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Partial derivative of a complex number

  1. Dec 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Given n=(x + iy)/2½L and n*=(x - iy)/2½L
    Show that ∂/∂n = L(∂/∂x - i ∂/∂y)/2½ and ∂/∂n = L(∂/∂x + i ∂/∂y)/2½
    2. Relevant equations
    n Ξ ∂/∂n, ∂x Ξ ∂/∂x, as well as y.

    3. The attempt at a solution
    n=(∂x + i ∂y)/2½L
    Apply complex conjugate on right side, ∂n=[(∂x + i ∂y)/2½L] * [(∂x - i ∂y)/(∂x - i ∂y)].
    ∂n=( (∂x)2 + (∂y)2) / 2½ L (∂x - i ∂y)
    I'm stuck, I think if (∂x)2 + (∂y)2=1 I can think of a way. But ugh... Any suggestions?
     
  2. jcsd
  3. Dec 11, 2015 #2
    Hey shinobi,

    Note that [itex]\partial_n[/itex] is not the derivative of [itex]n[/itex], but rather the derivative in the direction of [itex]n[/itex]. So suppose I have some function [itex]f(n)[/itex] then note that since [itex]n=n(x,y)[/itex] taking the derivative with respect to [itex]n[/itex] should respect the chain rule and so
    [itex]\partial_nf(n)=\partial_n(x)\partial_xf(n(x,y))+\partial_n(y)\partial_yf(n(x,y))[/itex]
    So [itex]\partial_n=\partial_n(x)\partial_x+\partial_n(y)\partial_y[/itex] now the problem comes down to calculating these coefficients. This you can do by inverting the derivate of [itex]n[/itex] with respect to [itex]x[/itex] and similarly for [itex]y[/itex].
     
    Last edited: Dec 11, 2015
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