- The next three calculations provide some insight into Cauchy’s theorem, which we treat in the next chapter.
- Evaluate the integrals
\[ \int_ {\gamma} z ^ {n} d z \]
for all integers \(n .\) . Here \(\gamma\) is any circle centered at the origin with the positive (counterclockwise) orientation.
Same question as before, but with \(\gamma\) any circle not containing the origin.
Show that if \(\vert a \vert < r < \vert b \vert\) , then
\[ \int_ {\gamma} \frac {1}{(z - a) (z - b)} d z = \frac {2 \pi i}{a - b}, \]
where \(\gamma\) denotes the circle centered at the origin, of radius \(r ,\) with the positive orientation.
Solution
1. (a) If \(\gamma\) is the positively oriented circle of radius \(R\) centered at \(0\), then \(\int_\gamma z^n\,dz = \boxed{0}\) for \(n \neq -1\) and \(\int_\gamma z^{-1}\,dz=\boxed{2\pi i}\).
Parametrize \(\gamma\) by \(z = R e^{i\theta}\), \(0 \le \theta \le 2\pi\), so \(dz = i R e^{i\theta}\,d\theta\) and \[\int_\gamma z^n\,dz = i R^{n+1} \int_0^{2\pi} e^{i(n+1)\theta}\,d\theta.\] If \(n \neq -1\), the integrand has the \(2\pi\)-periodic antiderivative \(e^{i(n+1)\theta}/(i(n+1))\), so the integral is \(0\). If \(n = -1\), the integrand is \(1\), and the integral is \(iR^0\cdot 2\pi = 2\pi i\).
2. (b) If the origin lies outside the closed disc bounded by \(\gamma\), then \(\int_\gamma z^n\,dz = \boxed{0}\) for every integer \(n\).
The closed disc bounded by \(\gamma\) is a compact set not containing \(0\), so it lies in a slightly larger open disc \(D\) with \(0\notin D\). The function \(z^n\) is holomorphic on \(D\) for every integer \(n\), and Cauchy’s theorem in the disc \(D\) gives \(\int_\gamma z^n\,dz=0\).
3. (c) If \(\abs a<r<\abs b\) and \(\gamma\) is the positively oriented circle \(\abs z=r\), then \(\int_\gamma\frac{dz}{(z-a)(z-b)}=\frac{2\pi i}{a-b}\).
Partial fractions give \(\frac{1}{(z-a)(z-b)}=\frac{1}{a-b}\qty{\frac{1}{z-a}-\frac{1}{z-b}}\). On \(\gamma\), \(\abs{a/z}=\abs a/r<1\) and \(\abs{z/b}=r/\abs b<1\), so the geometric series \[\frac{1}{z-a}=\sum_{k\ge0}\frac{a^k}{z^{k+1}},\qquad \frac{1}{z-b}=-\sum_{k\ge0}\frac{z^k}{b^{k+1}}\] converge uniformly on \(\gamma\) and may be integrated term by term. By step 1, only the term \(a^0z^{-1}\) has a nonzero integral, namely \(2\pi i\), so \(\int_\gamma\frac{dz}{z-a}=2\pi i\) and \(\int_\gamma\frac{dz}{z-b}=0\). Hence the integral is \(\frac{2\pi i}{a-b}\).