal argument for complex numbers) for it, but I cannot figure out how to give a constantly correct answer to the angle phi (or in some theta).

My book shows that it is the absolute value of the complex number z = a+bi, so sqrt(a^2+b^2).

Let's say I have $$ x=\sqrt[\Large 3]{8*\sqrt 3^3e^{\Large_{i\pi}}} $$

that would mean that by the above explained calculation, I would come up with something weird like $\sqrt 7$ (after calculating the 3rd root away and converting the e^pi*i into the factor -1), but this doesn't bring me any closer. The solution says that phi ends up being exactly pi. But no idea how or why. Sure, I know about the quadrants and have seem some pictures, explaining how it goes around in fractions of pi - but no idea how to consistently use all these things.

  • 1
    $\begingroup$ Well there are 3 solutions; not one.. And they come of dividing the angle $\pi $ plus any multiples of $2\pi $ by $3$. So the angles are $\frac \pi 3$ and $\frac \pi 3+\frac {2\pi}3=\pi $ and $\frac \pi 3+\frac {4\pi}3=\frac {5\pi}3$ $\endgroup$ – fleablood Feb 16 at 6:53
  • $\begingroup$ So $x=2\sqrt 3e^{i\frac \pi 3},-2\sqrt 3, -2\sqrt 3e^{i\frac \pi 3} $. $\endgroup$ – fleablood Feb 16 at 6:56
  • $\begingroup$ Thanks for the answer but I am looking for what you mentioned ultimately is the Addition of π. Why is it π and how do i calculate this without a calculator? $\endgroup$ – user1994928 Feb 17 at 7:20
  • $\begingroup$ $2\pi =360^{\circ}$ and $e^{i\theta} = e^{i(\theta + 2k\pi)}$ so the $n$ square roots of $re^{i\theta}$ are $\sqrt[n]{re^{i\theta}} = \sqrt[n]{re^{i(\theta + 2k\pi)}}=\sqrt[n]{r} e^{\frac{i(\theta + 2k\pi)}k}=\sqrt[n]{r}e^{i\frac {\theta}n + i\sqrt{2k\pi}n}$. And you don't need to calculate it. You can leave it as is. But if you wont $e^{i\theta}=\cos\theta+i\sin\theta$ so, $2\sqrt{3}e^{i\frac \pi 3} =2\sqrt{3}(\cos 60^\circ + i\sin 60^\circ)=2\sqrt{3}(\frac 12 + i\frac {\sqrt 3}2)=\sqrt{3}+ i$. $\endgroup$ – fleablood Feb 17 at 17:07

"The solution says that phi ends up being exactly pi."

The solution is wrong. There are $3$ third roots so there different values of $\phi$.

$x=\sqrt[\Large 3]{8*\sqrt 3^3e^{\Large_{i\pi}}}=\sqrt[3]{-8\sqrt{3}^3}= -2\sqrt{3}$

And $-2\sqrt{3} = 2\sqrt{3}e^{i\pi}$ so you knew one of the vales of $\phi$ what are the other two? (And why the heck does $e^{i\pi} = -1$ anyway?)

Capsule everything.

We can think of a complex number $a+bi$ as a point $(a,b)$ in a plane. And if that isn't $0=0+0i = (0,0)$ we can think of the point in the plane as having a length $r = \sqrt{a^2 + b^2}$ and an angle from the $x$ axis $\phi$.

Seen this way $a + bi = r\cos \phi + i r\sin \phi$. If we define $e^{i\phi}$ as $\cos \phi + i\sin \phi$ then we can define any non-zero $a + bi$ as $re^{i\phi} = r(\cos \phi + i \sin \phi)$ where $r = \sqrt{a^2 + b^2}$ and $a = \cos \phi$ and $b = \sin \phi$.

We do we define $e^{\sqrt{-1}\cdot angle}$ like that? Reasons.....

Now here's a neat thing and you can prove it with trig or by geometry.

If $z =a + bi$ and $w = c+di$ then $zw=(a+bi)(c+di)=(ac-bd) + (bc+ad)i$. but if $z=re^{i\phi}=r\cos \phi+ri\sin\phi$ and $w=se^{i\theta}=s\cos \theta+si\sin\theta$ then $zw= (rs\cos\phi\cos\theta -rs\sin\phi\cos\theta)+(rs\sin\phi\cos\theta +rs\cos\phi\sin\theta)i = rs(\cos(\phi + \theta) + i(\sin(\phi+\theta))$

So we have this law of multiplication of complex numbers:

$re^{i\phi} = se^{i\theta} = rse^{i(\phi + \theta)}$

And therefore $(r^{i\phi})^n =r^ne^{in\phi}$. And if we want to find the $n$ th root of $re^{i\phi}$ we find it is $\sqrt[n]{r} e^{i\frac \phi n}$.

But that is just ONE $n-th$ root.

If we notice:

$r(\cos \phi + i\sin phi) = $

$r(\cos (\phi + 2\pi) + i\sin(\phi + 2pi) =$

$r(\cos (\phi + 4\pi) + i\sin(\phi + 4pi) =$


It seems like all of the following will also be $n$th roots:

$\sqrt[n]re^{i\frac {\phi}n},\sqrt[n]re^{i\frac {\phi+2\pi}n},\sqrt[n]re^{i\frac {\phi+4\pi}n},.....$

One thing we note about this infinite list is that in will end and start repeating itself when we got to $\sqrt[n]re^{i\frac {\phi+2n\pi}n}=\sqrt[n]re^{i(\frac {\phi}n + 2\pi)}=\sqrt[n]re^{i\frac {\phi}n}$

So we conclude there are $n$, $n$ths roots and they are

$\sqrt[n]re^{i\frac {\phi}n},\sqrt[n]re^{i\frac {\phi+2\pi}n},\sqrt[n]re^{i\frac {\phi+4\pi}n},....., \sqrt[n]re^{i\frac{\phi + 2(n-1)\pi}n}$


So to answer what are the three cube roots of $-8\sqrt3$

we have $-8{\sqrt3}^3 = 8{\sqrt3}^3 e^{i\pi} = 8{\sqrt3}^3e^{3i\pi} = 8{\sqrt3}^3e^{5\pi }$

So $\sqrt[3]{-8{\sqrt3}^3}$ are

$2\sqrt 3e^{i\frac \pi 3}$ and $2\sqrt 3e^{i \pi } = -2\sqrt 3$ and $2\sqrt 3e^{i\frac {5\pi}3} = 2\sqrt 3e^{-i\frac {\pi}3}$


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