Question: Let vector field $\mathbb{X}$ be given by $\mathbb{X}(x,y)=(-y,x)$
Let $f(x,y)$ be the complex-valued function given by $f(x,y)=(x+iy)^m$ where $m>0$
Show that $L_\mathbb{X}f=imf$
My attempt: I know that I should use the hopotomy formula: $L_\mathbb{X}f=i_\mathbb{X}df+di_\mathbb{X}f$
I divide up my progress into: (1) find $i_\mathbb{X}df$, (2) find $di_\mathbb{X}f$
(1): To find $i_\mathbb{X}df$, start with computing $df=d((x+iy)^m)$
Is it correct to do: $d((x+iy)^m)\stackrel{?}{=}m(x+iy)^{m-1}+mi(x+iy)^{m-1}=m(1+i)(x+iy)^{m-1}$
Continuing I get: $i_\mathbb{X}df=i_{(-y,x)}m(1+i)(x+iy)^{m-1}$
I dont know how to compute this since I do not have $dx$ or $dy$ to use.
(2): I attempt to find $i_\mathbb{X}f$ and get $i_{(-y,x)}((x+iy)^m)$ this also doesn't contain $dx$ or $dy$ and so I once again am stuck.
How do I compute the contraction of functions in this form?
Any help would be greatly appreciated
Progress: After looking at this problem again I have found a few things
The definition of the Lie derivative is given by:
$$L_\mathbb{X}f=\left.\frac{\partial}{\partial t}\right|_{t=0}\Phi_t^*f=\left.\frac{\partial}{\partial t}\right|_{t=0}f\circ\Phi_t$$
and since $f$ is a $0$-form it is a function, hence, this can be given simply as:
$$L_\mathbb{X}f=\mathbb{X}\cdot\nabla f=\mathbb{X}^i\frac{\partial f}{\partial x^i}$$
Therefore we have $L_\mathbb{X}f=(-y)\frac{\partial f}{\partial x}+x\frac{\partial f}{\partial y}=-y\frac{\partial}{\partial x}(x+iy)^m+x\frac{\partial}{\partial y}(x+iy)^m$
$=-ym(x+iy)^{m-1}+xim(x+iy)^{m-1}$
$=(xi-y)m(x+iy)^{m-1}$
The problem has now been reduced to:
Show: $(xi-y)m(x+iy)^{m-1}=im[(x+iy)^m]$ for $m>0$
I have tried using induction, but for $m=1$, I get $(ix-y)=y$
Therefore doesn't hold for $m=1$
I feel that I may have made a mistake, going to check again.
Just realised the answer, I was looking for the imaginary part of the right hand side, when it was actually $i$ times $m$. Carrying on from my last line, I have
Show: $(xi-y)m(x+iy)^{m-1}=im(x+iy)^m$
RHS $=im(x+iy)^m=im(x+iy)(x+iy)^{m-1}=m(ix-y)(x+iy)^{m-1}=$ LHS
That $im$ notation is annoying, for all of my working I thought of it as the imaginary part.