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# Complex roots part 2 - Sum of the roots

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

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

Both of the above equations hold true as long as $$\boxed{n \neq 0}$$. But why does the top one hold true?

That's what we're going to be answering. So let's begin.

$\sum_{x = 0}^{n - 1}{Z_x} = Z_0 + Z_1 + \cdots + Z_{n - 1}$

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

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

$\text{cis}(a) + \text{cis}(b) + \text{cis}(c) + \cdots = \text{cis}(a)(1 + \text{cis}(b - a) + \text{cis}(c - a) + \cdots)$

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

$\text{cis}(nx) = (\text{cis}(x))^n$

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

$\text{Let } X = \text{cis}\left(\frac{2\pi}{n}\right)$

$\sum_{N = 0}^{n - 1}{X^N} = 1 + X + X^2 + \cdots + X^{n - 1}$

$\sum_{N = 0}^{n - 1}{X^N} = \frac{1 - X^{n}}{1 - X}$

$\sum_{x = 0}^{n - 1}{Z_x} = \sqrt[2n]{a^2 + b^2}\left(\text{ cis}\left(\frac{\arctan{\frac{b}{a}}}{n}\right)\frac{1 - X^{n}}{1 - X}\right)$

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

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

$\text{cis}(2\pi) = \cos(2\pi) + i\sin(2\pi) = 1 + 0i = 1$

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

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

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

Hope you enjoyed the note.

Note by Jack Rawlin
1 year, 5 months ago