Let \(\Omega\subset\mathbb C\) be a bounded domain whose boundary is a smooth Jordan curve \(\gamma\). Let \(f\) be holomorphic on a neighborhood of \(\overline\Omega\), assume \(f\ne0\) on \(\gamma\), and let \(z_1,\ldots,z_k\) be the zeros of \(f\) in \(\Omega\) with multiplicities \(n_1,\ldots,n_k\).
- Using Cauchy’s integral formula, prove \[ \frac1{2\pi i}\int_\gamma\frac{f'(z)}{f(z)}\,dz =\sum_{j=1}^k n_j. \]
- If \(f\) has exactly one zero \(z_1\) in \(\Omega\), and it is simple, find a boundary integral involving \(f\) whose value is \(z_1\).
Solution
Orient \(\gamma\) positively. Define \[ P(z)=\prod_{j=1}^k(z-z_j)^{n_j}. \]
1. There is a holomorphic function \(g\), nonvanishing on aneighborhood of \(\overline\Omega\), such that\[ f(z)=g(z)P(z) \]there.
The quotient \(f/P\) is holomorphic away from the points \(z_j\). Since \(n_j\) is exactly the order of the zero of \(f\) at \(z_j\), each apparent singularity is removable and the quotient extends holomorphically across \(z_j\). Call the extension \(g\).
The zeros \(z_1,\ldots,z_k\) are all the zeros of \(f\) in \(\Omega\), and \(f\) is nonzero on \(\gamma\), so \(g\) has no zeros on \(\overline\Omega\). By compactness, after shrinking the ambient neighborhood if necessary, \(g\) is nonvanishing on a neighborhood of \(\overline\Omega\).
2. Part 1 gives\[ \boxed{ \frac1{2\pi i} \int_\gamma\frac{f'(z)}{f(z)}\,dz = \sum_{j=1}^k n_j }. \]
By step 1, away from the zeros of \(f\), \[ \frac{f'(z)}{f(z)} = \frac{g'(z)}{g(z)} + \sum_{j=1}^k\frac{n_j}{z-z_j}. \] Since \(g\) is nonvanishing near \(\overline\Omega\), the function \(g'/g\) is holomorphic there, so Cauchy’s theorem gives \[ \int_\gamma\frac{g'(z)}{g(z)}\,dz=0. \] For each \(j\), Cauchy’s integral formula applied to the constant function \(1\) gives \[ \frac1{2\pi i} \int_\gamma\frac{1}{z-z_j}\,dz = 1. \] Integrating the logarithmic-derivative identity therefore yields the displayed formula.
3. More generally,\[ \frac1{2\pi i} \int_\gamma z\frac{f'(z)}{f(z)}\,dz = \sum_{j=1}^k n_jz_j. \]
Differentiating the factorization from step 1 and multiplying the resulting logarithmic-derivative identity by \(z\) gives \[ z\frac{f'(z)}{f(z)} = z\frac{g'(z)}{g(z)} + \sum_{j=1}^k n_j\frac{z}{z-z_j}. \] The first term is holomorphic near \(\overline\Omega\), so its boundary integral is zero. Cauchy’s integral formula applied to the holomorphic function \(h(z)=z\) gives \[ \frac1{2\pi i} \int_\gamma\frac{z}{z-z_j}\,dz = z_j. \] Summing over \(j\) proves the formula.
4. Under the hypothesis of part 2,\[ \boxed{ z_1 = \frac1{2\pi i} \int_\gamma z\frac{f'(z)}{f(z)}\,dz }. \]
If \(z_1\) is the only zero and it is simple, then \(k=1\) and \(n_1=1\). Step 3 reduces exactly to the displayed boundary integral.
5. Q.E.D.
Step 2 proves part 1, and step 4 gives the boundary integral requested in part 2.