Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 3 / 13E (Further Analysis)

The complex Taylor series theorem states that if $f$ is holomorphic in $|z-z_0|<R$, thenThe power series converges absolutely there and uniformly on each smaller closed disc; its coefficients are unique. For $|z-z_0|<r<R$, the Cauchy integral formula also gives $f^{(n)}(z_0)/n!=(2\pi i{)}^{-1}{\int }_{|\zeta -z_0|=r}f(\zeta )(\zeta -z_0{)}^{-n-1}d\zeta$. This is the analytic, rather than finite real-variable, form of Taylor theorem relevant here.

Let $h=f-g$, whose power series coefficients are $c_n=a_n-b_n$. Suppose some coefficient is nonzero, and let $N$ be the least such index. Inside the common convergence disc,The parenthesized analytic function tends to $c_N\ne 0$ as $z\to z_0$, so it does not vanish in some neighborhood. Hence $h$ has no zeros sufficiently near $z_0$ except possibly $z_0$ itself. This contradicts $h(z_k)=0$ with $z_k\ne z_0$ and $z_k\to z_0$. Thus all coefficients agree:This also proves the local isolation of zeros of a nonzero analytic function.

Put $h=f-g$. Choose a disc about $z_0$ contained in the domain $D$. Part (a) gives a convergent Taylor series there, and the accumulating zeros together with part (b) force all its coefficients to vanish. Thus $h=0$ near $z_0$.

To propagate this to the entire domain, defineThe set is nonempty by the preceding local argument. Each derivative is continuous, so $S$ is closed relative to $D$ as the intersection of their zero sets. If $z\in S$, its Taylor series is zero on a disc about $z$, and all derivatives vanish throughout that disc; hence $S$ is open relative to $D$. Connectedness forces $S=D$. Therefore $f=g$ throughout $D$, proving the identity theorem for holomorphic functions rather than merely invoking it.

Take the analytic functionIt vanishes at $z=1/n$ for every positive integer $n$, but $f(2)=1$, so it is not identically zero. The identity theorem does not apply at the accumulation point $0$, because that point lies outside the domain.

Let $f$ have the stated zeros but not be identically zero. The origin is an isolated singularity. If it were a removable singularity, the extended holomorphic function would have zeros accumulating at an interior point. The proof in part (c) would force the extension, and then $f$ on its domain, to be identically zero.

If the origin were a pole of order $m$, then $g(z)=z^{m}f(z)$ would have a holomorphic extension with $g(0)\ne 0$. Continuity would make $g$ nonzero on a small disc, so $f$ would have no zeros in its punctured disc, again a contradiction. The classification of isolated singularities leaves only an essential singularity at the origin. This argument applies to every function in part (d), not just the particular sine example.