I am trying to solve $z^6 = 1$ where $z\in\mathbb{C}$. So What I have so far is : $$z^6 = 1 \rightarrow r^6\operatorname{cis}(6\theta) = 1\operatorname{cis}(0 + \pi k)$$ $$r = 1,\ \theta = \frac{\pi k}{6}$$ $$k=0: z=\operatorname{cis}(0)=1$$ $$k=1: z=\operatorname{cis}\left(\frac{\pi}{6}\right)=\frac{\sqrt{3} + i}{2}$$ $$k=2: z=\operatorname{cis}\left(\frac{\pi}{3}\right)=\frac{1 + \sqrt{3}i}{2}$$ $$k=3: z=\operatorname{cis}\left(\frac{\pi}{2}\right)=i$$ According to my book I have a mistake since non of the roots starts with $\frac{\sqrt{3}}{2}$, also even if I continue to $k=6$ I get different (new) results, but I thought that there should be (by the fundamental theorem) only 6 roots. Can anyone please tell me where my mistake is? Thanks!
|
|
|||||||||||
|
|
You have to write your equation into the form $$z^6=e^{2in\pi}=1$$ then $$z=e^{in\frac{\pi}{3}}$$ and from this you can read off all the solutions. |
|||||||||||
|
|
It may be of interest to point out that this can also be solved using basic high school algebra factorization methods (i.e. no trig is needed). Begin by factoring as a difference of squares. Then factor the sum and difference of cubes that arise. Finally, put each of the resulting factors equal to $0$ and solve. $$x^6 - 1 \; = \; (x^3 - 1)(x^3 + 1) \; = \; (x-1)(x^2 + x + 1)(x+1)(x^2 - x + 1)$$ Now solve the following equations. Use the quadratic formula for two of them. $$x - 1 = 0$$ $$x^2 + x + 1 = 0$$ $$x+1=0$$ $$x^2 - x + 1 = 0$$ |
|||
|
|
|
You need $2\pi$ where you had $\pi$. Thus for $k=1$ you should get $$ \cos\left(\frac{2\pi\cdot1}{6}\right) + i \sin\left(\frac{2\pi\cdot1}{6}\right). $$ |
|||
|
|
