Question

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.

Answer 1

`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)

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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`