Assume \(f\) is continuous in the region: \[ 0 < \abs{z-a} \leq R,\quad 0 \leq \Arg(z-a) \leq \beta_0 \qquad \beta_0\in (0, 2\pi] .\]
and the following limit exists: \[ \lim_{z\to a}(z-a)f(z) = A .\] Show that \[\lim_{r \rightarrow 0} \int_{\gamma_r} f(z) dz = i A \beta_0 \; , \; \;\] where \[ \gamma_r : = \{ z \; | \; z = a + r e^{it}, \; 0 \leq t \leq \beta_0 \}. .\]
Solution
Goal: Let \(S = \{z \in \mathbb{C} : 0 < |z-a| \leq R, 0 \leq \arg(z-a) \leq \beta_0\}\) with \(0 < \beta_0 \leq 2\pi\). Let \(f: S \to \mathbb{C}\) be continuous with \(\lim_{z \to a, z \in S} (z-a) f(z) = A \in \mathbb{C}\). Prove that along the circular arcs \(\gamma_r(t) = a + r e^{it}\) (\(t \in [0, \beta_0]\)), \[\lim_{r \to 0^+} \int_{\gamma_r} f(z) \, dz = i A \beta_0.\]
Step 1: Parametrize the Integral and Isolate the Main Term
1. Parametrize \(\int_{\gamma_r} \frac{A}{z-a}\,dz\).
1.1. With \(\gamma_r(t) = a + r e^{it}\) for \(t \in [0, \beta_0]\), \(dz = i r e^{it} dt\).
Differentiating the parametrization \(\gamma_r(t) = a + r e^{it}\) with respect to \(t\) gives \(\gamma_r'(t) = i r e^{it}\), so \(dz = i r e^{it}\,dt\).
1.2. The integral of the simple pole term is: \[\int_{\gamma_r} \frac{A}{z-a} \, dz = \int_0^{\beta_0} \frac{A}{r e^{it}} \cdot i r e^{it} \, dt = i A \int_0^{\beta_0} dt = i A \beta_0.\]
Substituting \(z - a = r e^{it}\) and \(dz = i r e^{it}\,dt\), the factors \(r e^{it}\) cancel, leaving the integral of the constant \(iA\) over \([0, \beta_0]\), which is \(iA\beta_0\).
1.3. Q.E.D.
2. Express \(f(z)\) as \(\frac{A}{z-a} + \frac{g(z)}{z-a}\) where \(g(z) = (z-a)f(z) - A\).
2.1. Define \(g(z) = (z-a)f(z) - A\) for \(z \in S\).
This is the definition of \(g\): the difference between \((z-a)f(z)\) and its limit \(A\).
2.2. By the hypothesis \(\lim_{z \to a, z \in S} (z-a)f(z) = A\), we have \(\lim_{z \to a, z \in S} g(z) = 0\).
Subtracting the constant \(A\) from a function converging to \(A\) gives a function converging to \(0\).
2.3. For any \(z \in S\), \(f(z) = \frac{A + g(z)}{z-a} = \frac{A}{z-a} + \frac{g(z)}{z-a}\).
Since \(g(z) + A = (z-a)f(z)\), dividing both sides by \(z - a \neq 0\) (valid because \(z \in S\) has \(|z-a| > 0\)) gives the decomposition.
2.4. Q.E.D.
Step 2: Bound the Error Term
3. Show that \(\lim_{r \to 0^+} \int_{\gamma_r} \frac{g(z)}{z-a}\,dz = 0\).
3.1. Let \(\varepsilon > 0\) be given.
We prove the limit by the epsilon–delta definition: for an arbitrary \(\varepsilon > 0\) we exhibit a \(\delta\) such that the integral is bounded by \(\varepsilon\) whenever \(r < \delta\).
3.2. Since \(\lim_{z \to a, z \in S} g(z) = 0\), there exists \(\delta > 0\) (with \(\delta \leq R\)) such that for all \(z \in S\) with \(0 < |z-a| < \delta\), \(|g(z)| < \frac{\varepsilon}{\beta_0}\).
This is the definition of the limit \(\lim_{z \to a} g(z) = 0\) applied with the tolerance \(\frac{\varepsilon}{\beta_0}\).
3.3. For any \(r \in (0, \delta)\), the entire arc \(\gamma_r\) is contained in \(\{z \in S : 0 < |z-a| = r < \delta\}\).
For \(z = a + r e^{it}\) on \(\gamma_r\), we have \(|z - a| = |r e^{it}| = r < \delta\), and \(0 \le t \le \beta_0\) keeps \(z\) in \(S\).
3.4. On \(\gamma_r\), \(\left| \frac{g(z)}{z-a} \right| = \frac{|g(z)|}{r} < \frac{\varepsilon}{\beta_0 r}\).
Since \(|z - a| = r\) on \(\gamma_r\) and \(|g(z)| < \frac{\varepsilon}{\beta_0}\) by 2, dividing gives the bound.
3.5. The length of the arc \(\gamma_r\) is \(L(\gamma_r) = r \beta_0\).
The arc is a circular sector of radius \(r\) subtending angle \(\beta_0\), so its length is \(r \beta_0\).
3.6. By the \(ML\)-inequality: \[\left| \int_{\gamma_r} \frac{g(z)}{z-a} \, dz \right| \leq \left( \sup_{z \in \gamma_r} \left| \frac{g(z)}{z-a} \right| \right) \cdot L(\gamma_r) \leq \frac{\varepsilon}{\beta_0 r} \cdot (r \beta_0) = \varepsilon.\]
The \(ML\)-inequality bounds the modulus of a contour integral by the product of the maximum of the integrand and the length of the contour; substituting the bounds from 4 and 5 gives \(\varepsilon\).
3.7. Since \(\varepsilon > 0\) was arbitrary, \(\lim_{r \to 0^+} \int_{\gamma_r} \frac{g(z)}{z-a} \, dz = 0\).
For every \(\varepsilon > 0\) we found \(\delta\) such that \(r < \delta\) forces the integral to be at most \(\varepsilon\) in modulus, which is exactly the definition of the limit being \(0\).
3.8. Q.E.D.
Step 3: Conclusion
4. \(\lim_{r \to 0^+} \int_{\gamma_r} f(z)\,dz = i A \beta_0\).
4.1. By linearity of the line integral: \[\int_{\gamma_r} f(z) \, dz = \int_{\gamma_r} \frac{A}{z-a} \, dz + \int_{\gamma_r} \frac{g(z)}{z-a} \, dz = i A \beta_0 + \int_{\gamma_r} \frac{g(z)}{z-a} \, dz.\]
By <1>2.<2>3>, \(f(z) = \frac{A}{z-a} + \frac{g(z)}{z-a}\), so the integral splits by linearity; by <1>1.<2>2>, the first term equals \(iA\beta_0\).
4.2. Taking the limit as \(r \to 0^+\) on both sides: \[\lim_{r \to 0^+} \int_{\gamma_r} f(z) \, dz = i A \beta_0 + \lim_{r \to 0^+} \int_{\gamma_r} \frac{g(z)}{z-a} \, dz = i A \beta_0 + 0 = i A \beta_0.\]
The first term \(iA\beta_0\) is constant in \(r\), and by <1>3.<2>7> the second term tends to \(0\), so the limit is \(iA\beta_0\).
4.3. Q.E.D.