Complex Taylor series

The coefficients come from Cauchy's integral formula: \(a_n=\frac1{2\pi i}\oint_{|\zeta-z_0|=r}\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}d\zeta=\frac{f^{(n)}(z_0)}{n!}\). Expanding \(\frac1{\zeta-z}\) as a geometric series inside the integral proves the series converges to \(f(z)\) for every \(z\) in the disc. Thus holomorphic means analytic, which fails in the real case.

The radius of convergence is the distance from \(z_0\) to the nearest singularity of \(f\), not to a real singularity. That is why the series of \(1/(1+x^2)\) at \(0\) converges only for \(|x|<1\) even though the function is smooth on the whole line: the poles \(\pm i\) are at distance \(1\). Changing the centre changes the radius.

In practice one rarely differentiates \(n\) times: one starts from known series (\(e^z\), \(\sin z\), \(\cos z\), \(\frac1{1-z}\)) and substitutes, multiplies or differentiates term by term. The series is unique: any power series in \(z-z_0\) representing \(f\) near \(z_0\) is its Taylor series.

Zeros can be read from the series: if \(a_0=\cdots=a_{m-1}=0\) and \(a_m\neq0\), the zero has order \(m\). The case where \(f\) is holomorphic not on a disc but on an annulus is handled by the Laurent series. Typical mistake: taking the radius to be \(\infty\) because the function is differentiable on the real axis.