Complex roots part 2 - Sum of the roots

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

Zx=zn=a2+b22n cis(arctan⁡ban+2xπn)\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 n≠0\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.

∑x=0n−1Zx=Z0+Z1+⋯+Zn−1\sum_{x = 0}^{n - 1}{Z_x} = Z_0 + Z_1 + \cdots + Z_{n - 1}

∑x=0n−1Zx=a2+b22n cis(arctan⁡ban)+a2+b22n cis(arctan⁡ban+2πn)+⋯+a2+b22n cis(arctan⁡ban+2(n−1)πn)\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)

∑x=0n−1Zx=a2+b22n( cis(arctan⁡ban)+ cis(arctan⁡ban+2πn)+⋯+ cis(arctan⁡ban+(2n−2)πn))\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)

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

∑x=0n−1Zx=a2+b22n( cis(arctan⁡ban)(1+cis(2πn)+cis(4πn)+⋯+cis((2n−2)πn)))\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)

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

∑x=0n−1Zx=a2+b22n( cis(arctan⁡ban)(1+cis(2πn)+(cis(2πn))2+⋯+(cis(2πn))n−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)\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)

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

∑N=0n−1XN=1+X+X2+⋯+Xn−1\sum_{N = 0}^{n - 1}{X^N} = 1 + X + X^2 + \cdots + X^{n - 1}

∑N=0n−1XN=1−Xn1−X\sum_{N = 0}^{n - 1}{X^N} = \frac{1 - X^{n}}{1 - X}

∑x=0n−1Zx=a2+b22n( cis(arctan⁡ban)1−Xn1−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)

∑x=0n−1Zx=a2+b22n( cis(arctan⁡ban)1−(cis(2πn))n1−cis(2πn))\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)

∑x=0n−1Zx=a2+b22n( cis(arctan⁡ban)1−cis(2π)1−cis(2πn))\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)

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

∑x=0n−1Zx=a2+b22n( cis(arctan⁡ban)1−11−cis(2πn))\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)

∑x=0n−1Zx=a2+b22n( cis(arctan⁡ban)01−cis(2πn))\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)

∑x=0n−1Zx=0\sum_{x = 0}^{n - 1}{Z_x} = 0

Hope you enjoyed the note.

#Algebra

Note by Jack Rawlin
5 years, 3 months ago

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