In the book we are presented a proof for the inverse mapping theorem which I understand for a) and b) but fail to fully understand understand for c).
My approach:
The way I understand it is that after we let $w=z-z_0$ and define $F(w)= f(z)-f(z_0)= \sum a_nw^n$ . Then we set $w=0 =>F(w)=0$. From there we apply the previous discussion for $0$ and we arrive that $G(w)$ and $F(w)$ beeing inverses.
But from here I am puzzled, we haven't proved that $G(w)$ is analytic and Serge Lang also included the line " Let $w_0=f(z_0)$ and let $g(w)=G(w-w_o)+z_0$" ,which is the same as $g(w)=G(w)+z_o$, and I can't figure out why this line implies the result. Furthermore, in the theorem we have the bit "Suppose that $f'(z_0) \neq 0$" which I don't see beeing used anywhere in the proof. Can anybody please clarify the proof that Lang intended?
This proof is not really well written. I'll try to sketch what is going on. First make the reduction to $z_0 = f(z_0) = 0$. Being $f$ analytic we have that the $f$ has a power series expansion at $0$ that converges in some open neighborhood of $0$. From $f'(0) \neq 0$, we see that $f$ has a formal inverse $g$ by item (a). (Here is where he uses that the derivative does not vanish). By item (b), $g$ converges in some neighborhood of $0$ i.e. $g$ is also an analytic function.
Then he proceeds to prove that $g$ is in fact an inverse to $f$ for a good choice of neighborhoods $U_0$ and $V_0$.
Added: Here I am adding a clarification to the reduction step. I hope this is easier to follow as the coordinate changes are written as compositions of maps.
We have two translations $T_{z_0}(z)= z+z_0$ and $T_{w_0} =w+w_0$ where $w_0=f(z_0)$. Then $$F(z) = \left(T_{w_0}^{-1} \circ f \circ T_{z_0}\right) (z) = f(z+z_0)-w_0$$ From this we see that $F(0) = f(z_0)-w_0 =0$. Then Let $G$ be the inverse of $F$ and define $$g(w) = \left( T_{z_0} \circ G\circ T_{w_0}^{-1}\right)(w)= G(w-w_0)+z_0$$ By construction, $f(z) = \left(T_{w_0} \circ F\circ T_{z_0}^{-1}\right)(z)$. Hence
$$f(g(w)) = \left(T_{w_0} \circ F\circ T_{z_0}^{-1}\circ T_{z_0} \circ G\circ T_{w_0}^{-1}\right)(w) = \left(T_{w_0} \circ F\circ G\circ T_{w_0}^{-1}\right)(w) =\left(T_{w_0} \circ T_{w_0}^{-1}\right)(w) = w $$