Quadratic inequalities for complex variables?


by mathsciguy
Tags: complex, inequalities, quadratic, variables
mathsciguy
mathsciguy is offline
#1
Sep12-13, 09:29 AM
P: 132
Hello, I was looking at Riley's solution manual for this specific problem. Along the way, he ended up with a quadratic inequality:

how.bmp

If you looked at the image, he said it is given that λ is real, but he asserted that λ has no real roots because of the inequality. Doesn't that mean λ is imaginary or complex in some points then, contradicting his first statement? I reckon this has something to do with the properties of a quadratic inequality for complex variables.
Phys.Org News Partner Mathematics news on Phys.org
Math modeling handbook now available
Hyperbolic homogeneous polynomials, oh my!
Researchers help Boston Marathon organizers plan for 2014 race
pasmith
pasmith is offline
#2
Sep12-13, 10:59 AM
HW Helper
P: 775
Quote Quote by mathsciguy View Post
Hello, I was looking at Riley's solution manual for this specific problem. Along the way, he ended up with a quadratic inequality:

Attachment 61729

If you looked at the image, he said it is given that λ is real, but he asserted that λ has no real roots because of the inequality. Doesn't that mean λ is imaginary or complex in some points then, contradicting his first statement? I reckon this has something to do with the properties of a quadratic inequality for complex variables.
[itex]\lambda[/itex] doesn't have roots; it's an arbitrary real number. The quadratic [itex]P: z \mapsto az^2 + bz + c[/itex] has roots, which are those [itex]z \in \mathbb{C}[/itex] for which [itex]P(z) = 0[/itex]. The point is that if [itex]P(\lambda) > 0[/itex] for all real [itex]\lambda[/itex] then [itex]P[/itex] has no real roots, because if [itex]z[/itex] is real then [itex]P(z) \neq 0[/itex] and [itex]z[/itex] cannot be a root of [itex]P[/itex].
mathsciguy
mathsciguy is offline
#3
Sep13-13, 05:06 AM
P: 132
That's cool, I get it now. Then that means the roots are either purely imaginary or complex (but not purely real) right? Then why is it required that the discriminant be less than zero? Is it because it will make sure that part of the solution will have an imaginary part?

pasmith
pasmith is offline
#4
Sep13-13, 09:11 AM
HW Helper
P: 775

Quadratic inequalities for complex variables?


Quote Quote by mathsciguy View Post
That's cool, I get it now. Then that means the roots are either purely imaginary or complex (but not purely real) right? Then why is it required that the discriminant be less than zero? Is it because it will make sure that part of the solution will have an imaginary part?
Yes. A quadratic with real coefficients has no real roots if and only if the discriminant is negative.


Register to reply

Related Discussions
quadratic inequalities Precalculus Mathematics Homework 23
quadratic equations and inequalities / applications of quadratic functions question Precalculus Mathematics Homework 3
[SOLVED] Quadratic Equations and Inequalities question about properties of quadratic General Math 2
Quadratic Inequalities Precalculus Mathematics Homework 11
Quadratic Inequalities Precalculus Mathematics Homework 2