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Let $z_1, z_2,\dots,z_n$ be all the solution in $\mathbb{C}$ of $z^n=1$.

I have to prove that for odd power $n$ the multiplication of the roots equals $1$: $$z_1z_2\dots z_n=1.$$

I'm a bit stuck and would appreciate a tip how to progress.

$z^n=1 \Rightarrow z^n-1=0 $ and based on the basic theorem of algebra we know it has $n$ roots.

We assume $n,m\in\mathbb{Z}$ and $n=2m+1$ ($n$ is odd number)

I will use the trigonometric presentation:

$1 = \cos 2\pi + i \sin 2\pi$
$z^n=r^{2m+1}(\cos (2m+1)\alpha+i \sin(2m+1)\alpha)$

$\DeclareMathOperator{\cis}{cis}r^{2m+1}(\cis(2m+1)\alpha)=1 \cis 2\pi$

$ \Rightarrow r^{2m+1}=1 \Rightarrow r = 1$
$ \cis(\alpha (2m+1)) = \cis(2\pi + 2\pi k)$, $k \in\mathbb{Z}$, $k=0,1,2,\dots,n-1$
$ \Rightarrow \alpha (2m+1)=2\pi + 2\pi k$

How do I progress from this point showing that $z_1z_2\dots z_n=1$?

egreg
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Hector Isidor
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3 Answers3

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$z^n-1=(z-z_1)\cdots(z-z_n)$.

Set $z=0$ and get $-1=(-1)^nz_1\cdots z_n$ and so $z_1\cdots z_n=(-1)^{n+1}$.

When $n$ is odd, we get $z_1\cdots z_n=1$ because $n+1$ is even.

lhf
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    So nice, elegant and simple, The beauty of math, thank you so much for showing me this – Hector Isidor Dec 30 '16 at 20:43
  • why would you set z=0 ? you can eliminate z^n by transferring it to the other side of the equation, so you don't have to work with specific z=0 but stay at the general case, any z. – Hector Isidor Dec 30 '16 at 22:01
  • @HectorIsidor It's an equality between polynomials, so it's preserved for each value you substitute for $z$. – egreg Dec 30 '16 at 22:06
  • @egreg, got you, thanks for the clarification. Grazie – Hector Isidor Dec 30 '16 at 22:13
  • just another round to confirm my understanding: you mean that since the above equality is equality between polynomials it is true for any z, and as well for z=0, so then we take 0 as a convenient candidate to continue our "journey" with,am I right ? – Hector Isidor Dec 30 '16 at 22:33
3

If you think about the roots geometrically, they each have modulus $1$ and are spaced evenly around the unit circle. As you multiply two of them together, the distance from the origin in the complex plane stays at $1$; the only result is a rotation.

Geometrically interpreting the comment to think about pairing up each root with its complex conjugate, one of them rotates some number of degrees clockwise and multiplying by its conjugate has the effect of rotating it back by the same number of degrees counterclockwise.

And so all the rotations cancel out, leaving just the one real root: $1$, which is their final product.

(If the degree in the initial setup were even, then $-1$ would be among the multiplied factors, too; but it is not. And so you really do get $1$.)

Algebraically, each non-real root $a + bi$ with $a, b \in \mathbb{R}$ has complex conjugate $a - bi$, so that their product is $(a+bi)(a-bi) = a^2 + b^2$, which is $1$ since $a + bi$ is a root of unity. Again, paired off, you have non-real roots yielding a product of $1$s, and a single real root of $1$ to finish things off.

Benjamin Dickman
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0

Suppose $n>0$.

We have $z_k=e^{i\pi\frac{2(k-1)}{n}}$ for $k=1,2,...n$

then $$\prod_{k=2}^nz_k$$

$$\displaystyle{=e^{i\frac{2\pi}{n}\sum_{k=1}^n(k-1)}}$$

$$\displaystyle{=e^{i\frac{2\pi}{n}\frac{n(n-1)}{2}}}$$

$=1$ if $n$ is odd and $-1$ if $n$ is even.

for example, if $n=2$ then the roots are $1$ and $-1$.

hamam_Abdallah
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