Complex roots part 3 - Product of the roots

x=0n1Zx= ?\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.

Zx=zn=a2+b22ncis(arctanban+2xπn)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)

x=0n1Zx=Z0Z1Z2Zn1\prod_{x = 0}^{n - 1}{Z_x} = Z_0Z_1Z_2\cdots Z_{n - 1}

There are nn terms.

x=0n1Zx=a2+b22nncis(arctanban)cis(arctanban+2πn)cis(arctanban+4πn)cis(arctanban+2(n1)πn)\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)

x=0n1Zx=a2+b2cis(narctanban+2π+4π++2(n1)πn)\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)


2π+4π++2(n1)π=0+2π+4π++2(n1)π2\pi + 4\pi + \cdots + 2(n - 1)\pi = 0 + 2\pi + 4\pi + \cdots + 2(n - 1)\pi

There is now nn terms rather than n1n - 1 terms.

x=0n12xπ=n(2π)(n1)2\sum_{x = 0}^{n - 1}{2x\pi} = \frac{n(2\pi)(n - 1)}{2}

x=0n12xπ=nπ(n1)\sum_{x = 0}^{n - 1}{2x\pi} = n\pi(n - 1)

x=0n12xπ=n2πnπ\sum_{x = 0}^{n - 1}{2x\pi} = n^2\pi - n\pi


x=0n1Zx=a2+b2cis(arctanba+n2πnπn)\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)

x=0n1Zx=a2+b2cis(arctanba+nππ)\prod_{x = 0}^{n - 1}{Z_x} = \sqrt{a^2 + b^2} \cdot \text{cis}\left(\arctan\frac{b}{a} + n\pi - \pi\right)

x=0n1Zx=a2+b2cis(arctanba)cis(nππ)\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)


a2+b2cis(arctanba)=reθi=z\sqrt{a^2 + b^2} \cdot \text{cis}\left(\arctan \frac{b}{a}\right) = re^{\theta i} = z


x=0n1Zx=zcis(nππ)\prod_{x = 0}^{n - 1}{Z_x} = z\text{cis}(n\pi - \pi)

x=0n1Zx=zcis(nπ)cis(π)\prod_{x = 0}^{n - 1}{Z_x} = z\frac{\text{cis}(n\pi)}{\text{cis}(\pi)}


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


x=0n1Zx=zcis(nπ)1\prod_{x = 0}^{n - 1}{Z_x} = z\frac{\text{cis}(n\pi)}{-1}

x=0n1Zx=zcis(nπ)\prod_{x = 0}^{n - 1}{Z_x} = -z\text{cis}(n\pi)

x=0n1Zx=zcis(π)n\prod_{x = 0}^{n - 1}{Z_x} = -z\text{cis}(\pi)^n

x=0n1Zx=z(1)n\prod_{x = 0}^{n - 1}{Z_x} = -z(-1)^n

x=0n1Zx=(1)nz\prod_{x = 0}^{n - 1}{Z_x} = -(-1)^nz

x=0n1Zx=(1)n+1z\prod_{x = 0}^{n - 1}{Z_x} = (-1)^{n + 1}z

In conclusion, if nn is even then the product of the roots is z-z, if nn is odd then the product of the roots is zz.


When n is an even number: x=0n1Zx=z\text{When }n\text{ is an even number}:~\prod_{x = 0}^{n - 1}{Z_x} = -z

When n is an odd number: x=0n1Zx=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
3 years, 7 months ago

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Awesome article..Worth reading.

Amar Mavi - 3 years, 7 months ago

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