Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quadratic inequalities for complex variables?

  1. Sep 12, 2013 #1
    Hello, I was looking at Riley's solution manual for this specific problem. Along the way, he ended up with a quadratic inequality:

    View attachment 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.
     
  2. jcsd
  3. Sep 12, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    [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].
     
  4. Sep 13, 2013 #3
    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?
     
    Last edited: Sep 13, 2013
  5. Sep 13, 2013 #4

    pasmith

    User Avatar
    Homework Helper

    Yes. A quadratic with real coefficients has no real roots if and only if the discriminant is negative.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Quadratic inequalities for complex variables?
Loading...