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I'm wondering if someone could help me out.

I am asked to solve the equation: $z^6 =−1$ in part (a) of a question.

I have done this and so I now have a set of solutions: $z_0,z_1,z_2,z_3,z_4.$

I'm lost in part (b):

Let $z_0, z_1, z_2, z_3, z_4, z_5$ be the solutions that you found in part (a). Use the factorization

$z_6 +1 = (z−z_0)(z−z_1)(z−z_2)(z−z_3)(z−z_4)$

to determine the complex number that is obtained by multiplying together all the solutions of the equation $z^6 =−1$.

What I don't understand is why do I have to use the factorisation...? Can't I just multiple the separate solutions???

Rio Alvarado
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Matty
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  • You should have gotten $6$ solutions, $z_0,z_1,z_2,z_3,z_4,z_5$, and the factorization should be $z^6+1 = (z-z_0)(z-z_1)(z-z_2)(z-z_3)(z-z_4)(z-z_5)$. – JimmyK4542 Aug 10 '15 at 19:35
  • You are indeed correct, forgot to write it all out.. – Matty Aug 10 '15 at 21:23

2 Answers2

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You can multiply all of the $6$ complex solutions you got, but that might be tedious.

Instead, since $z^6+1 = (z-z_0)(z-z_1)(z-z_2)(z-z_3)(z-z_4)(z-z_5)$, we can plug in $z = 0$ to get $0^6+1 = (0-z_0)(0-z_1)(0-z_2)(0-z_3)(0-z_4)(0-z_5)$ $= (-1)^6z_0z_1z_2z_3z_4z_5$.

This gives you the answer much faster than multiplying out $6$ complex numbers.

JimmyK4542
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  • Oh I see, for some reason I was thinking about just multiplying the solutions together.. thank you for your help – Matty Aug 10 '15 at 21:25
  • Just a quick question on this, on the last part why is (−1)^6(z0)(z1)(z2)(z3)(z4)(z5) . This question has really given me a head ache!!! – Matty Aug 11 '15 at 16:30
  • $(0-z_0)(0-z_1)\cdots(0-z_5) = (-z_0)(-z_1)\cdots(-z_5)$ now pull out one $-1$ factor from each of the $6$ terms to get $(-1)^6z_0z_1\cdots z_5$. – JimmyK4542 Aug 11 '15 at 16:41
  • Penny has dropped.... thank you jimmy, I have been on this for 3 days trying to solve it my self! Can't believe how simple this was.... – Matty Aug 11 '15 at 16:54
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$$z^6=-1\Longleftrightarrow$$ $$z^6=e^{\left(\pi+2\pi k\right)i}\Longleftrightarrow$$ $$z=e^{\frac{1}{6}\left(\pi+2\pi k\right)i}\Longleftrightarrow$$ $$z=e^{\left(\frac{\pi}{6}+\frac{\pi k}{3}\right)i}$$

with $k \in \mathbb{Z}$ and $k$ goes from $0-5$

So the $6$ solutions are:

$$z_0=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 0}{3}\right)i}=e^{\frac{\pi}{6}i}$$ $$z_1=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 1}{3}\right)i}=i$$ $$z_2=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 2}{3}\right)i}=e^{\frac{5\pi}{6}i}$$ $$z_3=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 3}{3}\right)i}=e^{-\frac{5\pi}{6}i}$$ $$z_4=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 4}{3}\right)i}=-i$$ $$z_5=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 5}{3}\right)i}=e^{-\frac{\pi}{6}i}$$

Jan Eerland
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