- Find all possible values of
\[ \int _ { \gamma } { \frac { e ^ { \pi z } } { ( z - 1 ) ( z - i ) ^ { 2 } } } d z \]
where γ ranges over all simple closed smooth curves contained in \(\mathbb { C } \setminus \{ 1 , i \}\) . (A simple closed curve is a closed curve that does not intersect itself; i.e., it is a homeomorphic image of the circle.)
You do not need to give a proof for your answer to this problem, but show all your work.
Solution
Set \[ A=-\pi e^\pi,\qquad B=-\pi(\pi+1)+i\pi^2. \] The complete set of values is \[ \boxed{\{0,A,-A,B,-B,A+B,-A-B\}.} \]
1. The two poles give the contributions \(A\) and \(B\) for positive winding number one.
Write \(F(z)=e^{\pi z}/((z-1)(z-i)^2)\). At the simple pole \(1\), \[ \operatorname{Res}_{1}F=\frac{e^\pi}{(1-i)^2} =\frac{i e^\pi}{2}, \qquad 2\pi i\operatorname{Res}_{1}F=A. \] At the double pole \(i\), the derivative formula gives \[ \begin{aligned} \operatorname{Res}_{i}F &=\left.\frac{d}{dz}\frac{e^{\pi z}}{z-1}\right|_{z=i}\\ &=-\frac{\pi}{i-1}+\frac1{(i-1)^2} =\frac\pi2+\frac{i(\pi+1)}2, \end{aligned} \] where \(e^{\pi i}=-1\), \((i-1)^{-1}=(-1-i)/2\), and \((i-1)^{-2}=i/2\). Multiplication by \(2\pi i\) gives \(B\). These are the only poles [@SS03].
2. A simple closed curve allows exactly the displayed residue combinations.
A simple closed smooth curve has index zero at every point of its unbounded complementary component and a common index \(\varepsilon\in\{1,-1\}\) at every point of its bounded component, according to orientation [@SS03]. The two poles may lie outside, only \(1\) may lie inside, only \(i\) may lie inside, or both may lie inside. The residue theorem consequently gives, respectively, \[ 0,\qquad \varepsilon A,\qquad \varepsilon B, \qquad \varepsilon(A+B). \] In particular a curve cannot have index \(1\) at one enclosed pole and \(-1\) at the other while remaining simple. Thus no other combination is allowed.
3. Every listed value is attained.
The circle \(|z|=1/4\) encloses neither pole. The circles \(|z-1|=1/4\) and \(|z-i|=1/4\) enclose exactly \(1\) and \(i\), respectively, because their distance is \(\sqrt2\). The circle \(|z|=2\) encloses both. None passes through a pole, and all are smooth and simple. Counterclockwise orientation gives \(0,A,B,A+B\); reversing the last three orientations gives their negatives. This proves both attainability and exhaustion of the stated set.