Problem 5 - Complex Analysis for Mathematics and Engineering

Question

Let \(f\) be analytic and have a zero of order \(k\) at zo. Show that \(f^{\prime}\) has a zero of order \(k-1\) at \(z_{0}\) -

Step-by-Step Solution

Answer

The zero of \( f'(z) \) at \( z_0 \) is of order \( k-1 \).

Step 1: Express the function

Since the function \( f \) is analytic and has a zero of order \( k \) at \( z_0 \), we can express \( f(z) \) near \( z_0 \) as \( f(z) = (z - z_0)^k g(z) \), where \( g(z) \) is analytic and \( g(z_0) eq 0 \).

Step 2: Differentiate the function

Differentiate \( f(z) = (z - z_0)^k g(z) \) using the product rule: \[ f'(z) = \frac{d}{dz}[(z - z_0)^k] \, g(z) + (z - z_0)^k \, \frac{d}{dz}g(z). \]

Step 3: Apply product rule

Compute the derivatives:\[ \frac{d}{dz}[(z - z_0)^k] = k(z - z_0)^{k-1} \] and\[ f'(z) = k(z - z_0)^{k-1} g(z) + (z - z_0)^k g'(z). \]

Step 4: Factor out common terms

Factor \((z-z_0)^{k-1}\) from \(f'(z)\):\[ f'(z) = (z - z_0)^{k-1}[ k g(z) + (z-z_0) g'(z)]. \]

Step 5: Determine the order of the zero

Since the expression \( k g(z) + (z-z_0) g'(z) \) is analytic and does not vanish at \( z_0 \) (since \( g(z_0) eq 0 \)), \( f'(z) \) has a zero of order \( k-1 \) at \( z_0 \).

Key Concepts

Analytic Function

An analytic function is a function that is locally given by a power series. In simple terms, this means the function can be expressed as an infinite sum of powers of its variable that converges to the function within some neighborhood of every point in its domain.
A function is analytic at a point if it can be written in the form:

• Power series: If \, \(f(z)\, \) can be written as \, \( \sum_{n=0}^{\infty} a_n (z - z_0)^n\) at a point \, \(z_0\), it is analytic.
• Derivatives: All derivatives of an analytic function exist and are continuous.

The behavior of analytic functions is predictable and well-behaved near any point, which makes them a crucial part of complex analysis.

Zero of a Function

The zero of a function, also known as a root, is a point where the function evaluates to zero. For a complex function like \, \(f(z)\,\), a zero at a point \, \(z_0\,\) of order \, \(k\,\) indicates that the function not only has a value of zero at that point but also continues to do so through terms of up to \, \(k\,\) order.

• Mathematical expression: The function \, \(f(z)\,\) can be expressed as \, \((z - z_0)^k g(z)\,\) near \, \(z_0\).
• Analytic behavior: The function \, \(g(z)\,\) is analytic near \, \(z_0\) and \, \(g(z_0) eq 0\).

This conceptual understanding helps in examining the behavior of derivatives at points where a function has zeros.

Derivative of Analytic Function

The derivative of an analytic function captures how the function changes at every point. For any function \, \(f(z)\), differentiating it involves using calculus to find \, \(f'(z)\).

• Derivative existence: Since analytic functions are smooth and differentiable everywhere in their domain, \, \(f'(z)\,\) is also analytic.
• Behavior near zeros: If \, \(f(z)\) has a zero of order \, \(k\) at \, \(z_0\), then \, \(f'(z)\) will have a zero of order \, \(k-1\) at \, \(z_0\).

The derivative of an analytic function behaves predictably, maintaining the structure of zeros, though with reduced order.

Product Rule in Differentiation

The product rule is a fundamental technique in calculus for finding derivatives of products of functions. It states that the derivative of a product of two functions \, \(u(z)\) and \, \(v(z)\) is given by:\[ (uv)' = u'v + uv' \]In the context of complex functions, this rule allows us to differentiate expressions involving products of analytic functions.

• Application: For \, \(f(z) = (z - z_0)^k g(z)\), you differentiate using the product rule to find \, \(f'(z)\).
• Combined derivatives: Differentiate each part separately and combine: \, \(k(z - z_0)^{k-1} g(z) + (z - z_0)^k g'(z)\).

This rule is indispensable for understanding how combined functions behave under differentiation.