Finding the n-th root of a complex number 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!
 A: 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.
A: 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.
$$x - 1 = 0$$
$$x^2 + x + 1 = 0$$
$$x+1=0$$
$$x^2 - x + 1 = 0$$
A: You need $2\pi$ where you had $\pi$.  Thus for $k=1$ you should get
$$
\cos\left(\frac{2\pi\cdot1}{6}\right) + i \sin\left(\frac{2\pi\cdot1}{6}\right).
$$
A: I will solve $z^6=1$ where $z∈\mathbb{C}$. First I will take off all data: $r=1$, $\theta = 0$, $n=6$ and $k=0,1,2,3,4,5$ then
For $k=0: w_1=\sqrt[6]{1}[\cos{\frac{2\pi (0)}{6}}+ i\sin{\frac{2\pi (0)}{6}}]=1[\cos{0}+i \sin{0}]$ therefore $z_1=\cos{0}+i\sin{0}$ finally $z_1=1$
For $k=1: w_2=\sqrt[6]{1}[\cos{\frac{2\pi (1)}{6}}+ i\sin{\frac{2\pi (1)}{6}}]=1[\cos{\frac{\pi}{3}}+i \sin{\frac{\pi}{3}}]$ therefore $z_2=\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}}$ finally $z_2=\frac{1}{2}+\frac{\sqrt{3}}{2}i$
For $k=2: w_3=\sqrt[6]{1}[\cos{\frac{2\pi (2)}{6}}+ i\sin{\frac{2\pi (2)}{6}}]=1[\cos{\frac{2\pi}{3}}+i \sin{\frac{2\pi}{3}}]$ therefore $z_3=\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}$ finally $z_3=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$
For $k=3: w_4=\sqrt[6]{1}[\cos{\frac{2\pi (3)}{6}}+ i\sin{\frac{2\pi (3)}{6}}]=1[\cos{\pi}+i \sin{\pi}]$ therefore $z_4=\cos{\pi}+i\sin{\pi}$ finally $z_4=-1$
For $k=4: w_5=\sqrt[6]{1}[\cos{\frac{2\pi (4)}{6}}+ i\sin{\frac{2\pi (4)}{6}}]=1[\cos{\frac{4\pi}{3}}+i \sin{\frac{4\pi}{3}}]$ therefore $z_5=\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}$ finally $z_5=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$
For $k=5: w_6=\sqrt[6]{1}[\cos{\frac{2\pi (5)}{6}}+ i\sin{\frac{2\pi (5)}{6}}]=1[\cos{\frac{5\pi}{3}}+i \sin{\frac{5\pi}{3}}]$ therefore $z_6=\cos{\frac{5\pi}{3}}+i\sin{\frac{5\pi}{3}}$ finally $z_6=\frac{1}{2}-\frac{\sqrt{3}}{2}i$
I hope this help you!!!
A: Hint: Imagine that there is a unit circle on the Argand Plane. Now the roots will be the  6 equidistant points on that circle each having the argument as multiple of $\frac{2\pi}{6}=\frac{\pi}{3}$
