Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2007 / 5 / Solution

For a holomorphic function on $|z-a|<R$, the complex Taylor theorem stateswith 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 seriesconverges uniformly. Substitute it into Cauchy integral formula and integrate term by term to getIf $M_{\rho }={\max }_{|\zeta -a|=\rho }|f(\zeta )|$, 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 $n=0$, giveThe 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: