The question provides some insight into Cauchy’s theorem. Solve the problem without using the Cauchy theorem.
Evaluate the integral \(\displaystyle{\int_{\gamma} z^n dz}\) for all integers \(n\). Here \(\gamma\) is any circle centered at the origin with the positive (counterclockwise) orientation.
Same question as (a), but with \(\gamma\) any circle not containing the origin.
Show that if \(|a|<r<|b|\), then \(\displaystyle{\int_{\gamma}\frac{dz}{(z-a)(z-b)} dz=\frac{2\pi i}{a-b}}\). Here \(\gamma\) denotes the circle centered at the origin, of radius \(r\), with the positive orientation.
Solution
Goal: Evaluate the given contour integrals purely by direct parametrization and power series expansions, without using Cauchy’s theorem.
Part 1: Integral of \(z^n\) on a Circle Centered at 0
1. Parametrize \(\gamma(t) = R e^{it}\) for \(t \in [0, 2\pi]\) (\(R > 0\)).
1.1. \(dz = i R e^{it} dt\), and \(z^n = R^n e^{int}\).
Differentiating \(\gamma(t) = R e^{it}\) gives \(\gamma'(t) = i R e^{it}\), so \(dz = i R e^{it}\,dt\); substituting \(z = R e^{it}\) gives \(z^n = R^n e^{int}\).
1.2. The integral becomes: \[\int_\gamma z^n \, dz = \int_0^{2\pi} R^n e^{int} \cdot i R e^{it} \, dt = i R^{n+1} \int_0^{2\pi} e^{i(n+1)t} \, dt.\]
Substituting \(z = R e^{it}\) and \(dz = i R e^{it}\,dt\) into the definition of the line integral, and combining the powers of \(R\) and the exponentials.
1.3. Case \(n = -1\): \(n+1 = 0 \implies \int_0^{2\pi} e^0 \, dt = 2\pi\).Thus \(\int_\gamma z^{-1} \, dz = i R^0 (2\pi) = 2\pi i\).
When \(n = -1\), the exponent \(n + 1 = 0\), so the integrand is the constant \(1\) and its integral over \([0, 2\pi]\) is \(2\pi\); multiplying by \(i R^0 = i\) gives \(2\pi i\).
1.4. Case \(n \neq -1\): \(n+1 \neq 0\), so: \[\int_0^{2\pi} e^{i(n+1)t} \, dt = \left[ \frac{e^{i(n+1)t}}{i(n+1)} \right]_0^{2\pi} = \frac{e^{i(n+1)2\pi} - 1}{i(n+1)} = \frac{1 - 1}{i(n+1)} = 0.\]
The antiderivative of \(e^{i(n+1)t}\) is \(\frac{e^{i(n+1)t}}{i(n+1)}\); evaluating at the endpoints gives \(\frac{e^{i(n+1)2\pi} - 1}{i(n+1)}\), and \(e^{i 2\pi k} = 1\) for the integer \(k = n + 1 \neq 0\), so the difference is \(0\).
1.5. Therefore: \[\int_\gamma z^n \, dz = \begin{cases} 2\pi i & \text{if } n = -1, \\ 0 & \text{if } n \in \mathbb{Z} \setminus \{-1\}. \end{cases}\]
Combining the two cases 3 and 4 gives the stated result.
1.6. Q.E.D.
Part 2: Integral of \(z^n\) on a Circle Not Containing 0
2. Let \(\gamma\) be a circle whose closed interior does not contain the origin \(0\). Then \(\int_\gamma z^n \, dz = 0\) for all \(n \in \mathbb{Z}\).
2.1. Case \(n \neq -1\): For every \(n \neq -1\), \(z^n\) has a global single-valued primitive on \(\mathbb{C} \setminus \{0\}\), namely \(F(z) = \frac{z^{n+1}}{n+1}\).
Differentiating gives \(\frac{d}{dz}\left(\frac{z^{n+1}}{n+1}\right) = z^n\), so \(F\) is a primitive of \(z^n\) valid on \(\mathbb{C} \setminus \{0\}\).
2.2. The integral of a function with a continuous primitive along any closed loop is zero: \(\int_\gamma z^n \, dz = F(\gamma(2\pi)) - F(\gamma(0)) = 0\).
By the Fundamental Theorem of Calculus for line integrals, the integral of a derivative along a closed curve equals the difference of the primitive at the two (equal) endpoints, which is \(0\).
2.3. Case \(n = -1\): Since the circle \(\gamma\) does not enclose the origin, the winding number of \(\gamma\) around \(0\) is \(\text{Ind}_\gamma(0) = 0\).
The origin lies in the unbounded connected component of \(\mathbb{C} \setminus \gamma\), so the winding number of \(\gamma\) about \(0\) is \(0\).
2.4. Equivalently, there exists a ray from the origin (e.g. through the opposite side of \(\gamma\)) on which a continuous single-valued branch of \(\log z\) is defined that contains the entire circle \(\gamma\).Thus \(\frac{1}{z} = \frac{d}{dz}\log z\) has a primitive on an open neighborhood of \(\gamma\).
A slit plane \(\mathbb{C} \setminus L\) obtained by deleting a ray \(L\) from the origin that misses \(\gamma\) admits a holomorphic branch of the logarithm, whose derivative is \(\frac{1}{z}\).
