Find product of roots of complex equation $z^{12}=-i$

Question

Let's apply modulus on both sides: $|z^{12}| = |-i|$
It turns out, that $|z|^{12}=1 \implies |z|=1 = r$
Now, use the polar form: $z=re^{i\phi}$

The argument is: $$\phi = \frac{-\pi+2k\pi}{12}$$

Roots are in form (because $r=1$): $$z=e^{i\phi}$$ So, the product of roots is: $$e^{i\phi_{1}} \cdot e^{i\phi_{2}} \cdot \ldots \cdot e^{i\phi_{12}} = e^{i(\phi_{1}+\phi_{2}+\ldots+\phi_{12})}$$

Sum of arguments, using the formula for sum of members the arithmetic progression: $\frac{n(a_1 + a_n)}{2}$ $$\sum\limits_{k=1}^{12} \frac{-\pi+2k\pi}{12} = (\frac{\pi}{12} + \frac{23\pi}{12})\cdot\frac{12}{2} = 12\pi$$

Finally: $$e^{i\phi_{1}} \cdot e^{i\phi_{2}} \cdot \ldots \cdot e^{i\phi_{12}} = e^{i12\pi}=1$$

Is that solution correct? Also, is there faster method to solve this problem?

Edit: I made a mistake. Argument should be equal to: $\phi = \frac{- \frac{\pi}{2}+2k\pi}{12}$ Then the result is equal to $e^{\frac{\pi}{2}i}=i$

Note that the $\phi$ that you've found is a root of $z^{12}=-1$. You want $\frac{-\pi}{24}$ as the base root for $z^{12}=-i$ — Thomas Andrews, Feb 19 '14 at 03:16
@ThomasAndrews: Well, now sum of angles is equal to $\frac{\pi}{2}$, so it seems that solution is $e^{i\frac{\pi}{2}}=i$ — stil, Feb 19 '14 at 03:26

Thomas Andrews · Answer 1 · 2014-02-19T04:29:33.050

3

That can't be the right answer.

The product of the solutions (with multiplicities) of $$x^n+a_{n-1}x^{n-1}\dots + a_1x + a_0=0$$ is $(-1)^n a_0$. Since $x^{12}+i$ does not have any repeated roots, the product of the roots is...

Fundamentally your answer $\phi$ is wrong - it is $e^{-\pi/2}=-i$, so the general solution should be:

$$\theta =\frac{1}{12}\left(\frac{-\pi}{2}+2\pi k\right)=\frac{\pi}{24}\left(4k-1\right)\\\phi=e^{i\theta}$$

edited Feb 19 '14 at 04:29

answered Feb 19 '14 at 03:08

Thomas Andrews

177,126

There is a typo there. – chubakueno Feb 19 '14 at 03:15
Thx, @chubakueno – Thomas Andrews Feb 19 '14 at 03:17
$a_{0} = i$, so the product is equal to $i$... Thank you very much. I should have used Vieta's formula and not struggle with calculating sum of angles. – stil Feb 19 '14 at 03:33
Yeah, I completely left out the $e$ stuff, fixed by calling it $\theta$ and adding $\phi$. Thanks @DonAntonio – Thomas Andrews Feb 19 '14 at 04:30

score 3 · Answer 2 · answered Feb 19 '14 at 03:13

If your polynomial has roots $\alpha_1,\dots,\alpha_n$, your polynomial can take the form $$(x-\alpha_1)\dots(x-\alpha_n).$$

So when you expand that polynomial, it is clear to see that the constant term is $(-1)^n\alpha_1\dots\alpha_n$, which is the product of all the roots. So, what is your constant term in the polynomial?

Find product of roots of complex equation $z^{12}=-i$

2 Answers2