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#