Partial derivative of a complex number

[shinobi20]

Homework Statement

Given n=(x + iy)/2^½L and n*=(x - iy)/2^½L
Show that ∂/∂n = L(∂/∂x - i ∂/∂y)/2^½ and ∂/∂n = L(∂/∂x + i ∂/∂y)/2^½

Homework Equations

∂_n Ξ ∂/∂n, ∂_x Ξ ∂/∂x, as well as y.

The Attempt at a Solution

∂_n=(∂_x + i ∂_y)/2^½L
Apply complex conjugate on right side, ∂n=[(∂_x + i ∂_y)/2^½L] * [(∂_x - i ∂_y)/(∂_x - i ∂_y)].
∂n=( (∂x)² + (∂y)²) / 2^½ 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?

 

[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 $y$.