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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.
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