Is there a simple way to define the $n$-th roots of the unity?

Question

Is there a simple way to calculate the $n$-th roots of the unity? I gotta solve the equation $$\frac{z+1}{z-1}=\sqrt[n]{1}.$$

Perhaps you mean "write down" or "calculate"? Travis's answer is the standard definition, but perhaps not very helpful for your situation. — pjs36, Jul 5, 2015 at 15:28
To solve the equation, it might help to note that any root of unit has modulus $1$, so taking norms of both sides of the equation gives $\left\vert\frac{z + 1}{z - 1}\right\vert = 1$. But the l.h.s. is $\frac{|z + 1|}{|z - 1|}$, and this is equal to $1$ precisely for $z$ equidistant from $1$ and $-1$, that is, imaginary $z$. — Travis Willse, Jul 5, 2015 at 15:31
Is it correct to say that if 1 is not a solution to this equation (assuming z is a complex number), then there are no other solutions? Because "the nth roots of unity are at the vertices of a regular n-sided polygon inscribed in the unit circle, with one vertex at 1". — Franciele Daltoé, Jul 5, 2015 at 16:52

Travis Willse · Accepted Answer · 2015-07-05 15:22:16Z

3

Yes, by definition, for $n \in \Bbb Z$, $\zeta$ is an $n$th root of unity iff $$\zeta^n = 1.$$

answered Jul 5, 2015 at 15:22

Travis Willse

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$\begingroup$ Perhaps the downvoter would explain their objection? This answer's OP's question precisely, in probably the shortest and easiest terms possible. $\endgroup$
– Travis Willse
Jul 5, 2015 at 15:23
1

$\begingroup$ I don't understand these trigger happy down voters. I'll neutralize with a +1 $\endgroup$
– Mark Viola
Jul 5, 2015 at 15:40

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Henry · Accepted Answer · 2015-07-05 15:25:12Z

1

One possibility is $$e^{2\pi i k /n}$$ for integer $k$.

answered Jul 5, 2015 at 15:23

Henry

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Mark Viola · Accepted Answer · 2015-07-05 15:37:31Z

1

The $n$ roots of unity $w_n$ can be written

$$w_n=e^{i2\ell \pi/n}$$

for $\ell=0,1,2,\cdots,n-1$.

For the problem $\frac{z+1}{z-1}=w_n$ we have

$$z=\frac{w_n+1}{w_n-1}=-i\cot(\ell\pi/n)$$

for $\ell=1,2,\cdots,n-1$.

answered Jul 5, 2015 at 15:37

Mark Viola

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$\begingroup$ Is it correct to say that if 1 is not a solution to this equation (assuming z is a complex number), then there are no other solutions? Because "the nth roots of unity are at the vertices of a regular n-sided polygon inscribed in the unit circle, with one vertex at 1". $\endgroup$
– Franciele Daltoé
Jul 5, 2015 at 17:43
$\begingroup$ @FrancieleDaltoé The only solutions to $\left(\frac{z+1}{z-1}\right)^n=1$ are purely imaginary. $\endgroup$
– Mark Viola
Jul 5, 2015 at 18:04
$\begingroup$ By writing $z=a+bi$, and calculating the absolute value of $(a+1)+bi$ and $(a-1)+bi$, I get that $a=0$, which means the solutions are purely imaginary. Is that enough to answer the question or can I calculate these numbers? $\endgroup$
– Franciele Daltoé
Jul 5, 2015 at 18:53
2

$\begingroup$ @FrancieleDaltoé The solutions are in my answer. If you proceed as described, you will find that $a=0$, but you will not have any information about $b$. That information is lost once the magnitude is taken. $\endgroup$
– Mark Viola
Jul 5, 2015 at 20:19

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3 Answers 3