Problem 17
Question
Determine the order of the poles for the given function. \(f(z)=\tan z\)
Step-by-Step Solution
Verified Answer
The poles of \( \tan z \) are of order 1 at \( z = \frac{\pi}{2} + n\pi \), where \( n \) is an integer.
1Step 1: Understanding the Function
We need to determine the order of the poles of the function \( f(z) = \tan z \). The tangent function can be expressed as \( \tan z = \frac{\sin z}{\cos z} \). This expression indicates that \( f(z) \) will have poles where \( \cos z = 0 \).
2Step 2: Locating the Poles
The cosine function \( \cos z \) is zero at the points \( z = \frac{\pi}{2} + n\pi \), where \( n \) is any integer. At these points, the denominator is zero and creates poles in the function \( \tan z \).
3Step 3: Determining the Order of the Poles
To determine the order of the poles, examine the behavior of \( \tan z \) near \( z = \frac{\pi}{2} + n\pi \). Near these points, \( \cos z \) behaves linearly as it crosses zero. This indicates that \( \tan z \) behaves like \( \frac{1}{x} \), suggesting a simple pole (order 1).
4Step 4: Conclusion
Therefore, the function \( f(z) = \tan z \) has simple poles (order 1) at \( z = \frac{\pi}{2} + n\pi \) for integer \( n \).
Key Concepts
Tangent FunctionPoles of a FunctionOrder of a Pole
Tangent Function
The tangent function is a fundamental concept in trigonometry and complex analysis. In its simplest form, the tangent of an angle in a right triangle is the ratio of the length of the opposite side to the adjacent side. However, in complex analysis, the tangent function is expressed as:\[\tan z = \frac{\sin z}{\cos z}\]This representation highlights its composition as the quotient of the sine and cosine functions. The properties of the tangent function are directly related to these trigonometric functions. A distinctive feature of \(\tan z\) is that it is undefined wherever \(\cos z = 0\). This aspect forms the basis for understanding poles of the function. The tangent function also has a periodic nature, repeating its values every \(\pi\) units along the real axis. This periodicity results in a series of regularly spaced poles, impacting the behavior of complex functions based on \(\tan z\).
Poles of a Function
In complex analysis, poles play a crucial role in understanding the behavior of meromorphic functions. A pole of a function is a specific type of singularity where the function approaches infinity at some point.
- A function \( f(z) \) has a pole at a point \( z_0 \) if \( f(z) \) goes to infinity as \( z \) approaches \( z_0 \).
- In the case of tangent function, \( \tan z = \frac{\sin z}{\cos z} \), poles occur where the cosine function, the denominator, equals zero.
- These zero points of \( \cos z \) correspond to poles of \( \tan z \), located at \( z = \frac{\pi}{2} + n\pi \).
Order of a Pole
The order of a pole is an important concept in complex analysis. It characterizes how a meromorphic function behaves near a singularity. Specifically, the order of a pole at \( z_0 \) is determined by how many times the function "blows up" or becomes infinite as it approaches \( z_0 \).
- A pole of order 1 is known as a simple pole. Here, the function resembles \( \frac{1}{(z-z_0)} \) near \( z_0 \).
- Higher-order poles like order 2 or 3, would behave like \( \frac{1}{(z-z_0)^2} \) or \( \frac{1}{(z-z_0)^3} \) respectively.
- For the tangent function \( \tan z \), the poles occur at \( z = \frac{\pi}{2} + n\pi \), and these are simple poles.
Other exercises in this chapter
Problem 17
Evaluate the Cauchy principal value of the given improper integral. \(\int_{-\infty}^{\infty} \frac{2 x^{2}-1}{x^{4}+5 x^{2}+4} d x\)
View solution Problem 17
Use Cauchy's residue theorem, where appropriate, to evaluate the given integral along the indicated contours. \(\oint_{C} \frac{1}{(z-1)(z+2)^{2}} d z\) (a) \(|
View solution Problem 17
Expand \(f(z)=\frac{z}{(z+1)(z-2)}\) in a Laurent series valid for the indicated annular domain. \(0
View solution Problem 17
In Problems 11-30, evaluate the Cauchy principal value of the given improper integral. $$ \int_{-\infty}^{\infty} \frac{2 x^{2}-1}{x^{4}+5 x^{2}+4} d x $$
View solution