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B Complex Replacement: Justification?

  1. May 11, 2016 #1

    Is there a proof that complex replacement is a valid way to solve a differential equation? I'm lacking some intuition on the idea that under any algebraic manipulations the real and imaginary parts of an expression don't influence each other.

    For example, if I'm given:

    $$p(D) x = cos(t)$$ where p(D) is some differential operator, how can I be sure that the real part of the solution to $$p(D) z = e^{it}$$ will be a solution to my original differential equation? How can I be sure that I can manipulate any of these expressions with complex numbers and will still be able to take the real part?
  2. jcsd
  3. May 11, 2016 #2

    Paul Colby

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    Gold Member

    You can if ##p(D)## is linear in that ##p(D) (a z_1+b z_2) = a p(D)z_1 + b p(D) z_2##

    Okay, this may be stupid, if ##p(D)z_r = cos(t)## and ##p(D)z_i = i sin(t)## then it may work
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