Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 2F (Topics in Analysis)

The complex Taylor theorem says that an analytic function has the expansionthroughout every disk centred at $z_0$ contained in $D$. In particular it converges for $|z-z_0|<\mathrm{dist}(z_0,\mathbb{C}\setminus D)$, and uniformly on every smaller closed disk. The coefficients are uniquely its derivatives divided by factorials. The convergence radius can be larger if the function extends beyond the boundary of $D$; the theorem does not say that the given domain determines its exact radius.

An elementary geometric-series construction suffices, without assuming a general polynomial-approximation theorem. Rotate the square by a multiple of $\pi /2$ so that $\Re w>1$. Put $c=-R$ with $R>0$. Every point of the square then satisfies $|z-c{|}^2\le (R+1{)}^2+1$, whereasfor sufficiently large $R$. Thus the whole square lies inside a disk centred at $c$ whose radius is less than $|w-c|$. The other three ways $w$ can lie outside the square are handled by the indicated rotations, after which rotating back preserves polynomial form.

Now expandLet $\rho ={\max }_{z\in K}|z-c|/|w-c|<1$. The tail after degree $m$ is uniformly bounded by ${\rho }^{m+1}/[|w-c|(1-\rho )]$, tending to zero. Hence the displayed partial sums are the required uniform polynomial approximations. This is the Taylor series about a carefully chosen distant centre, rather than a potentially inadequate series about $0$.