# Quadratic equation with complex coefficients

## Homework Statement

z^2 + 4(1 + i(3^0.5))z - 16 = 0

## The Attempt at a Solution

I think I've done this correctly, I just wanted to verify.
I've only done the solution for k=0

http://i129.photobucket.com/albums/p201/elimenohpee182/Capture-1.png

verty
Homework Helper
I'm interested to know, why did you use polar coordinates? Would it not be easier to let z = a + b.i, then solve for a and b?

I thought using polar coordinates would be easiest to eliminate the square root of the complex number.

I don't know if its right or not, thats why I wanted someone to check it.

Mentallic
Homework Helper
You want to get rid of the square root of the determinant, so let $$\sqrt{-96+32i}=a+ib$$ on squaring both sides, we solve $$-96+32i=a^2-b^2+2abi$$

Thus you have two equations to solve, $$-96=a^2-b^2$$ and $$32=2ab$$ since the real and imaginary coefficients must be equal.

But first you may want to check if you can simplify -96+32i. Notice 96=32*3