How do I solve this complex numbers problem?

Given that 2+i is the root of z z^3 -11z+20 = 0. Find the remaining roots.

1 Answer
May 9, 2018

z=2+i
z=2-i
z=-4

Explanation:

All cubics will have three roots. One of the other roots will be 2-i, since this is the complex conjugate.

Revision:

To find the other root, suppose the three roots of the equation are z=alpha, z=beta and z=gamma, z in CC Therefore:

(z-alpha)(z-beta)(z-gamma)=0
(z^2-alphaz-betaz+alphabeta)(z-gamma)=0
(z^2-(alpha+beta)z+alphabeta)(z-gamma)=0
z^3-(alpha+beta)z^2+alphabetaz-gammaz^2+gamma(alpha+beta)-alphabetagamma=0

z^3-(alpha+beta+gamma)z^2+(alphabeta+alphagamma+betagamma)z-alphabetagamma=0

So for a cubic with equation az^3+bz^2+cz+d=0:
z^3+b/az^2+c/az+d/a=0:

Sigmaalpha=-b/a (sum of the roots)
Sigmaalphabeta=c/a (sum of the pairs)
alphabetagamma=-d/a (product)


We can use these facts to form an equation.
Probably the easiest of these to use is the sum of the roots; the others can get fiddly. Therefore:

Since x^2 coefficient is 0;
Sigmaalpha=0

Let our other root be a

a+(2+i)+(2-i)=0
a+4=0
a=-4

So the other root is -4.

So the roots of the equation are:
z=2+i
z=2-i
z=-4