Let \(D=\{z\in\mathbb C:|z|<1\}\), and let \(f:D\to\mathbb C\) be holomorphic with simple zeros at \(1/3\), \(2/3\) and \(i/4\) and no other zeros. Evaluate \[ \int_{|z|=1/2}(z^2-1)e^z\frac{f'(z)}{f(z)}\,dz, \] with counterclockwise orientation.
Solution
The integral is \[ \boxed{-2\pi i\left(\frac89e^{1/3}+\frac{17}{16}e^{i/4}\right).} \]
1. Each simple zero contributes the value of the holomorphic weight as a residue.
Set \(W(z)=(z^2-1)e^z\). At a simple zero \(a\) of \(f\), the Taylor expansion gives \(f(z)=(z-a)h(z)\) with \(h\) holomorphic and \(h(a)\ne0\). In a sufficiently small disk, \[ \frac{f'(z)}{f(z)}=\frac1{z-a}+\frac{h'(z)}{h(z)}. \] The second term is holomorphic, so \[ \operatorname{Res}_{a}\left(W\frac{f'}f\right)=W(a). \] Away from the prescribed zeros, the same integrand is holomorphic. These facts determine all its possible poles on the unit disk.
2. The two enclosed zeros give the stated sum.
The zeros \(1/3\) and \(i/4\) lie inside \(|z|=1/2\), while \(2/3\) lies outside. None lies on the contour. The closed radius-one-half disk has a neighborhood contained in \(D\), so the residue theorem applies to this counterclockwise circle [@SS03]. Using step 1, it gives \[ \int_{|z|=1/2}W(z)\frac{f'(z)}{f(z)}\,dz =2\pi i\bigl(W(1/3)+W(i/4)\bigr). \] Finally, \[ W(1/3)=-\frac89e^{1/3},\qquad W(i/4)=\left(-\frac1{16}-1\right)e^{i/4} =-\frac{17}{16}e^{i/4}. \] Substitution yields the displayed answer.