# 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!

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So what is written in your book? – draks ... Feb 6 '12 at 13:46
$cis=cos+i\cdot sin$? – Rasmus Feb 6 '12 at 14:01
yes its is. why do i have to write 12 chars? – yotamoo Feb 6 '12 at 14:10
You want $\theta=\frac{2\pi k}6 = \frac{\pi k}3$. – Thomas Andrews Feb 6 '12 at 14:12
What ever "start with ...", I think there is a 2 missing. Shouldn't it be ${\rm cis}(x)=e^{ix}=e^{i(x+2k\pi)}$? This would also drop the solution that "starts with $\sqrt{3}/2$", which is a root of -1 anyhow. – draks ... Feb 6 '12 at 14:15

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.

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But isn't that what he did? ${\rm cis} (x) = e^{ix}$ – draks ... Feb 6 '12 at 13:48
@draks: You are right. Sorry for the repetition. – Jon Feb 6 '12 at 13:53
That's because you divided wrong, @Jon. $\frac{2\pi n i}6 \neq \frac{\pi n i}6$ – Thomas Andrews Feb 6 '12 at 14:14
@ThomasAndrews: Thank you a lot! I will fix this embarrassing mistake. – Jon Feb 6 '12 at 14:18

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.

$$x - 1 = 0$$

$$x^2 + x + 1 = 0$$

$$x+1=0$$

$$x^2 - x + 1 = 0$$

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You need $2\pi$ where you had $\pi$. Thus for $k=1$ you should get $$\cos\left(\frac{2\pi\cdot1}{6}\right) + i \sin\left(\frac{2\pi\cdot1}{6}\right).$$

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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 $z_6=\frac{1}{2}-\frac{\sqrt{3}}{2}i$

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 $\frac{2\pi}{6}=\frac{\pi}{3}$