1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complex equations

  1. Feb 9, 2009 #1
    If i am solving for a complex equation f(z) = 0. can i assume that if i solve [f(z)]* = 0, I'll get the solution to the first one??

    I mean, does f(z) = 0 imply [f(z)]* = f(z*) = 0???

    I think it should be, but its giving me vague answers for a question i'm trying to solve. pls help
  2. jcsd
  3. Feb 9, 2009 #2
    Well, obviously f(z)* = 0 if and only if f(z) = 0, since 0* = 0. On the other hand, it is not necessarily true that f(z*) = 0. For instance, if f(z) = i+z, then f(i*) = f(-i) = 0, but f(i) = 2i ≠ 0.

    Now, it may be the case that you can prove that f(z*) = f(z)* for every z -- for instance, if f was a polynomial in z with real coefficients, this equation would hold, and then solving for f(z*) = 0 would get you a solution to the original equation. But it's not true in general.
  4. Feb 10, 2009 #3
    ya...i had realised that....but the question i am trying to solve, has real coefficients....and also...

    [exp(z)]* = exp(z*) right????

    I dont know why i am getting stuck.... neway. i'll post it in the homework section if i cant do it
  5. Feb 10, 2009 #4
    Yes, that is correct.
  6. Feb 10, 2009 #5
    Yes, you can expand the exp function into powerseries. Plugging the properties a*+b*=(a+b)* and a*∙b*=(a∙b)* into these powerseries will lead you pretty quickly to [exp(z)]* = exp(z*)

    This work for any holomorphic function you can expand into powerseries with real coefficients.
  7. Feb 10, 2009 #6
    ok... I found my mistake. I can't do what I have been trying to do. Turns our there's an imaginary coefficient after all.

    Thanks anyway.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook