Question

How do you solve `2x^3+4+2i=0`?

Answer 1

The roots are:

`x_1 = root(3)(2+i)`

`x_2 = omega root(3)(2+i)`

`x_3 = omega^2 root(3)(2+i)`

Explanation:

Given:

`2x^3+4+2i=0`

Divide through by `2` to get:

`x^3+2+i = 0`

Subtract `2+i` from both sides to get:

`x^3 = 2+i`

This has roots:

`x_1 = root(3)(2+i)`

`x_2 = omega root(3)(2+i)`

`x_3 = omega^2 root(3)(2+i)`

where `omega = -1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.

If you prefer these roots in `a+bi` form, then it can be done with trigonometric functions:

First note that:

`abs(2+i) = sqrt(2^2+1^2) = sqrt(5)`

`Arg(2+i) = tan^(-1)(1/2)`

Hence, using de Moivre's formula:

`x_1 = root(6)(5)cos(1/3 tan^(-1) (1/2)) + i root(6)(5)sin(1/3 tan^(-1) (1/2))`

`x_2 = root(6)(5)cos(1/3 tan^(-1) (1/2)+(2pi)/3) + i root(6)(5)sin(1/3 tan^(-1) (1/2)+(2pi)/3)`

`x_3 = root(6)(5)cos(1/3 tan^(-1) (1/2)+(4pi)/3) + i root(6)(5)sin(1/3 tan^(-1) (1/2)+(4pi)/3)`