5.2 Cauchy’s Theorem

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We compute integrals of complex functions around closed curves.

A contour is called closed if its initial and terminal points coincide. Further it is called simple if those are the only points which coincide.

Cauchy’s Theorem If $f$ is analytic along a simple closed contour $C$ and also analytic inside $C$ , then $\int _C f(z) \, dz = 0$

Let $C$ be a simple closed contour that does not pass through $z_0$ or contain $z_0$ in its interior. Compute $\dis \int _C \frac {1}{z-z_0} \, dz$ .
The function $\dis f(z) = \frac {1}{z-z_0}$ is analytic everywhere except at $z_0$ . From the description of the contour $C$ , $f(z)$ is analytic along $C$ and in its interior. Hence, by Cauchy’s Theorem, $\int _C \frac {1}{z-z_0} \, dz = 0$

If the contour $C$ in the previous example passed through the point $z_0$ or contained $z_0$ in its interior, Cauchy’s Theorem would not apply and the integral could be non-zero.

(problem 1a) Let $C = C(z_0, r)$ traversed clockwise.
$\int _C e^z \, dz = \answer {0}$

(problem 1b) Let $C = C(z_0, r)$ traversed clockwise.
$\int _C \sin (z) \, dz = \answer {0}$

(problem 1c) Let $C = C(1+i, 1)$ traversed counter-clockwise.
$\int _C \Log (z) \, dz = \answer {0}$

(problem 1d) Let $C = C(1+i, 1)$ traversed counter-clockwise.
$\int _C \frac {1}{z} \, dz = \answer {0}$

A positively oriented simple closed contour is one in which the interior of the contour is to the left as the contour is traced out. In the case of a circle, positive orientation is counter-clockwise.

Deformation of Contour Suppose $C_1$ and $C_2$ are positively oriented, simple closed contours such that $C_1$ lies entirely inside $C_2$ . If $f$ is analytic along $C_1$ and $C_2$ and also analytic in the region between them, then $\int _{C_1} f(z) \, dz = \int _{C_2} f(z) \, dz$

The Deformation of Contour theorem is typically used to replace arbitrary positively oriented, simple closed curves with circles traversed counter-clockwise.

example 2 Let $C$ be a positively oriented, simple closed contour containing the origin in its interior. Compute $\dis \int _C \frac {1}{z} \, dz$ .
Since the origin is interior to $C$ , there exists a radius $r$ such that the circle $C(0, r)$ lies in the interior of $C$ . By the Deformation of Contour theorem, $\int _C \frac {1}{z} \, dz = \int _{C(0,r)} \frac {1}{z} \, dz$ where $C(0,r)$ is traversed in the positive direction, i.e., counter-clockwise. Using the parametrization $\gamma (t) = re^{it}, \, 0\leq t \leq 2\pi$ for $C(0, r)$ , we have $\int _{C(0,r)} \frac {1}{z} \, dz = \int _0^{2\pi } \frac {\gamma '(t)}{\gamma (t)} \, dt = \int _0^{2\pi } i \, dt = 2\pi i$ Hence, $\int _C \frac {1}{z} \, dz = 2\pi i$ as well.

(problem 2a) Let $C$ be a positively oriented simple closed curve containing the point $z_0$ in its interior.
$\int _C \frac {1}{z-z_0} \, dz = \answer {2\pi i}$

(problem 2b) Let $C$ be a positively oriented simple closed curve containing the point $z_0$ in its interior.
$\int _C \frac {1}{(z-z_0)^2} \, dz = \answer {0}$

(problem 2c) Let $C$ be a positively oriented simple closed curve containing the point $z_0$ in its interior and let $n > 2$ .
$\int _C \frac {1}{(z-z_0)^n} \, dz = \answer {0}$

Here is a video solution to problem 2c:

Extended Deformation of Contour Suppose $C_1, C_2, ..., C_n$ are non-intersecting positively oriented, simple closed contours each lying inside of another positively oriented, simple closed contour $C$ . If $f$ is analytic along $C, C_1, C_2, ..., C_n$ and also analytic in the region between them (i.e. interior to $C$ but exterior to each of $C_1, C_2, ..., C_n$ ), then $\int _C f(z) \, dz = \sum _{k=1}^n \int _{C_k} f(z) \, dz$

Cauchy’s Integral Formula Suppose $f$ is analytic along and inside a positively oriented, simple closed contour $C$ containing the point $z_0$ in its interior. Then $f^{(n)}(z_0) = \frac {n!}{2\pi i} \int _C \frac {f(z)}{(z-z_0)^{n+1}} \, dz$

