91Ó°ÊÓ

To prove that the sum of the nth roots of any complex number is zero.

Complex numbers possess real numbers and imaginary numbers; a complex can be written in the form of:

$z=x+iy$

Here x and y are real numbers, and i is the imaginary number which is known as iota, whose value is $\sqrt{-}1$ .

Let's assume a complex number z then z = r$×\exp \left({\mathrm{θ}}_i\right)$ ……. (1) Write the Roots and angle of the number.

$z_k={\left[rexp\left({\mathrm{θ}}_k\right)\right]}^{\frac{1}{n}}$ ……. (2)

${\mathrm{θ}}_k=\frac{\mathrm{θ}+2n\mathrm{π}k}{n}$ ……. (3)

Substitution in (2) from (3) as:

$$\begin{gathered} z_k=r^{\frac{1}{n}}\left[exp\left(\frac{\mathrm{θ}i+2n\mathrm{π}ki}{n}\right)\right] \\ =\left[r^{\frac{1}{n}}exp\left(\frac{\mathrm{θ}i}{n}\right).exp\left(\frac{2n\mathrm{π}ki}{n}\right)\right] \\ =Cexp\left(\frac{2n\mathrm{π}ki}{n}\right)Wherek=0,1,2,3,... \\ \mathrm{Take}\underset{}{∑\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{as}:} \end{gathered}$$

$$\begin{gathered} \underset{}{∑}{z}_k=C\underset{k=0}{\overset{n-1}{∑}}exp{\left(\frac{2n\mathrm{Ï€}ki}{n}\right)}^n \\ =C+Cexp\left(\frac{2n\mathrm{Ï€}ki}{n}\right)+Cexp{\left(\frac{2n\mathrm{Ï€}ki}{n}\right)}^2+... \\ =C\left(1+exp\left(\frac{2n\mathrm{Ï€}ki}{n}\right)+exp{\left(\frac{2n\mathrm{Ï€}ki}{n}\right)}^2+...\right) \\ =C\left(1+\mathrm{Ó\neg }+{\mathrm{Ó\neg }}^2+{\mathrm{Ó\neg }}^3+...\right) \\ \underset{}{∑}{z}_k=C\underset{k=0}{\overset{n-1}{∑}}exp{\left(\frac{2n\mathrm{Ï€}ki}{n}\right)}^n \\ =C+Cexp\left(\frac{2n\mathrm{Ï€}ki}{n}\right)+Cexp{\left(\frac{2n\mathrm{Ï€}ki}{n}\right)}^2+... \\ =C\left(1+exp\left(\frac{2n\mathrm{Ï€}ki}{n}\right)+exp{\left(\frac{2n\mathrm{Ï€}ki}{n}\right)}^2+...\right) \\ 1+\mathrm{Ó\neg }+{\mathrm{Ó\neg }}^2+{\mathrm{Ó\neg }}^3+...\mathrm{is}\mathrm{roots}\mathrm{of}\mathrm{number}1. \end{gathered}$$

The above series is a geometrics series, put in the form of:

$\underset{}{∑}{z}_k=C\left(\frac{{\mathrm{Ó\neg }}^n-1}{\mathrm{Ó\neg }-1}\right)$ …â¶Ä¦.(6)

$$\begin{gathered} \mathrm{Ó\neg }=\sqrt[n]{1} \\ \mathrm{Ó\neg }=exp\left(2\mathrm{Ï€}i/n\right) \\ {\mathrm{Ó\neg }}^n=1 \end{gathered}$$

Substitution in (6) yield$\underset{}{∑}{z}_k=0$ .

Hence, proved that the sum of the n nth roots of any complex number is zero.