Problem 10 - Advanced Engineering Mathematics

Question

Since \(\alpha_{1}=3 \pi / 2=\alpha_{2}, f^{\prime}(z)=A(z+1)^{1 / 2}(z-1)^{1 / 2}=A\left(z^{2}-1\right)^{1 / 2} .\) Therefore, $$f(z)=A\left[\frac{z\left(z^{2}-1\right)^{1 / 2}}{2}-\frac{1}{2} \operatorname{Ln}(z+\left(z^{2}-1\right)^{1 / 2}\right]+B.$$ but \(f(-1)=-a i\) and \(f(1)=a i .\) It follows that $$a i=f(1)=B, \quad-a i=f(-1)=A\left(-\frac{\pi i}{2}\right)+B$$ and so \(B=a i\) and \(A=4 a / \pi .\) Therefore, $$f(z)=\frac{4 a}{\pi}\left[\frac{z\left(z^{2}-1\right)^{1 / 2}}{2}-\frac{1}{2} \operatorname{Ln}\left(z+\left(z^{2}-1\right)^{1 / 2}\right)\right]+a i=\frac{4 a}{\pi}\left[\frac{z\left(z^{2}-1\right)^{1 / 2}}{2}-\frac{1}{2} \cosh ^{-1} z\right]+a i.$$

Step-by-Step Solution

Answer

The function is \( f(z) = \frac{4a}{\pi} \left( \frac{z\sqrt{z^2-1}}{2} - \frac{1}{2} \cosh^{-1}(z) \right) + ai \).

Step 1: Analyze Given Function Derivative

We begin with the function's derivative given by \( f'(z) = A(z+1)^{1/2}(z-1)^{1/2} = A(z^2-1)^{1/2} \). This form suggests integration involving square roots in terms of \( z \).

Step 2: Integrate the Function Derivative

Integrate \( f'(z) \) to determine \( f(z) \). The integral of \( \left(z^2-1\right)^{1/2}, \) typically results in a combination of algebraic and logarithmic forms: \[ f(z) = A\left[\frac{z\sqrt{z^2-1}}{2} - \frac{1}{2} \ln(z+\sqrt{z^2-1})\right] + B \] where \( B \) is an integration constant.

Step 3: Apply Boundary Conditions

Use the conditions \( f(-1) = -ai \) and \( f(1) = ai \) to solve for the constants \( A \) and \( B \). From \( f(1) = ai = B \), it follows that \( B = ai \). Substituting into \( f(-1) = -ai \), calculate \( -ai = A\left(-\frac{\pi i}{2}\right) + ai \) which simplifies to \( A = \frac{4a}{\pi} \).

Step 4: Formulate the Final Expression

Substitute the values of \( A \) and \( B \) into \( f(z) \). Thus, \[ f(z) = \frac{4a}{\pi}\left[\frac{z\sqrt{z^2-1}}{2} - \frac{1}{2} \ln(z+\sqrt{z^2-1})\right] + ai \] which can be simplified further to: \[ f(z) = \frac{4a}{\pi}\left[\frac{z\sqrt{z^2-1}}{2} - \frac{1}{2} \cosh^{-1}(z)\right] + ai \].

Key Concepts

Integration of Complex Functions

Integrating complex functions is an extension of integration from real analysis to the complex plane. Instead of only dealing with real numbers, we integrate functions involving complex numbers, where a complex number is expressed as \( z = x + yi \), with \( x \) and \( y \) being real numbers, and \( i \) representing the imaginary unit. The derivative of a complex function, like \( f'(z) = A(z^2-1)^{1/2} \), typically suggests a need for integration techniques that can handle square roots and may result in complex algebraic and logarithmic expressions.

• Integrals involving square roots like \( \int (z^2-1)^{1/2} dz \) often result in complex expressions, involving inverse hyperbolic functions or logarithms.
• In the example provided, integration is applied to a complex derivative to derive \( f(z) \), leading to a combination of terms including logarithms.

Breaking down complex integrals requires understanding of similar rules as in real number integration but extended into the complex plane with a focus on maintaining analytic properties.

Boundary Conditions

Boundary conditions are crucial in solving differential equations and integrals that involve determining specific solutions based on given conditions. These conditions provide limits at which a function is known, enabling calculation of particular constants, such as \( A \) and \( B \) in the given problem.

• In this exercise, the boundary conditions \( f(-1) = -ai \) and \( f(1) = ai \) help to determine these constants.
• The boundary condition \( f(1) = ai = B \) implies that \( B \) is directly derived from the value of \( f(z) \) at \( z = 1 \).
• Using \( f(-1) = -ai \), further manipulations allow us to solve for \( A \), illustrating the importance of boundary conditions in narrowing down to precise solutions.

Understanding and applying boundary conditions makes it possible to tailor general integral solutions into specific forms required for a problem.

Hyperbolic Functions

Hyperbolic functions appear naturally in the context of complex analysis, especially when dealing with square roots and logarithmic solutions of integrals. Similar to circular functions, hyperbolic functions such as \( \cosh(z) \), \( \sinh(z) \), and their inverses, are used to express complex integral solutions and transformations.

• The expression \( \frac{4a}{\pi} \left[ \frac{z \sqrt{z^2-1}}{2} - \frac{1}{2} \cosh^{-1}(z) \right] + ai \) shows the use of the inverse hyperbolic cosine \( \cosh^{-1}(z) \) in simplifying complex integrals.
• Hyperbolic functions are useful tools in complex analysis due to their connection with exponential functions, making them apt for expressing the results of intricate integrations.

Understanding hyperbolic functions can enrich one's grasp of complex function integration, offering simplified paths and expressions in otherwise complicated integer equations.