2.5. By the FTC for line integrals, \(\int_\gamma \frac{1}{z} \, dz = 0\).
Since \(\frac{1}{z}\) is the derivative of \(\log z\) on a domain containing \(\gamma\), the integral along the closed curve \(\gamma\) vanishes.
2.6. Therefore, \(\int_\gamma z^n \, dz = 0\) for all \(n \in \mathbb{Z}\).
Combining the case \(n \neq -1\) (<2>2) and the case \(n = -1\) (<2>5) gives the result for all integers \(n\).
2.7. Q.E.D.
Part 3: Evaluation of \(\int_\gamma \frac{dz}{(z-a)(z-b)}\) for \(|a| < r < |b|\)
3. Partial fraction decomposition of the integrand.
3.1. Algebraic partial fractions: \[\frac{1}{(z-a)(z-b)} = \frac{1}{a-b} \left( \frac{1}{z-a} - \frac{1}{z-b} \right).\]
Compute \(\frac{1}{a-b}\left(\frac{1}{z-a} - \frac{1}{z-b}\right) = \frac{1}{a-b}\left(\frac{(z-b)-(z-a)}{(z-a)(z-b)}\right) = \frac{1}{a-b}\left(\frac{a-b}{(z-a)(z-b)}\right) = \frac{1}{(z-a)(z-b)}\).
3.2. Thus: \[\int_\gamma \frac{dz}{(z-a)(z-b)} = \frac{1}{a-b} \left( \int_\gamma \frac{dz}{z-a} - \int_\gamma \frac{dz}{z-b} \right).\]
The integral is linear, so it distributes over the difference of the two terms.
3.3. Q.E.D.
4. Evaluate \(\int_\gamma \frac{dz}{z-a}\) using geometric series since \(|a| < r = |z|\).
4.1. On \(\gamma\), \(|z| = r > |a|\), so \(|a/z| = |a|/r < 1\).
This is the hypothesis \(|a| < r\) together with \(|z| = r\) on \(\gamma\).
4.2. Expand \(\frac{1}{z-a} = \frac{1}{z} \frac{1}{1 - a/z} = \frac{1}{z} \sum_{k=0}^\infty \left(\frac{a}{z}\right)^k = \sum_{k=0}^\infty a^k z^{-(k+1)} = \frac{1}{z} + \sum_{k=1}^\infty a^k z^{-(k+1)}\).
Factor \(\frac{1}{z-a} = \frac{1}{z}\frac{1}{1 - a/z}\) and expand \(\frac{1}{1 - a/z}\) as a geometric series, valid and uniformly convergent on \(|z| = r\) because \(|a/z| < 1\).
4.3. Integrating term-by-term on \(\gamma\) using Part 1 (<1>1.<2>5): \[\int_\gamma \frac{dz}{z-a} = \int_\gamma \frac{dz}{z} + \sum_{k=1}^\infty a^k \int_\gamma z^{-(k+1)} \, dz = 2\pi i + \sum_{k=1}^\infty a^k \cdot 0 = 2\pi i.\]
By Part 1, \(\int_\gamma z^{-1}\,dz = 2\pi i\) and \(\int_\gamma z^{-m}\,dz = 0\) for all \(m \ge 2\); uniform convergence justifies term-by-term integration.
4.4. Q.E.D.
5. Evaluate \(\int_\gamma \frac{dz}{z-b}\) using geometric series since \(|b| > r = |z|\).
5.1. On \(\gamma\), \(|z| = r < |b|\), so \(|z/b| = r/|b| < 1\).
This is the hypothesis \(r < |b|\) together with \(|z| = r\) on \(\gamma\).
5.2. Expand \(\frac{1}{z-b} = -\frac{1}{b} \frac{1}{1 - z/b} = -\sum_{k=0}^\infty \frac{z^k}{b^{k+1}}\), which converges uniformly on \(|z| = r\).
Factor \(\frac{1}{z-b} = -\frac{1}{b}\frac{1}{1 - z/b}\) and expand \(\frac{1}{1 - z/b}\) as a geometric series, valid and uniformly convergent on \(|z| = r\) because \(|z/b| < 1\).
5.3. Integrating term-by-term on \(\gamma\) using Part 1 (<1>1.<2>5): \[\int_\gamma \frac{dz}{z-b} = -\sum_{k=0}^\infty \frac{1}{b^{k+1}} \int_\gamma z^k \, dz = -\sum_{k=0}^\infty \frac{1}{b^{k+1}} \cdot 0 = 0.\]
By Part 1, \(\int_\gamma z^k\,dz = 0\) for all \(k \ge 0\); uniform convergence justifies term-by-term integration.
5.4. Q.E.D.
6. Combine the results.
6.1. Substituting <1>4.<2>3 and <1>5.<2>3 into <1>3.<2>2: \[\int_\gamma \frac{dz}{(z-a)(z-b)} = \frac{1}{a-b} (2\pi i - 0) = \frac{2\pi i}{a-b}.\]
Substituting the two evaluated integrals into the partial-fraction expression gives \(\frac{1}{a-b}(2\pi i - 0) = \frac{2\pi i}{a-b}\).
6.2. Q.E.D.