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A function on a domain invariant under $x\mapsto -x$ is even when $f(-x)=f(x)$. Its graph is symmetric about the vertical axis. Examples include $x^2$, $\cos x$ and the centered Gaussian function $e^{-x^2}$. On a symmetric finite interval, an integrable even function satisfies ${\int }_{-a}^{a}f=2{\int }_0^{a}f$.

When a Taylor series about zero represents the function, its odd coefficients vanish: substituting $-x$ and comparing coefficients gives $(-1{)}^n{a}_n=a_n$. The analogous statement holds for a Laurent series on an annulus centered at zero. Parity does not itself guarantee such a series representation. For instance, the even smooth function $e^{-1/x^2}$, extended by zero at zero, has every Taylor coefficient zero there while being positive away from zero.

For $f(z)=z+{\sum }_{n\ge 2}{a}_n{z}^n$ in the normalized univalent function class, the integral mean bound for an odd square-root transform gives $(2\pi {)}^{-1}\int |f(r{e}^{i\theta })|d\theta \le r/(1-r)$. The Cauchy integral formula for Taylor coefficients therefore yields $|a_n|\le r^{1-n}/(1-r)$. Choosing $r=(n-1)/n$ and applying the logarithm tangent bound gives the displayed bound for every $n\ge 2$.

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:

The identity theorem for holomorphic functions says that a holomorphic function on a connected open set which has distinct zeros accumulating at a point of the set is identically zero. Equivalently, two holomorphic functions agreeing on such a set of points agree everywhere in that connected open set.

To prove it, expand at an accumulation point $a$. If the first nonzero Taylor coefficient had index $m$, the function would factor as $(z-a{)}^{m}h(z)$ with $h(a)\ne 0$. Continuity would make $h$ nonzero near $a$, so no other zeros could approach $a$, a contradiction. Thus all coefficients vanish and the function vanishes on a neighborhood of $a$. Let $E$ be the set of points having such a zero neighborhood. It is nonempty and open. If a point in the domain is a limit point of $E$, it is either already in $E$ or is an accumulation point of distinct zeros, and the same Taylor argument puts it in $E$. Thus $E$ is relatively closed as well. Connectedness forces $E$ to be the whole domain. Apply this result to the difference of two functions for the equivalent formulation.

The identity theorem for holomorphic functions says that a holomorphic function on a connected open set which has distinct zeros accumulating at a point of the set is identically zero. Equivalently, two holomorphic functions agreeing on such a set of points agree everywhere in that connected open set.

To prove it, expand at an accumulation point $a$. If the first nonzero Taylor coefficient had index $m$, the function would factor as $(z-a{)}^{m}h(z)$ with $h(a)\ne 0$. Continuity would make $h$ nonzero near $a$, so no other zeros could approach $a$, a contradiction. Thus all coefficients vanish and the function vanishes on a neighborhood of $a$. Let $E$ be the set of points having such a zero neighborhood. It is nonempty and open. If a point in the domain is a limit point of $E$, it is either already in $E$ or is an accumulation point of distinct zeros, and the same Taylor argument puts it in $E$. Thus $E$ is relatively closed as well. Connectedness forces $E$ to be the whole domain. Apply this result to the difference of two functions for the equivalent formulation.

The complex Hahn-Banach theorem says that every bounded complex-linear functional on a linear subspace of a normed complex space extends to the whole space with the same norm.

For the separation assertion choose $x_0\notin M$ and $d=\mathrm{dist}(x_0,M)>0$. On $M+\mathbb{C}{x}_0$ define $g_0(m+\lambda x_0)=\lambda d$. The inequality $\Vert m+\lambda x_0\Vert \ge |\lambda |d$ makes its norm at most one; approximating the distance $d$ shows the norm is exactly one. Hahn-Banach gives an extension $g$ of norm one annihilating $M$.

Each geometric sequence satisfies $\Vert x_k{\Vert }_1=(1-|{\beta }_k|{)}^{-1}\le (1-\delta {)}^{-1}$. Let $M$ be their closed linear span. If $M\ne {\ell }^1$, the separation just proved gives a nonzero bounded functional annihilating all $x_k$. Represent it by $b\in {\ell }^{\infty }$ and formThe series converges locally uniformly and defines a holomorphic function. It vanishes at every ${\beta }_k$. These distinct points lie in the compact disk $|z|\le \delta <1$, so have an accumulation point inside the domain. The identity theorem gives $F=0$, hence every Taylor coefficient $b_j=0$, contradicting the functional's nonzero norm. Therefore

The constant and linear Taylor terms cancel in the Newton step, leavingThus residuals strictly decrease whenever the current residual is nonzero, and the steps tend to zero doubly exponentially. By the ultrametric inequality, the distance between any two sufficiently late iterates is bounded by the largest intervening step. The sequence is Cauchy and converges in the complete valuation ring to $a$. Continuity gives $f(a)=0$, andThe literal infinite chain of strict inequalities needs the root-hit exception: if $f(a_0)=0$, or a later iterate is already a root, it cannot keep strictly decreasing below zero. The construction is eventually constant in that case; all existence and distance conclusions remain valid.

For the equality case, suppose $|f(a_0)|=d^2$ with $0<d<1$. The first Newton step has norm $|h_0|=d$. Write the Taylor coefficient of $h^2$ as $c_2(a_0)=f^{\prime \prime }(a_0)/2$. Its norm is strictly below one by hypothesis; all higher coefficients have norm at most one. HenceThe derivative changes by $2c_2(a_0)h_0+O(h_0^2)$, of norm strictly below $d$, so it still has norm $d$. The strict Hensel argument applies from this first iterate onward, and the root remains within distance $d=|f(a_0)|/d$ of $a_0$. If characteristic two is admitted, interpret the displayed half-second-derivative as the second Taylor coefficient ${\sum }_j\left(\genfrac{}{}{0ex}{}{j}{2}\right)c_j{b}^{j-2}$; this is the division-free form of the required hypothesis.

For the final local equation put $F(t)=t^3+3t$. Its values modulo nine are $0,4,5$, so the sum of two such values is confined to $0,1,4,5,8$. Conversely choose initial pairs for these five classes asrespectively. Fix the second coordinate $y_0$ and apply the equality-case Hensel result to $g(X)=F(X)+F(y_0)-n$ at the first coordinate $x_0$. Its residual is divisible by nine, $g^{\prime }(x_0)=3(x_0^2+1)$ has 3-adic norm $1/3$, and $g^{\prime \prime }(b)/2=3b$ has norm below one for every $b\in {\mathbb{Z}}_3$. The strict or equality case therefore supplies a root in ${\mathbb{Z}}_3$. Thusis the necessary and sufficient condition.

Use $h(\sqrt{r}{e}^{i\theta }{)}^2=\varphi (r{e}^{2i\theta })$. As $\theta$ runs around the circle, the doubled angle goes around it twice, so iv givesThe Cauchy integral formula for the Taylor coefficients now impliesFor $n\ge 2$, choosing $r=(n-1)/n$ gives the linear coefficient bound for normalized univalent functions,The last inequality follows from $\log (1+x)\le x$ for $x>-1$.