x=2+sqrt(3)-i is a root of x^2+2ix-4[2+sqrt(3)]=0, what is the other root, what is the characteristic of them can be found?

You may be learning techniques I never studied, but here's how I'd approach it:

By completing the square, x+2ix = (x + i)² + 1, so the equation rewrites as:

(x + i)² = 4(2 + √3) - 1

The right side doesn't matter if you're not finding the roots "from scratch." The roots will be:

x + i = ± √(something)

x = (-i) ± √(something)

Apparently that +√(something) is 2 + √3, so the other root must be

x = -i - √(something) = -i - 2 - √3

Both that and the given actually do work if plugged back into the original equation. I was surprised that √[ 4(2 + √3) - 1 ] = √(7 + 4√3) simplifies to 2+√3..

By the synthetic division:

1..................2i...............-8-4sqrt(3) | [2+sqrt(3)-i]

.....2+sqrt(3)-i................8+4sqrt(3) |

-------------------------------------------------

1...2+sqrt(3)+i

=>

x= -2-sqrt(3)-i is the other root

These 2 non-real roots are symetrical

to the imaginary y-axis in the complex

plane.