Problem 11 - Complex Analysis

Question

Let \(D=\mathbb{R}\) or \(D=\mathbb{C} .\) Let \(C \in \mathbb{C}\) be a constant and \(f: D \rightarrow \mathbb{C}\) differentiable with $$ f^{\prime}(z)=C f(z) \quad \text { for all } \quad z \in D $$ If \(A=f(0)\), then $$ f(z)=A \exp (C z) \quad \text { for all } \quad z \in D $$

Step-by-Step Solution

Answer

The function is \( f(z) = A \exp(Cz) \) for all \( z \in D \).

Step 1: Understanding the Problem Statement

We are given a function \(f: D \rightarrow \mathbb{C}\) with the property \(f'(z) = C f(z)\) for all \(z \in D\), where \(D\) can be either \(\mathbb{R}\) or \(\mathbb{C}\). We are asked to show that if \(A=f(0)\), then \(f(z)=A \exp (C z)\) for all \(z \in D\).

Step 2: Using the Differential Equation

Since \(f'(z) = C f(z)\), this is a first-order linear differential equation. We recognize this form as having an exponential solution.

Step 3: Solving the Differential Equation

The standard approach for a differential equation of the type \(f'(z) = C f(z)\) is \(f(z) = A e^{Cz}\), where \(A\) is a constant. This represents the general form of the solution, given the exponential function \(e^{Cz}\) satisfies this type of differential equation.

Step 4: Applying Initial Condition

We are given that \(A = f(0)\). Evaluating \(f(z) = A e^{Cz}\) at \(z = 0\) gives \(f(0) = A e^{0} = A\). This confirms \(A = f(0)\) is consistent with our solution.

Step 5: Conclusion

Given the differential equation \(f'(z) = C f(z)\) and that \(A = f(0)\), the function \(f(z) = A e^{Cz}\) satisfies both the differential equation and the initial condition.

Key Concepts

Exponential Solutions

In differential equations, exponential solutions are common, especially when dealing with first-order linear equations. When we have a differential equation like \(f'(z) = C f(z)\), the solution often takes the form of an exponential function, \(f(z) = A e^{Cz}\). This is because the exponential function inherently possesses the property that its derivative is proportional to itself.
Understanding exponential solutions is crucial because they describe many real-world phenomena, ranging from population growth to electric currents in circuits. The parameter \(C\) in the exponent \(e^{Cz}\) affects the growth or decay rate and alters the function's behavior.

• If \(C\) is positive, the solution represents exponential growth, indicating that the function increases as \(z\) increases.
• If \(C\) is negative, the solution represents exponential decay, indicating that the function decreases as \(z\) increases.

Recognizing and solving for exponential solutions help in predicting long-term behavior in various fields of science and engineering.

Complex Analysis

Complex analysis provides a set of tools and concepts that enable us to study functions of complex variables. The exercise we are exploring is rooted in this field, utilizing the concept of differentiable functions with complex outputs.
Within complex analysis, functions like \(f(z) = A e^{Cz}\) are examined in the context of the complex plane, with \(z\) typically having both real and imaginary components. This is powerful since complex exponential functions can represent oscillatory behavior and growth/decay simultaneously.

• The real part of \(C\) affects the magnitude or decay/growth.
• The imaginary part of \(C\), often expressed as \(C = a + bi\), contributes to oscillations (waves) when \(f(z)\) is viewed as part of a sinusoidal function.

By applying techniques from complex analysis, we can solve and understand differential equations in both real and complex domains. This is essential for upper-level math and numerous applications in physics and engineering, such as electrical engineering and signal processing.

Initial Conditions

When solving differential equations, initial conditions are vital as they determine the specific solution that satisfies the given conditions of a problem. Knowing that \(A = f(0)\) allows us to solve for the constant of integration in the general solution \(f(z) = A e^{Cz}\).
Initial conditions specify point values for the solution, which are essential in customizing the general solution to fit the specifics of a given problem. Applying these conditions ensures that the mathematical model accurately reflects the initial scenario under investigation.

• In our exercise, \(f(0) = A\) acts as the anchor point determining the function's shape.
• The condition applies constraints that tailor the exponential function to the problem's context, effectively narrowing down from an infinite set of possible solutions to the single correct one with these specifications.

Thus, initial conditions help finalize the solution path of a differential equation and ensure that it aligns with the physical or theoretical scenario described by the original situation.