I am dealing with a Lagrangian that depends on 6 complex fields plus their complex conjugate, hence, in total 12 complex fields. The Lagrangian is very long and I do not think it is necessary to show you how it looks like term by term. I want to expand this Lagrangian up to second order around the equilibrium values of those 6+6 complex fields. I was advised to separate all the fields in their real and imaginary parts, and then Taylor expand with respect to them. I did it, but since it is a very very long expression, I think it is better if I Taylor expand with respect to the complex fields without separating them into real and imaginary parts.
Are these two ways of Taylorexpansion equivalent?
To be concrete, let's say, we have a two dimensional Lagrangian $L(z,w,z^*,w^*)$, where $z$ and $w$ are complex fields. And the equilibrium values are $z_0$ and $w_0$. A multidimensional Taylor expansion around this points would be: $$ L(z,w,z^*,w^*)= L(z_0,w_0,z^*_0,w^*_0)+\frac{\partial L}{\partial z}\lvert_{z=z_0}(z-z_0)+\frac{\partial L}{\partial z^*}\lvert_{z^*=z^*_0}(z^*-z^*_0)+\frac{\partial L}{\partial w}\lvert_{w=w_0}(w-w_0)+\frac{\partial L}{\partial w^*}\lvert_{w^*=w^*_0}(w^*-w^*_0)+.... $$ After separating the field into their real and imaginary parts, the Lagrangian is $L(z,w)=L(z^r,z^i,w^r,w^i)$, where $r/i$ denote real and imaginary parts. Now, it is a four dimensional Taylor expansion $$ L(z^r,z^i,w^r,w^i)= L(z^r_0,z^i_0,w^r_0,w^i_0)+\frac{\partial L}{\partial z^r}\lvert_{z^r=z^r_0}(z^r-z^r_0)+\frac{\partial L}{\partial z^i}\lvert_{z^i=z^i_0}(z^i-z^i_0)+\frac{\partial L}{\partial w^r}\lvert_{w^r=w^r_0}(w^r-w^r_0)+\frac{\partial L}{\partial w^i}\lvert_{w^i=w^i_0}(w^i-w^i_0)+....\qquad . $$ Are these two expansions equivalent? I'd be happy to see plausible reasons, if not a proof. Thank you in advance.