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

  • 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 $e^{\frac{\pi}{2}i}=i$

  • 2
    $\begingroup$ 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 $z^{12}=-i$ $\endgroup$ Feb 19 '14 at 3:16
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Vieta's_formulas $\endgroup$ Feb 19 '14 at 3:23
  • $\begingroup$ @ThomasAndrews: Well, now sum of angles is equal to $\frac{\pi}{2}$, so it seems that solution is $e^{i\frac{\pi}{2}}=i$ $\endgroup$
    – stil
    Feb 19 '14 at 3:26

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?


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:

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

  • $\begingroup$ There is a typo there. $\endgroup$
    – chubakueno
    Feb 19 '14 at 3:15
  • $\begingroup$ Thx, @chubakueno $\endgroup$ Feb 19 '14 at 3:17
  • $\begingroup$ $a_{0} = i$, 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. $\endgroup$
    – stil
    Feb 19 '14 at 3:33
  • $\begingroup$ Yeah, I completely left out the $e$ stuff, fixed by calling it $\theta$ and adding $\phi$. Thanks @DonAntonio $\endgroup$ Feb 19 '14 at 4:30

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