Problem 9
Question
From \(\quad f(z)=z\left(1-\cos z^{2}\right)=z\left(-\frac{z^{4}}{2 !}+\frac{z^{8}}{4 !}-\cdots\right)=z^{5}\left(-\frac{1}{2 !}+\frac{z^{4}}{4 !}-\cdots\right)\) we see that \(z=0\) is a zero of order five.
Step-by-Step Solution
Verified Answer
The function has a zero at \( z=0 \) of order 5.
1Step 1: Express the Function
Start with the original function: \( f(z) = z(1 - \cos z^2) \). Substitute the Maclaurin series expansion of \( \cos z^2 \), which is \( 1 - \frac{(z^2)^2}{2!} + \frac{(z^2)^4}{4!} - \cdots \).
2Step 2: Simplify the Series Expansion
Simplify the series: \( f(z) = z \left(-\frac{z^4}{2!} + \frac{z^8}{4!} - \cdots \right) \). This step shows the rearrangement and simplification of terms from the substitution of the series expansion.
3Step 3: Factor out the Greatest Power of z
Factor \( z^5 \) out of the expression: \( f(z) = z^5 \left(-\frac{1}{2!} + \frac{z^4}{4!} - \cdots \right) \). This step makes it clear that the smallest power of \( z \) in the series is 5.
4Step 4: Determine the Order of the Zero
Since \( f(z) \) is expressed as \( z^5 \cdot g(z) \) with \( g(z) = -\frac{1}{2!} + \frac{z^4}{4!} - \cdots \), the function has a zero at \( z = 0 \) occurring at the smallest power of \( z \) in \( z^5 \), which is 5. Therefore, \( z = 0 \) is a zero of order 5.
Key Concepts
Maclaurin SeriesOrder of ZeroPower Series Expansion
Maclaurin Series
The Maclaurin series is a special type of power series expansion. It's designed around simplifying complex functions by expressing them as infinite series of their derivatives at zero. In mathematical terms, the Maclaurin series for any function \(f(x)\) is given by:
For our exercise, the function \(\cos(z^2)\) is expanded using its Maclaurin series. This is essential because it helps to express \(f(z)\) in a simpler form that reveals the order of zeros, which is typically relevant in complex analysis. Each term in the series represents a derivative order, providing an approximation that becomes more precise as additional terms are included.
- \(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots\)
For our exercise, the function \(\cos(z^2)\) is expanded using its Maclaurin series. This is essential because it helps to express \(f(z)\) in a simpler form that reveals the order of zeros, which is typically relevant in complex analysis. Each term in the series represents a derivative order, providing an approximation that becomes more precise as additional terms are included.
Order of Zero
The concept of the 'order of zero' is fundamental in understanding the behavior of functions in complex analysis. It tells us how many times a function becomes zero at a given point, often denoted as \(z = c\). More formally, if a function can be expressed as \((z-c)^n g(z)\) where \(g(z)\) is non-zero at \(z = c\), then \(z = c\) is a zero of order \(n\).
In our example, we observe the function \(f(z) = z(1-\cos(z^2))\). When simplified through series expansion, it becomes clear that \(z = 0\) is a zero of order five. This is because the smallest power in the expanded series, after factoring, is \(z^5\). Understanding the order of a zero helps in studying the root's multiplicity and how sharply or smoothly the function reaches zero at that point.
In our example, we observe the function \(f(z) = z(1-\cos(z^2))\). When simplified through series expansion, it becomes clear that \(z = 0\) is a zero of order five. This is because the smallest power in the expanded series, after factoring, is \(z^5\). Understanding the order of a zero helps in studying the root's multiplicity and how sharply or smoothly the function reaches zero at that point.
Power Series Expansion
A power series expansion helps express a function as a sum of its powers. This kind of expansion is vital because it can simplify many intricate calculations, allowing complex functions to be handled more straightforwardly as polynomials of potentially infinite degree.
The use of power series is a powerful technique in complex analysis, as it facilitates the evaluation of zeros, derivatives, and even integrals, with greater ease and flexibility. It turns intricate expressions into manageable parts, paving the way for deeper insights into the behavior and properties of functions.
- In concise terms, any function \(f(z)\), when expanded as a power series around point \(a\) (often \(a = 0\)), is written as \(\sum_{n=0}^{\infty} c_n (z-a)^n\), where \(c_n\) are coefficients.
The use of power series is a powerful technique in complex analysis, as it facilitates the evaluation of zeros, derivatives, and even integrals, with greater ease and flexibility. It turns intricate expressions into manageable parts, paving the way for deeper insights into the behavior and properties of functions.
Other exercises in this chapter
Problem 8
The zeros of \(f\) are the zeros of \(\sin z,\) that is, \(n \pi, n=0,\pm 1,\pm 2, \ldots .\) From \(f^{\prime}(z)=2 \sin z \cos z\) we see \(f^{\prime}(n \pi)=
View solution Problem 8
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View solution Problem 9
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View solution Problem 10
From \(\quad f(z)=z-\sin z=\frac{z^{3}}{3 !}-\frac{z^{5}}{5 !}+\cdots=z^{3}\left(\frac{1}{3 !}-\frac{z^{2}}{5 !}+\cdots\right)\) we see that \(z=0\) is a zero o
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