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Derivate of complex numbers

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Why do you need to use Cauchy-Riemann in the derivate of [itex] x^3 + i(1-y)^3[/itex], while not in the derivate of [itex] \frac { 1 } {z^2 + 1}[/itex] in using the quotient rule?

    where z = x + iy.

    I derivated the first expression implicitly in the exam which resulted in zero points of the exercise.

    3. The attempt at a solution

    I would derivate the latter by expanding the denominator such that

    [tex] \frac{1} { (x+iy)^2 +1} =...= \frac {x^2 - y^2 +1} {x^4 + y^4 - 6x^2 y^2 +1} - \frac {2xyi} {x^4 + y^4 -6x^2 y^2 +1} [/tex]

    and then derivate as real and then as complex.
    The expression is rather challenging at the moment.
    I am not sure what is the best way to derivate the latter statement correctly.
     
    Last edited: Nov 8, 2009
  2. jcsd
  3. Nov 8, 2009 #2
    I'm guessing that to get:
    [tex]\frac{1}{ (x+iy)^2 +1} =...= \frac{x^2 - y^2 +1}{x^4 + y^4 - 6x^2 y^2 +1} - \frac{2xy}{x^4 + y^4 -6x^2 y^2 +1}[/tex]

    you multiplied by the conjugate, so you multiplied top and bottom by:
    [x^{2}-y^{2}+1]-2xyi
    right?

    if that's right, i dont agree with what you got. can you please show one more intermediate step?
     
  4. Nov 8, 2009 #3
    I did multiply by the conjugate.

    The denominator is

    [tex] (x + iy)^2 +1 = x^2 + 2xyi -y^2 +1 [/tex]
    [tex] = (x^2 - y^2 +1) + 2xyi [/tex]

    Multiplying by the conjugate

    [tex] \frac{1} { (x^2 - y^2 +1) + 2xyi } = \frac { (x^2 - y^2 +1) - 2xyi } { (x^2 - y^2 +1)^2 + (2xy)^2 } [/tex]

    Let [itex] A = x^2 - y^2 [/itex]. Considering the denominator
    [tex] (A +1)^2 = A^2 +2A +1 [/tex]
    [tex] = (x^2 - y^2)^2 + 2(x^2 - y^2) +1 [/tex]
    [tex] = x^4 - 2x^2 y^2 + y^4 + 2x^2 - 2y^2 +1 [/tex]

    Making the projection: [itex]B |-> B + (2xy)^2[/itex], we get for the denominator

    [tex] x^4 + y^4 + 2x^2 -2y^2 +1 [/tex],

    which is the same denominator as in my question.

    The numerator is simply the conjugate [itex] (x^2 - y^2 +1) - 2xyi [/itex].
     
  5. Nov 8, 2009 #4
    you mean, making the projection;
    we get for the denominator:

    [tex]B+(2xy)^{2}=x^4-2x^2y^2+y^4+2x^2-2y^2+1+4x^2y^2=x^4+2x^2y^2+y^4+2x^2-2y^2+1[/tex]

    ?
     
  6. Nov 8, 2009 #5

    Yes, I mean what you say.
    I did not use apparently the term "projection" correctly.
    Perhaps, a more appropriate term would be a mapping or simply a plus between the two.
     
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