The case $n=0$ corresponds to $f(z_0) = \frac {1}{2\pi i} \int _C \frac {f(z)}{z-z_0} \, dz$

example 3a Let $C = C(0, 1)$ traversed counter-clockwise. Compute the contour integral: $\dis \int _C \frac {\sin z}{z^2} \, dz$
We use Cauchy’s Integral Formula. To do so, we first need to identify $z_0, n$ and $f(z)$ and verify that $f(z)$ is analytic in an on $C$ . We have $z_0 = 0, n=1,$ and $f(z) = \sin z$ which is entire. Thus, we can apply the formula and we obtain $\int _C \frac {\sin z}{z^2} \, dz = \frac {2\pi i}{1!} f'(0) = 2\pi i \cos 0 = 2 \pi i$

example 3b Let $C = C(2, 1)$ traversed counter-clockwise. Compute the contour integral: $\dis \int _C \frac {\sin z}{z(z-2)} \, dz$
We use Cauchy’s Integral Formula. To do so, we first need to identify $z_0, n$ and $f(z)$ and verify that $f(z)$ is analytic in an on $C$ . We have $z_0 = 2, n=0,$ and $f(z) = \frac {\sin z}{z}$ which is analytic in and on $C(2,1)$ . Thus, we can apply the formula and we obtain $\int _C \frac {\sin z}{z(z-2)} \, dz = \frac {2\pi i}{0!} f(2) = 2\pi i \frac {\sin 2}{2} = \pi i \sin 2$

(problem 3a) Let $C = C(0, 1)$ traversed counter-clockwise. Compute the contour integral: $\dis \int _C \frac {e^z}{z^2} \, dz$
$z_0 = \answer {0} \quad n = \answer {1} \quad f(z) = \answer {e^z}$ $\int _C \frac {e^z}{z^2} \, dz = \answer {2\pi i}$

(problem 3b) Let $C = C(0, 1)$ traversed counter-clockwise. Compute the contour integral: $\dis \int _C \frac {e^z}{z^4} \, dz$
$z_0 = \answer {0} \quad n = \answer {3} \quad f(z) = \answer {e^z}$ $\int _C \frac {e^z}{z^4}\, dz = \answer {i pi /3}$

(problem 3c) Let $C = C(i, 1)$ traversed clockwise. Compute the contour integral: $\dis \int _C \frac {\cos z}{z-i} \, dz$
$z_0 = \answer {i} \quad n = \answer {0} \quad f(z) = \answer {\cos z}$ $\int _C \frac {\cos z}{z-i}\, dz = \answer {-2\pi i \cosh 1}$

(problem 3d) Let $C = C(4, 2)$ traversed counter-clockwise. Compute the contour integral: $\dis \int _C \frac {\sin z}{(z-1)(z-3)} \, dz$
$z_0 = \answer {3} \quad n = \answer {0} \quad f(z) = \answer {\frac {\sin z}{z-1}}$ $\int _C \frac {\sin z}{(z-1)(z-3)} \, dz = \answer {\pi i \sin 3}$

Here is a video solution to problem 3b:

example 4 Let $C = C(0, 2)$ traversed counter-clockwise. Compute the contour integral: $\dis \int _C \frac {e^z}{z^2 +1} \, dz$
The integrand has singularities at $\pm i$ , so we use the Extended Deformation of Contour Theorem before we use Cauchy’s Integral Formula. By the Extended Deformation of Contour Theorem we can write $\int _C \frac {e^z}{z^2 +1} \, dz = \int _{C_1} \frac {e^z}{z^2 +1} \, dz + \int _{C_2} \frac {e^z}{z^2 +1} \, dz$ where $C_1 = C(i, 1/2)$ traversed counter-clockwise and $C_2 = C(-i, 1/2)$ traversed counter-clockwise. Now by Cauchy’s Integral Formula with $n = 0$ , we have $\int _{C_1} \frac {e^z}{z^2 +1} \, dz = \int _{C_1} \frac {e^z}{(z+i)(z-i)} \, dz = 2\pi i f(i)$ where $f(z) = \frac {e^z}{z+i}$ . Hence $\int _{C_1} \frac {e^z}{z^2 +1} \, dz = 2\pi i \cdot \frac {e^i}{2i} = \pi e^i$ Similarly, $\int _{C_2} \frac {e^z}{z^2 +1} \, dz = \int _{C_2} \frac {e^z}{(z+i)(z-i)} \, dz = 2\pi i f(-i)$ where $f(z) = \frac {e^z}{z-i}$ . Hence $\int _{C_1} \frac {e^z}{z^2 +1} \, dz = 2\pi i \cdot \frac {e^{-i}}{-2i} = -\pi e^{-i}$ Combining these we get $\int _C \frac {e^z}{z^2 +1} \, dz = \pi e^i - \pi e^{-i} = 2\pi \sin (1)$

(problem 4) Let $C = C(0, 2)$ traversed counter-clockwise. Compute the contour integral:
$\int _C \frac {e^z}{z^2 -1} \, dz \answer {\pi i (e-1/e)}$

Here is a video solution to problem 4:

2024-09-27 14:03:34