Tricky complex numbers question

[thomas49th]

Homework Statement

Harder: given that
√(−15 − 8i) = ±(1 − 4i) obtain the two solutions of the equation
z² + (−3 + 2i)z + 5 − i = 0

Homework Equations

I can easily prove √(−15 − 8i) = ±(1 − 4i) but that's not important

The Attempt at a Solution

I would of thought that a compex solution would be a + b and a - b, but a quick glance at the answers shows 2 completely different complex numbers - no complex conjugates.

Well seperating the equation into real and imaginary parts then solving for z:
real:
(z² - 3z + 5) = 0
=> z = \frac{3\pm i \sqrt{11}}{2}

imag:
(2z - 1) = 0
=> z = 0.5

This isn't taking me anywhere nice...

Ideas!? :)

Thanks

 

[Dick]

You can't separate it into real and imaginary parts like that, z itself is probably complex. Just use the quadratic formula on the original equation.