Register to reply

Tricky complex numbers question

by thomas49th
Tags: complex, numbers, tricky
Share this thread:
thomas49th
#1
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:
real:
(z - 3z + 5) = 0
=> [tex] z = \frac{3\pm i \sqrt{11}}{2}[/tex]

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

This isn't taking me anywhere nice...

Ideas!? :)

Thanks
Phys.Org News Partner Science news on Phys.org
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
Dick
#2
Feb15-10, 02:37 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
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.


Register to reply

Related Discussions
Tricky complex numbers problem Precalculus Mathematics Homework 5
One Question....Complex Numbers Calculus & Beyond Homework 4
Complex numbers question. Calculus & Beyond Homework 9
Complex numbers question Precalculus Mathematics Homework 2
Tricky complex numbers proof Calculus & Beyond Homework 3