Hello readers,

This post is related to calculate sum and product of roots of an equation in the complex variable z^n = z_0 where z_0 is given the complex number. You can refer to the following link to study how to calculate powers of complex number and roots of the complex variable equation z^n = z_0.

Powers of Complex number and roots of equation z^n = z_0

Now,  look at how to calculate the sum of roots of equation z^n=z_0.

Sum of roots of  z^n=z_0 :

Consider equation z^n = z_0.

This equation has n distinct roots say c, cw, cw^2, cw^3,………….cw^(n-1)

where c = (r_0)^(1/n) e^((iθ_0)/n)  and w = e^(i2Π/n) .

hence sum of roots = c+ cw+ cw^2+…….cw^(n-1)

=c( 1 – w^n )/ ( 1- w)

= c (0),

= 0

since w^n = 1 (as w^n = e^(i2Πn/n)= e^(i2Π)= cos 2Π + i sin 2Π=1)

We have proved that sum of roots of z^n = z_0 is always zero.

The product of roots of z^n = z_0.

the product of roots = c (cw) ( cw^2)…….(cw^(n-1))

= c ^n ( w ^{1+2+3+….+(n-1)})

= (r_0)^(n/n) e^((inθ_0)/n)  ( w^{(n-1)n/2})

= (r_0) e^(iθ_0)  e^{((i2Π)/n)(n-1)n/2}

= z_0 e^{(i(n-1)Π)}

= z_0 (-1)^(n-1)

 

Example: find the product of roots of equation z^100 = 1+i

Solution:  In this example n= 100, z_0 = 1+i

hence product of roots = z_0 (-1)^(n-1)

= (1+i) (-1)^(100-1)

= (1+i) (-1)^(99)

= -1-i

 

 

Thank you for reading

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Keep on reading, stay learnt.

Dr. Namita Tiwari