Let:$$z^3 = \cos\theta + i\sin\theta = \mathop{\mathrm{cis}}\theta$$ prove that the sum of the three roots is always zero regardless of $\theta$. I've already proved it, but in a very long proof, about 6 – 7 lines, and during it sometimes the expression ran over the page. I won't bother writing it but will describe briefly. What I did was I used De Moivre's theorem to find the three solutions for z. Then I summed the solutions together, in doing so, I applied the compound angle rule for the two solutions where the argument was $(\theta/3) + (2\pi/3)$ and $(\theta/3) + (4\pi/3)$. Parts of the expression containing $\cos(2\pi/3)$ and other likewise parts were able to be simplified, but it was still a long expression that eventually cancelled out leaving $0$ as the answer. The proof took me over 10 minutes and was very tiresome. I've looked for a long time and I don't see any quicker or simpler way to prove it. I checked the book I used for the answer and it turns out 'proof' was written as the answer. Uh. So I couldn't compare my proof to the authors. If there's anyone that can help me out, much would be appreciated. The book is called Level 3 Calculus study guide, published by ESA, written by Frances Hinchliffe and Tom Sidebotham. It's a New Zealand book written for the NZ curriculum. The question can be found at the end of page 169 if anyone happens to have the book on hand. If anyone would like me to show my proof, I would gladly do so, however, it would be very long!
Thanks!
Best Answer
By de Movire's theorem, the three roots are $\mathop{\mathrm{cis}}(\frac{\theta+2\pi n}{3})$ for $n=0,1,2$. Using the sum-to-product formulae, the sum is
\begin{align*} \sum_{n=0}^2\mathop{\mathrm{cis}}(\tfrac{\theta+2\pi n}{3})&=\cos\tfrac\theta3+\cos(\tfrac{\theta+2\pi}3)+\cos(\tfrac{\theta+4\pi}3)+i\big(\sin\tfrac\theta3 + \sin(\tfrac{\theta+2\pi}3)+\sin(\tfrac{\theta+4\pi}3)\big)\\ &= \cos(\tfrac{\theta+2\pi}3)-2\cos(\tfrac{\theta+2\pi}3)\cos(\tfrac{4\pi}{3}) + i\big(\sin(\tfrac{\theta+2\pi}3)-2\sin(\tfrac{\theta+2\pi}3)\cos(\tfrac{4\pi}{3})\big)\\ &=0, \end{align*} as required.