/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Write \(f(z)=\left(z^{2}+9\right... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 6

Write \(f(z)=\left(z^{2}+9\right) / z=(z-3 i)(z+3 i) / z\) to see that \(3 i\) and \(-3 i\) are zeros of \(f .\) Now \(f^{\prime}(z)=1-9 / z^{2}\) and \(f^{\prime}(3 i)=f^{\prime}(-3 i)=2 \neq 0 .\) This indicates that each zero is of order one.

Short Answer

Expert verified
Zeros are \(3i\) and \(-3i\), both of order one.

Step by step solution

01

Define the function

The function given is \( f(z) = \frac{z^2 + 9}{z} \) which can also be expressed as \( f(z) = \frac{(z - 3i)(z + 3i)}{z} \).
02

Identify the zeros of the numerator

To find the zeros of \( f(z) \), set the numerator \( (z - 3i)(z + 3i) = 0 \). This gives the solutions \( z = 3i \) and \( z = -3i \). Thus, \( 3i \) and \( -3i \) are zeros of \( f(z) \).
03

Differentiate the function

The derivative of \( f(z) \) is given as \( f'(z) = 1 - \frac{9}{z^2} \).
04

Substitute the zeros into the derivative

Evaluate the derivative at the zeros: \( f'(3i) = 1 - \frac{9}{(3i)^2} = 1 - \frac{9}{-9} = 1 + 1 = 2 \) and \( f'(-3i) = 1 - \frac{9}{(-3i)^2} = 1 - \frac{9}{-9} = 1 + 1 = 2 \).
05

Determine the order of the zeros

Since the derivative \( f'(z) \) evaluated at both zeros \( 3i \) and \( -3i \) is \( 2 eq 0 \), each zero \( 3i \) and \( -3i \) is of order one.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Zeros of Holomorphic Functions
In complex analysis, understanding zeros of holomorphic functions is fundamental. Holomorphic functions are complex functions that are differentiable at every point in their domain. A zero of a function is a point where the function value is zero. For our function, given as \( f(z) = \frac{z^2 + 9}{z} \), we first need to find the zeros of the numerator, which in this case is \((z - 3i)(z + 3i)\).
To find these zeros, we solve the equation \((z - 3i)(z + 3i) = 0\). This results in two solutions: \(z = 3i\) and \(z = -3i\). Hence, these are the zeros of the function.
Identifying zeros is crucial as they are the points where the function touches the complex plane's horizontal axis, and they often provide insight into the function's behavior and properties.
Differentiation of Complex Functions
Complex differentiation is similar to real differentiation but with complex numbers. For the given function \( f(z) = \frac{z^2 + 9}{z} \), the derivative is found using standard differentiation rules: here, \( f'(z) = 1 - \frac{9}{z^2} \).
A pivotal step in solving complex function problems involves finding the derivative. Differentiate carefully, ensuring each function component adheres to the rules of calculus, especially concerning division and multiplication of complex numbers.
The derivative helps us study the function's behavior around its zeros. By substituting the zeros \(3i\) and \(-3i\) into the derivative, we can evaluate the effect of the zeros on the function. Specifically, if the derivative at a particular zero is not zero, it implies a certain behavior regarding the order of that zero.
Order of Zeros
The order of a zero of a function is tied to the zero's algebraic multiplicity. A zero \(z_0\) of a function \(f(z)\) has order \(n\) if \((z - z_0)^n\) is a factor of \(f(z)\) but \((z - z_0)^{n+1}\) is not. In simpler terms, it's how many times the zero appears as a root.
For the function \(f(z)\), zeros \(3i\) and \(-3i\) have been determined to be of order one. This determination is made by evaluating the derivative of the function at these zeros. Because \(f'(3i)\) and \(f'(-3i)\) both equal 2, which is non-zero, we conclude each zero is of order one. This means the zeros are simple zeros.
The order of zeros can influence the graph and analytic properties of the function, affecting how it behaves near these points. Understanding this concept helps in predicting the nature of complex functions, which is essential in advanced mathematical contexts such as analytic continuation.

