Q31P - Mathematical Methods in Physical Sciences

Question

Show that the sum of the n nth roots of any complex number is zero.

Step-by-Step Solution

Answer

The sum of the n nth roots of any complex number is zero

Step 1: Given Information

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

Step 2: Definition of the complex number

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  $\sqrt{-}1$ .

Step 3- Find the roots and prove the statement

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

 $z_k={\left[r exp\left({\theta }_k\right)\right]}^{\frac{1}{n}}$                                                                                          ……. (2)

 

${\theta }_k=\frac{\theta +2n\pi k}{n}$                                                                                              ……. (3)

Substitution in (2) from (3) as:

 $$\begin{gathered} z_k=r^{\frac{1}{n}}\left[exp\left(\frac{\theta i+2n\pi ki}{n}\right)\right] \\ =\left[r^{\frac{1}{n}}exp\left(\frac{\theta i}{n}\right).exp\left(\frac{2n\pi ki}{n}\right)\right] \\ =C exp\left(\frac{2n\pi ki}{n}\right) Where k=0,1,2,3,... \\ \mathrm{Take} \underset{ }{\sum \mathrm{of} \mathrm{the} \mathrm{above} \mathrm{equation} \mathrm{as}:} \end{gathered}$$

$$\begin{gathered} \sum _{ }{z}_k=C\sum _{k=0}^{n-1}exp{\left(\frac{2n\pi ki}{n}\right)}^n \\ =C+C exp\left(\frac{2n\pi ki}{n}\right)+C exp{\left(\frac{2n\pi ki}{n}\right)}^2+... \\ =C\left(1+exp\left(\frac{2n\pi ki}{n}\right)+exp{\left(\frac{2n\pi ki}{n}\right)}^2+...\right) \\ =C\left(1+\omega +{\omega }^2+{\omega }^3+...\right) \\ \sum _{ }{z}_k=C\sum _{k=0}^{n-1}exp{\left(\frac{2n\pi ki}{n}\right)}^n \\ =C+C exp\left(\frac{2n\pi ki}{n}\right)+C exp{\left(\frac{2n\pi ki}{n}\right)}^2+... \\ =C\left(1+exp\left(\frac{2n\pi ki}{n}\right)+exp{\left(\frac{2n\pi ki}{n}\right)}^2+...\right) \\ 1+\omega +{\omega }^2+{\omega }^3+... \mathrm{is} \mathrm{roots} \mathrm{of} \mathrm{number} 1. \end{gathered}$$

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

 

  $\sum _{ }{z}_k=C\left(\frac{{\omega }^n-1}{\omega -1}\right)$                                                                            …….(6)    

 

$$\begin{gathered} \omega =\sqrt[n]{1} \\ \omega =exp\left(2\pi i/n\right) \\ {\omega }^n=1 \end{gathered}$$ 

Substitution in (6) yield $\sum _{ }{z}_k=0$  .

 

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