Problem 35
Question
Suppose the analytic function \(f(z)\) has a zero of order \(n\) at \(z=z_{0} .\) Prove that the function \([f(z)]^{m}, m\) a positive integer, has a zero of order \(m n\) at \(z=z_{0}\).
Step-by-Step Solution
Verified Answer
[f(z)]^m has a zero of order mn at z=z_0.
1Step 1: Identify the Problem
We need to prove that if an analytic function \(f(z)\) has a zero of order \(n\) at \(z=z_0\), then the function \([f(z)]^m\) has a zero of order \(mn\) at the same point.
2Step 2: Express f(z) Near Zero
If \(f(z)\) has a zero of order \(n\) at \(z=z_0\), it can be expressed as \(f(z) = (z-z_0)^n g(z)\), where \(g(z)\) is analytic and \(g(z_0) eq 0\).
3Step 3: Substitute into [f(z)]^m
Substitute the expression of \(f(z)\) into \([f(z)]^m\). We have: \([f(z)]^m = [(z-z_0)^n g(z)]^m = (z-z_0)^{mn} [g(z)]^m\).
4Step 4: Identify the Order of the Zero
The expression \((z-z_0)^{mn} [g(z)]^m\) demonstrates that \([f(z)]^m\) has a zero of order \(mn\) at \(z=z_0\), since \(g(z_0)\) is non-zero.
Key Concepts
Analytic FunctionsZeros of FunctionsOrder of a Zero
Analytic Functions
In complex analysis, an analytic function, also known as a holomorphic function, is a function that is locally given by a convergent power series.
These functions are incredibly important due to their smooth and continuous nature on their domain.
They not only have derivatives at every point in their domain but also have derivatives of all orders.
Analytic functions can be thought of as an extension of polynomials or Taylor series for complex numbers. A function is defined as analytic on a region if it can be locally expanded into a power series:
- They are smooth on their domain and have derivatives of all orders.
- They can be expressed using power series around any point in their domain.
- Analytic functions include elementary functions like exponentials, logarithms, and polynomials.
Zeros of Functions
Zeros of functions occur when the function value equals zero at a given point. For an analytic function, these zeros are particularly important as they can be tied to many properties of the function. Finding zeros helps in understanding the function's nature and its behavior in certain domains. When talking specifically about zeros, the terms used are:
- Simple Zeros: Where the function value is zero, and around it, the function crosses the x-axis just once.
- Multiple Zeros: Occur when the x-axis is tangent to the function at the point, meaning the zero has more influence over the function's behavior.
Order of a Zero
The order of a zero refers to the number of times the function can be divided by \((z-z_0)\) before getting a non-zero value. This reveals how the function behaves when approaching that point. For a function \( f(z) \) with a zero of order \( n \) at \( z_0 \), we can write: \[ f(z) = (z-z_0)^n \cdot g(z) \] where \( g(z) \) is analytic and \( g(z_0) eq 0 \). Understanding the order of a zero is crucial:
- Higher Order Zeros: Represent points where the derivative and some higher derivatives of the function are also zero.
- Impacts Behavior: The order indicates the behavior of the function at or near that point and affects the function's local topology.
Other exercises in this chapter
Problem 34
Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. $$ \oint_{C} e^{4 /(z-2)} d z, C:|z-1|=3 $$
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Show that the power series \(\sum_{k=1}^{\infty} k z^{k}\) diverges at every point on its circle of convergence.
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(a) Use series to show that \(z=0\) is a zero of order 2 of \(1-\cos z\). (b) In view of part (a), \(z=0\) is a pole of order two of the function \(f(z)=e^{z} /
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By considering the series \(\sum_{k=0}^{\infty} r^{k} e^{i k \theta}, 0
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