= Solution
For a <holomorphic function> on $|z-a|<R$, the complex <Taylor theorem> states
$$
f(z)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(z-a)^n\qquad(|z-a|<R),
$$
with uniform convergence on each smaller closed <disc>. Here is a proof. Choose $r<\rho<R$. For $|z-a|\le r$ and $|\zeta-a|=\rho$, the <geometric series>
$$
\frac1{\zeta-z}=\sum_{n=0}^{\infty}\frac{(z-a)^n}{(\zeta-a)^{n+1}}
$$
converges uniformly. Substitute it into <Cauchy integral formula> and integrate term by term to get
$$
f(z)=\sum_{n=0}^{\infty}c_n(z-a)^n,\qquad
c_n=\frac1{2\pi i}\int_{|\zeta-a|=\rho}\frac{f(\zeta)}{(\zeta-a)^{n+1}}\,d\zeta.
$$
If $M_\rho=\max_{|\zeta-a|=\rho}|f(\zeta)|$, then $|c_n|\le M_\rho\rho^{-n}$. These bounds justify differentiating the <power series> any fixed number of times on a smaller closed <disc>, by comparison with convergent differentiated <geometric series>. Evaluation at $a$ gives $f^{(n)}(a)=n!c_n$. Since every $|z-a|<R$ belongs to some such smaller <disc>, the formula holds throughout the stated <domain>.
The <identity theorem> says that two <holomorphic functions> on a connected open subset of $\mathbb C$ are equal everywhere if their equality set has an accumulation point in that subset. Equivalently, a <holomorphic function> with an interior accumulation point of zeros is identically zero there.
The proposed <Taylor coefficients> at $i$, including the specified value for $n=0$, give
$$
\sum_{n=0}^{\infty}\frac{(-1)^n(z-i)^n}{i^{n+1}}
=\frac{1/i}{1+(z-i)/i}=\frac1z\qquad(|z-i|<1).
$$
The function $1/z$ is <holomorphic> on $\mathbb C\setminus\{0\}$ and has exactly the given <derivatives>, so it supplies existence. Any other admissible <holomorphic function> has the same <Taylor series> on $|z-i|<1$. The punctured plane is connected, so the <identity theorem> gives uniqueness:
$$
\boxed{f(z)=\frac1z\quad(z\ne0).}
$$
Back to article page
