12 Introduction to Complex Analysis
12.1 2-D Problems in Potential Theory Solved Using Complex Analysis
2-D problems in potential theory can be elegantly solved using complex analysis, the theory of differentiable complex-valued functions of complex variables.
The following tools from mathematics are used:
- Cauchy’s Integral Theorem
- Cauchy’s Integral Formula
- Cauchy-Riemann Differential Equations
- Residue Theorem of Complex Analysis
12.2 Space of Complex Numbers \(\mathbb{C}\)
\(\mathbb{C}\) is a two-dimensional real vector space with the canonical basis \((1, i)\). A point \(z \in \mathbb{C}\) has real Cartesian coordinates \(x := \mathrm{Re}(z)\), \(y := \mathrm{Im}(z) \in \mathbb{R}\), briefly expressed as \(z = x + i y\).
A complex-valued function \(f: U \subset \mathbb{C} \to \mathbb{C}\) on an open subset of \(\mathbb{C}\) can therefore be expressed by decomposing it into its real and imaginary parts \(f(x + i y) = u(x, y) + i v(x, y)\), viewed as an \(\mathbb{R}^2\)-valued function of two real variables \((x, y) \in \tilde{U} := \{(a, b) \in \mathbb{R}^2 \ | \ a + i b \in U\}\).
12.3 Cauchy’s Integral Theorem
In real analysis, the value of an integral depends on the integration bounds. For complex functions, the path also matters.
Let \(U \subset \mathbb{C}\) be a simply connected (no holes) open domain (elementary domain), and \(f: U \mapsto \mathbb{C}\) a [[Holomorphy|holomorphic]] function. Furthermore, let \(\gamma : [a, b] \to U\) be a smooth, closed curve. Then: \[ \oint_{\gamma} f(z) \, \mathrm{d}z = 0. \] Proof: With \(f = u + i v\) and \(\mathrm{d}z = \mathrm{d}x + i \mathrm{d}y\), it follows: \[ \oint_{\gamma} f(z) \, \mathrm{d}z = \oint_{\gamma} (u + i v) (\mathrm{d}x + i \mathrm{d}y) = \oint_{\gamma} (u \mathrm{d}x - v \mathrm{d}y) + i \oint_{\gamma} (v \mathrm{d}x + u \mathrm{d}y). \] Using Green’s Theorem: \[ \iint_D \left(\frac{\partial g}{\partial x}(x,y) - \frac{\partial f}{\partial y}(x,y)\right)\, \mathrm{d}x \, \mathrm{d}y = \oint_{C} \left(f(x,y)\, \mathrm{d}x + g(x,y)\, \mathrm{d}y\right), \] the line integrals are replaced by surface integrals over the region \(D\) enclosed by the curve \(\gamma\): \[ \begin{align} \oint_\gamma (u \, \mathrm{d}x - v \, \mathrm{d}y) & = \iint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right) \, \mathrm{d}x\,\mathrm{d}y, \\ \oint_\gamma (v \, \mathrm{d}x + u \, \mathrm{d}y) & = \iint_D \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) \, \mathrm{d}x\,\mathrm{d}y. \end{align} \] For the real and imaginary parts \(u\) and \(v\) of holomorphic functions in the domain \(D\), the Cauchy-Riemann differential equations hold: \[ \begin{align} u_x & = +v_y, \\ u_y & = -v_x. \end{align} \] Thus, both integrands—and consequently the integral—are zero: \[ \iint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right) \, \mathrm{d}x\,\mathrm{d}y = \iint_D \left( \frac{\partial u}{\partial y} -\frac{\partial u}{\partial y} \right) \, \mathrm{d}x\,\mathrm{d}y = 0, \] and \[ \iint_D \left( \frac{\partial u}{\partial x} -\frac{\partial v}{\partial y} \right) \, \mathrm{d}x\,\mathrm{d}y = \iint_D \left( \frac{\partial u}{\partial x} -\frac{\partial u}{\partial x} \right) \, \mathrm{d}x\,\mathrm{d}y = 0. \] Hence: \[ \oint_{\gamma} f(z) \, \mathrm{d}z = 0 \qquad \square \]
12.4 Cauchy’s Integral Formula (CIF)
Let \(U\) be an open subset of the complex plane \(\mathbb{C}\). Consider the closed disk \(D\) fully contained within \(U\), defined as: \[ D = \{ z : |z - z_{0}| \le r \}. \] Let \(f: U \mapsto \mathbb{C}\) be a holomorphic function, and \(\gamma\) a counterclockwise-oriented circle along the boundary of \(D\). Then, for any point \(a\) inside \(D\): \[ f(a) = \frac{1}{2 \pi i} \oint_{\gamma} \frac{f(z)}{z - a} \, \mathrm{d}z. \] The CIF states that the values of a holomorphic function \(f\) inside a disk are fully determined by its values on the boundary of that disk.
12.5 Residue Theorem
If \(f(z)\) is holomorphic in the neighborhood of a point \(z_{0}\), then: \[ \oint_{C} f(z) \, \mathrm{d}z = 0. \] However, if \(f(z)\) has an isolated singularity at \(z_{0}\), the integral is generally nonzero and satisfies: \[ \frac{1}{2 \pi i} \oint_{C} f(z) \, \mathrm{d}z = \mathrm{Res} f(z). \] For a simple pole: \[ \mathrm{Res} f(z) = \lim_{ z \to z_{0} } (z - z_{0}) f(z). \] Instead of computing contour integrals, one only needs to calculate residues, which is often simpler.