# Complex roots part 3 - Product of the roots

$\prod_{x = 0}^{n - 1}{Z_x} =~ ?$

The last note proved that the sum of all the roots is equal to zero. This note is going to explore what answer the product of all the roots gives.

Let's start.

$Z_x = \sqrt[n]{z} = \sqrt[2n]{a^2 + b^2} \cdot \text{cis}\left(\frac{\arctan \frac{b}{a}}{n} + \frac{2x\pi}{n}\right)$

$\prod_{x = 0}^{n - 1}{Z_x} = Z_0Z_1Z_2\cdots Z_{n - 1}$

There are $n$ terms.

$\prod_{x = 0}^{n - 1}{Z_x} = \sqrt[2n]{a^2 + b^2}^n \cdot \text{cis}\left(\frac{\arctan\frac{b}{a}}{n}\right)\text{cis}\left(\frac{\arctan\frac{b}{a}}{n} + \frac{2\pi}{n}\right)\text{cis}\left(\frac{\arctan\frac{b}{a}}{n} + \frac{4\pi}{n}\right)\cdots \text{cis}\left(\frac{\arctan\frac{b}{a}}{n} + \frac{2(n - 1)\pi}{n}\right)$

$\prod_{x = 0}^{n - 1}{Z_x} = \sqrt{a^2 + b^2} \cdot \text{cis}\left(n\frac{\arctan\frac{b}{a}}{n} + \frac{2\pi + 4\pi + \cdots + 2(n - 1)\pi}{n}\right)$

$2\pi + 4\pi + \cdots + 2(n - 1)\pi = 0 + 2\pi + 4\pi + \cdots + 2(n - 1)\pi$

There is now $n$ terms rather than $n - 1$ terms.

$\sum_{x = 0}^{n - 1}{2x\pi} = \frac{n(2\pi)(n - 1)}{2}$

$\sum_{x = 0}^{n - 1}{2x\pi} = n\pi(n - 1)$

$\sum_{x = 0}^{n - 1}{2x\pi} = n^2\pi - n\pi$

$\prod_{x = 0}^{n - 1}{Z_x} = \sqrt{a^2 + b^2} \cdot \text{cis}\left(\arctan\frac{b}{a} + \frac{n^2\pi - n\pi}{n}\right)$

$\prod_{x = 0}^{n - 1}{Z_x} = \sqrt{a^2 + b^2} \cdot \text{cis}\left(\arctan\frac{b}{a} + n\pi - \pi\right)$

$\prod_{x = 0}^{n - 1}{Z_x} = \sqrt{a^2 + b^2} \cdot \text{cis}\left(\arctan\frac{b}{a}\right)\text{cis}(n\pi - \pi)$

$\sqrt{a^2 + b^2} \cdot \text{cis}\left(\arctan \frac{b}{a}\right) = re^{\theta i} = z$

$\prod_{x = 0}^{n - 1}{Z_x} = z\text{cis}(n\pi - \pi)$

$\prod_{x = 0}^{n - 1}{Z_x} = z\frac{\text{cis}(n\pi)}{\text{cis}(\pi)}$

$\text{cis}(\pi) = \cos(\pi) + i\sin(\pi) = -1 + 0i = -1$

$\prod_{x = 0}^{n - 1}{Z_x} = z\frac{\text{cis}(n\pi)}{-1}$

$\prod_{x = 0}^{n - 1}{Z_x} = -z\text{cis}(n\pi)$

$\prod_{x = 0}^{n - 1}{Z_x} = -z\text{cis}(\pi)^n$

$\prod_{x = 0}^{n - 1}{Z_x} = -z(-1)^n$

$\prod_{x = 0}^{n - 1}{Z_x} = -(-1)^nz$

$\prod_{x = 0}^{n - 1}{Z_x} = (-1)^{n + 1}z$

In conclusion, if $n$ is even then the product of the roots is $-z$, if $n$ is odd then the product of the roots is $z$.

$\text{When }n\text{ is an even number}:~\prod_{x = 0}^{n - 1}{Z_x} = -z$

$\text{When }n\text{ is an odd number}:~\prod_{x = 0}^{n - 1}{Z_x} = z$

Hope you enjoyed the note Note by Jack Rawlin
4 years, 7 months ago

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