Let \(f(z)\) be complex analytic in the disc \(|z| < \pi\). Assume that the only zero of \(f\) in the closed unit disc \(\overline{D} = \{z : |z| \leq 1\}\) is a simple zero at the origin. Let \(C\) be the unit circle, oriented counterclockwise. Evaluate \[ \int_C \frac{dz}{f(z)}, \] in the sense that no integration symbols should appear in the answer.
Let \(f\) be as in part (a), except assume that \(f\) has a 2nd-order (i.e., double) zero at the origin. Verify or give a counterexample: \[ \text{Assertion:}\quad \int_C \frac{dz}{f(z)} = 0. \]
Solution
1. In part (a), the integral is \(2\pi i/f'(0)\).
Because the zero at the origin is simple, write \[ f(z)=z g(z) \] with \(g\) holomorphic near the closed unit disk and \(g(0)=f'(0)\ne0\). The hypothesis that zero is the only zero of \(f\) in the closed unit disk implies that \(g\) has no zero there. Hence \(1/f\) is meromorphic inside and on \(C\), with its only pole at zero, and that pole is simple. Its residue is \[ \operatorname{Res}_{0}\frac1{f(z)} =\lim_{z\to0}\frac{z}{f(z)} =\frac1{g(0)}=\frac1{f'(0)}. \] Since \(C\) is counterclockwise, the residue theorem gives \[ \boxed{\int_C\frac{dz}{f(z)}=\frac{2\pi i}{f'(0)}}. \]
2. The assertion in part (b) is false.
Take \[ f(z)=z^2e^z. \] This function is entire and has exactly one zero, a double zero at the origin, so it satisfies the hypotheses of part (b). Its reciprocal has Laurent expansion \[ \frac1{f(z)}=\frac{e^{-z}}{z^2} =\frac1{z^2}-\frac1z+\frac12+\cdots. \] Thus \[ \operatorname{Res}_{0}\frac1f=-1. \] There are no other poles inside the unit circle, so the residue theorem yields \[ \int_C\frac{dz}{f(z)}=-2\pi i\ne0. \] Therefore the stated assertion is false.