let $f$ be a holomorphic function on $D=\mathbb{D}(0,1)$
Let $f(z)=\sum \frac{f^{(n)}(0)}{n!}z^n$ be its Taylor expansion.
If I use the Taylor McLaurin inequality:
$||f(z) - f(0) +f'(0)z|| \le Mz^2/2 $ where $M=Sup ||f''(z')||$ for $z' \in D(0,z)$?
Is it still valid for holomorphic function like in real analysis?
Thank you for your help.
Well, you need an absolute value on the $z$ ($z$ could be negative, of course). The integral form of the remainder remains valid, $$ f(z) - \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(z-a)^k = \frac{1}{n!} \int_a^z (z-w)^n f^{(n+1)}(w) \, dw, $$ where the path can be chosen to be the line segment joining $a$ to $z$, but need not be. Then the right-hand side is bounded by $$ \frac{\lvert z-a \rvert^{n+1}}{(n+1)!} \sup_{w \in [a,z]}{\lvert f^{(n+1)}(w) \rvert} $$ by a simple integral estimate, which is in turn bounded by $$ \frac{\lvert z-a \rvert^{n+1}}{(n+1)!} \sup_{\lvert w-a \rvert < \lvert z-a \rvert}{\lvert f^{(n+1)}(w) \rvert} = \frac{\lvert z-a \rvert^{n+1}}{(n+1)!} \sup_{\lvert w-a \rvert = \lvert z-a \rvert}{\lvert f^{(n+1)}(w) \rvert} $$ by the Maximum Modulus Theorem.