Let \(C\) be the closed curve defined by two pieces: the first piece is given by the set of all \(z\) satisfying \(|z - 1| = 3\) and \(\operatorname{Re}(z - 1) \geq 0\). The second piece is the straight line segment from \(1 + 3i\) to \(1 - 3i\). Orient \(C\) in the counterclockwise direction, and let \(\Omega\) be the region enclosed by \(C\). Suppose \(f\) is holomorphic in a neighborhood of \(\overline{\Omega}\) with no zeros on \(C\). Suppose also that:
\[\frac{1}{2\pi i} \int_C \frac{zf'(z)}{f(z)} \, dz = 3 \qquad \text{and} \qquad \frac{1}{2\pi i} \int_C \frac{z^2 f'(z)}{f(z)} \, dz = \frac{5}{2}.\]
Determine all the zeros of \(f\) in \(\Omega\) explicitly.
Solution
The zeros in \(\Omega\), counted with multiplicity, are \[ \boxed{\frac32+i\quad\text{and}\quad\frac32-i}. \]
1. The two contour integrals are the first two power sums of the zeros.
Let the zeros of \(f\) in \(\Omega\), repeated according to multiplicity, be \(\zeta_1,\dots,\zeta_N\). There are finitely many because \(f\) is holomorphic on a neighborhood of \(\overline\Omega\), has no zero on \(C\), and is not identically zero.
If \(\zeta\) is a zero of multiplicity \(m\), then locally \(f(z)=(z-\zeta)^m h(z)\) with \(h(\zeta)\ne0\), so \[ \frac{f'(z)}{f(z)}=\frac{m}{z-\zeta}+\frac{h'(z)}{h(z)}. \] Hence the residues of \(z f'/f\) and \(z^2f'/f\) at \(\zeta\) are respectively \(m\zeta\) and \(m\zeta^2\). The residue theorem therefore gives \[ \sum_{j=1}^N\zeta_j=3, \qquad \sum_{j=1}^N\zeta_j^2=\frac52. \]
2. The geometry of \(\Omega\) forces \(N=2\).
The contour consists of the right semicircle centered at \(1\) of radius \(3\) and the vertical segment from \(1+3i\) to \(1-3i\). Thus every point of the interior \(\Omega\) satisfies \[ \operatorname{Re}z>1. \] Taking real parts in the first power-sum identity gives \[ 3=\sum_{j=1}^N\operatorname{Re}\zeta_j>N. \] Hence \(N\le2\). If \(N=1\), the first identity would give \(\zeta_1=3\), but then the second would give \(\zeta_1^2=9\), contradicting \(5/2\). Thus \(N=2\).
3. The two power sums determine the two zeros uniquely.
Let the two zeros be \(\zeta_1,\zeta_2\). Then \[ \zeta_1+\zeta_2=3, \qquad \zeta_1^2+\zeta_2^2=\frac52. \] Using \((\zeta_1+\zeta_2)^2=\zeta_1^2+\zeta_2^2+2\zeta_1\zeta_2\) gives \[ \zeta_1\zeta_2=\frac{9-5/2}{2}=\frac{13}{4}. \] Therefore \(\zeta_1,\zeta_2\) are the roots of \[ t^2-3t+\frac{13}{4}=0, \] namely \[ \frac{3\pm\sqrt{-4}}2=\frac32\pm i. \] Both lie in \(\Omega\) because their real part is \(3/2>1\) and \[ \left|\left(\frac32\pm i\right)-1\right|^2 =\frac14+1=\frac54<9. \] Thus these are exactly all the zeros requested.