Problem 10 - Complex Analysis

Question

By splitting \(f\) into its real and imaginary parts, represent the complex line integral \(\int_{\alpha} f(z) d z\) in terms of real integrals. Result: If \(f=u+\mathrm{i} v, \alpha(t)=x(t)+\mathrm{i} y(t), t \in[a, b]\), then $$ \begin{aligned} \int_{\alpha} f(z) d z=\int_{\alpha}(u d x-v d y)+\mathrm{i} \int_{\alpha}(v d x+u d y) \\ =& \int_{a}^{b}\left[u(x(t), y(t)) x^{\prime}(t)-v(x(t), y(t)) y^{\prime}(t)\right] d t \\ &+\mathrm{i} \int_{a}^{b}\left[v(x(t), y(t)) x^{\prime}(t)+u(x(t), y(t)) y^{\prime}(t)\right] d t \end{aligned} $$

Step-by-Step Solution

Answer

Split the complex integral into real and imaginary integrals using real and imaginary parts of the function and path.

Step 1: Define the Components of the Integral

Given a function \(f(z) = u(x, y) + iv(x, y)\) with a curve \(\alpha(t) = x(t) + iy(t)\), identify the real and imaginary parts of both the function and the parameterized curve. Note that \(x(t)\) and \(y(t)\) are real-valued functions representing the path of integration.

Step 2: Express the Complex Integral

The complex line integral \(\int_\alpha f(z) dz\) can be rewritten considering the components of \(f\) and \(\alpha\). The differential \(dz\) is given by \(\frac{dz}{dt} = x'(t) + iy'(t)\), therefore, \(dz = dx + i\,dy = x'(t) dt + i\,y'(t) dt\).

Step 3: Split into Real and Imaginary Integrals

Separate the expression into real and imaginary parts:1. For the real part: \(\int_a^b [u(x(t), y(t)) x'(t) - v(x(t), y(t)) y'(t)] dt\).2. For the imaginary part: \(i\int_a^b [v(x(t), y(t)) x'(t) + u(x(t), y(t)) y'(t)] dt\).

Step 4: Identify the Real and Imaginary Integrals

The real integral results from the combination of the real part of \(f\) with the real part of \(dz\) and the imaginary part with the imaginary part of \(dz\):\(\int_\alpha (u dx - v dy)\).The imaginary integral results similarly:\(i \int_\alpha (v dx + u dy)\).

Step 5: Combine Integrals into Final Expression

Combine the real and imaginary integrals to represent the original complex integral:\[\int_{\alpha} f(z) dz = \int_{a}^{b}\left[u(x(t), y(t)) x'(t) - v(x(t), y(t)) y'(t)\right] dt + i \int_{a}^{b}\left[v(x(t), y(t)) x'(t) + u(x(t), y(t)) y'(t)\right] dt\]

Key Concepts

Real and Imaginary Parts

In complex analysis, every complex function can be expressed in terms of its real and imaginary parts. If we have a complex function of the form, \( f(z) = u(x, y) + iv(x, y) \), where \( u(x, y) \) and \( v(x, y) \) are real-valued functions, it represents two separate components: the real part, \( u(x, y) \), and the imaginary part, \( v(x, y) \). These components are crucial because they allow us to understand the behavior of the function by examining what happens along the real axis and the imaginary axis separately.

• **Real part** \( u(x, y) \): Often represents real-world values in applications and is responsible for quantifiable changes.
• **Imaginary part** \( v(x, y) \): Typically accounts for phenomena that aren't directly observable but are nonetheless impactful, such as phase shifts in signals.

When dealing with line integrals, separating a complex function into its real and imaginary parts enables us to handle each part distinctly. This separation simplifies solving complex integrals since each part can be integrated using standard real calculus techniques, then combined back to form the complex solution.

Parameterized Curve

The concept of a parameterized curve is central to understanding complex line integrals. When we define a curve using a parameter \( t \) between limits \( a \) and \( b \), we can describe the path through which integration occurs. For instance, a curve \( \alpha(t) = x(t) + iy(t) \) helps define the path by expressing each point \( z \) on the complex plane individually.

• **Parameter \( t \):** Think of \( t \) as a slider that moves along the curve from its start at \( a \) to its end at \( b \).
• **Functions \( x(t) \) and \( y(t) \):** These are real-valued functions that specify the curve's path in the real and imaginary directions respectively.

Using a parameterized curve ensures that every point on the curve is accounted for during integration. It turns something abstract like the path of integration in the complex plane into something tangible and computationally manageable. By handling derivatives of these functions, one manages to calculate changes along the curve efficiently, letting us perform line integrals with accuracy.

Differential Calculus

Differential calculus plays a crucial role in evaluating complex line integrals. We utilize it to handle the differentials \( dx \) and \( dy \), which represent small changes along the curve. These are determined from the parameterized expressions \( x(t) \) and \( y(t) \) as \( x'(t) \) and \( y'(t) \) respectively.

• **Derivative \( x'(t) \):** It measures the rate of change of the real part \( x(t) \) with respect to the parameter \( t \).
• **Derivative \( y'(t) \):** This measures the rate of change of the imaginary part \( y(t) \) concerning \( t \).
• **Differential Expression:** The differentials needed for integration, such as \( dz = dx + i \, dy \), where \( dx = x'(t) dt \) and \( dy = y'(t) dt \), stem directly from applying differential calculus.

Employing derivatives in this manner is key because it converts the abstract notion of a path into a precise quantity. This allows us to evaluate the line integrals efficiently by working within a familiar framework, that of real-valued calculus, simplifying complex integration tasks into manageable pieces.