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Given a polynomial such as:

$x^3 + 8i$

How can I solve this?

The first obvious step is to move the $8i$ over, so you get:

$x^3 = -8i$

From there, I need to create a complex number of some sort. I'm not sure how I can do this.

Vaughan Hilts
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  • Try De Moivre's formula. http://en.wikipedia.org/wiki/De_Moivre%27s_formula – Christopher Liu Dec 12 '13 at 03:34
  • Do you know the identity $e^{i \theta}=cos(\theta)+isin(\theta)$? If so, can you write $-8i$ in terms of this? You'll get an exponential term, which can easily be manipulated in terms of taking powers. – Lost Dec 12 '13 at 03:35
  • @Lost Sorry, I'm not sure I understand. Are you suggesting I write the right hand side like this? $8cis(-pi/2)$ is what I'd end up getting. Plugging this in, I obviously get $exp((-pi/2)i)$. I'm not sure how that's helping me. – Vaughan Hilts Dec 12 '13 at 03:37
  • Take for example $x^2 = i$. We want $\theta = \pi/2$ which gives us $i=e^{i \pi/2}$. Thus, we have $x^2=e^{i \pi/2}$. Since exp($i\theta$) is periodic in $2 \pi$, we have exp($i\theta$) = exp($i(\theta + 2\pi k$) where $k \in \mathbb{Z}$. Of course, taking the square root of exp divides the power by 2, so we get exp($i(\theta/2 + 2\pi k/2$). Notice that, now we must have $k=2$ for periodicity, thus $k=0$ and $k=1$ give us two solutions exp($i\pi/4$) and exp$(i(\pi/4 + \pi))$. Turn this back into cos and sin in order to evaluate. Apply this idea here. – Lost Dec 12 '13 at 03:52
  • Related : http://math.stackexchange.com/questions/192742/how-to-solve-x3-1 – lab bhattacharjee Dec 12 '13 at 15:52

1 Answers1

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You can express $-8i$ in terms of the complex exponential. If $e^{i \theta} = \cos \theta + i \sin \theta$, then

$$-8i = 8(-i) = 8[\cos (3 \pi / 2) + i \sin (3 \pi / 2)] = 8e^{(3i \pi / 2)+2 \pi n} = x^3,$$

with $n = 0, 1, 2.$ Then

$$x = 2e^{(i \pi/2)+2 \pi n/3},$$

with $n = 0, 1, 2$.

For $n = 0$, $x = 2e^{i \pi/2} = 2(\cos \pi / 2 + i \sin \pi / 2) = 2i.$

You can get the other two roots similarly for $n = 1, 2$.

John
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