Given $n=(x + iy)/\sqrt{2}L$ and $\overline n=(x - iy)/\sqrt{2}L$.
$\partial_n=\partial/\partial n$, $\partial_x=\partial/\partial x$, $\partial_y=\partial/\partial y$
Show that $\partial_n$ = $L(\partial_x - i \partial_y)/\sqrt{2}$ and $\partial_\overline n$ = $L(\partial_x + i \partial_y)/\sqrt{2}$.
I'm not sure on how to start proving these but should I use the definition for the Wirtinger derivatives?
If ever I use that, how can $\sqrt2 L$ be in the numberator? Any suggestions?
You do not need the Wirtinger derivatives, but do know that the Wirtinger derivatives are derived the same way when we do not have the $\sqrt{2}L$ factor present. Anyways, notice that we can redefine $x$ and $y$ in terms of $n$ and $\bar{n}$:
$$x(n,\bar{n}) = \frac{L}{\sqrt{2}}(n + \bar{n})$$
$$y(n,\bar{n}) = \frac{L}{\sqrt{2}i}(n - \bar{n})$$
We are going to need the partial derivatives:
$$\partial_nx = \frac{L}{\sqrt{2}}, \hspace{.5cm} \partial_\bar{n}x = \frac{L}{\sqrt{2}}$$
$$\partial_ny = -\frac{iL}{\sqrt{2}}, \hspace{.5cm} \partial_\bar{n}y = \frac{iL}{\sqrt{2}}$$
Now take any function in terms of $x$ and $y$ and parametrize it:
$$f(x,y) = f(x(n,\bar{n}),y(n,\bar{n}))$$
and find the derivatives using the chain rule:
$$\partial_nf = \partial_xf \partial_nx + \partial_yf \partial_ny = \frac{L}{\sqrt{2}}\partial_xf - i\frac{L}{\sqrt{2}}\partial_yf = \frac{L}{\sqrt{2}}(\partial_x - i\partial_y)f$$
$$\partial_\bar{n}f = \partial_xf \partial_\bar{n}x + \partial_yf \partial_\bar{n}y = \frac{L}{\sqrt{2}}\partial_xf + i\frac{L}{\sqrt{2}}\partial_yf = \frac{L}{\sqrt{2}}(\partial_x + i\partial_y)f$$
The last step is to notice that we are operating on the same scalar function on both sides. Therefore, the operators acting on both sides must be equivalent giving the desired result:
$$\partial_n = \frac{L}{\sqrt{2}}(\partial_x - i\partial_y)$$
$$\partial_\bar{n} = \frac{L}{\sqrt{2}}(\partial_x + i\partial_y)$$