Tricky complex numbers question

by thomas49th
Tags: complex, numbers, tricky
thomas49th is offline
Feb15-10, 01:50 PM
P: 656
1. The problem statement, all variables and given/known data
Harder: given that
√(−15 − 8i) = (1 − 4i) obtain the two solutions of the equation
z + (−3 + 2i)z + 5 − i = 0

2. Relevant equations

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

3. 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:
(z - 3z + 5) = 0
=> [tex] z = \frac{3\pm i \sqrt{11}}{2}[/tex]

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

This isn't taking me anywhere nice...

Ideas!? :)

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Dick is offline
Feb15-10, 02:37 PM
Sci Advisor
HW Helper
P: 25,174
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.

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