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Most popular questions from this chapter

$$\operatorname{Re}\left(z_{n}\right)=\frac{8 n^{2}+n}{4 n^{2}+1} \rightarrow 2 \text { as } n \rightarrow \infty, \text { and } \operatorname{Im}\left(z_{n}\right)=\frac{6 n^{2}-4 n}{4 n^{2}+1} \rightarrow \frac{3}{2} \text { as } n \rightarrow \infty$$.

$$f(z)=\frac{1}{z^{2}} \cdot \frac{1}{1-\frac{3}{z}}=\frac{1}{z^{2}}\left[1+\frac{3}{z}+\frac{3^{2}}{z^{2}}+\frac{3^{3}}{z^{3}}+\dots\right]=\frac{1}{z^{2}}+\frac{3}{z^{3}}+\frac{3^{2}}{z^{1}}+\frac{3^{3}}{z^{5}}+\cdots$$

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.

(a) \(R=1,\) which is the distance from the origin to \(z=-1\) (b) Using Taylor's Theorem [or integrating the series for \(1 /(1+z)]\) we obtain for \(R=1\) $$\operatorname{Ln}(1+z)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} z^{k}$$ (c) By replacing \(z\) in part $$\operatorname{Ln}(1+z)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} z^{k}$$ (c) By replacing \(z\) in part (b) by \(-z\) we obtain for \(R=1\) $$\operatorname{Ln}(1-z)=-\sum_{k=0}^{\infty} \frac{z^{k}}{k}$$ (d) One way of obtaining the Maclaurin series for \(\operatorname{Ln}\left(\frac{1+z}{1-z}\right)\) is to use Taylor's Theorem. Alternatively, let us write $$\operatorname{Ln}\left(\frac{1+z}{1-z}\right)=\operatorname{Ln}(1+z)-L(1-z)$$ and subtract the series in parts (b) and (c). This gives for the common circle of convergence \(|z|=1\) $$\operatorname{Ln}\left(\frac{1+z}{1-z}\right)=2 z+\frac{2}{3} z^{3}+\frac{2}{5} z^{5}+\frac{2}{7} z^{7}+\cdots=2 \sum_{k=0}^{\infty} \frac{1}{(2 k+1)} z^{2 k+1}$$ But recall that in general \(\operatorname{Ln}\left(z_{1} / z_{2}\right) \neq \operatorname{Ln} z_{1}-\operatorname{Ln} z_{2}\) since \(\operatorname{Ln} z_{1}\) and \(\operatorname{Ln} z_{2}\) could differ by a constant multiple of \(i .\) That is, \(\operatorname{Ln} z_{1}-\operatorname{Ln} z_{2}=C i\) for some \(C .\) So $$\operatorname{Ln}\left(\frac{1+z}{1-z}\right)=\operatorname{Ln}(1+z)-\operatorname{Ln}(1-z)-C i$$ When \(z=0\) we obtain \(\operatorname{Ln} 1=\operatorname{Ln} 1-\operatorname{Ln} 1-C i .\) since \(\operatorname{Ln} 1=0\) we get \(C=0\)

$$\begin{aligned} &\int_{-\infty}^{\infty} \frac{x e^{i x}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x=2 \pi i[\operatorname{Res}(f(z), i)+\operatorname{Res}(f(z), 2 i)]=2 \pi i\left[\frac{1}{6} e^{-1}-\frac{1}{6} e^{-2}\right]=\frac{\pi}{3}\left(e^{-1}-e^{-2}\right) i\\\ &\int_{-\infty}^{\infty} \frac{x \sin x}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x=\operatorname{Im}\left(\int_{-\infty}^{\infty} \frac{x e^{i x}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x\right)=\frac{\pi}{3}\left(e^{-1}-e^{-2}\right) \end{aligned}$$ Therefore, $$\int_{0}^{\infty} \frac{x \sin x}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x=\frac{1}{2}\left[\frac{\pi}{3}\left(e^{-1}-e^{-2}\right)\right]=\frac{\pi}{6}\left(e^{-1}-e^{-2}\right)$$
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