# Partial derivative of a complex number

1. Dec 10, 2015

### shinobi20

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. Dec 11, 2015

### conquest

Hey shinobi,

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

Last edited: Dec 11, 2015
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