https://file.aiquickdraw.com/mathbot/file/1728482383333w1auo33s.jpg

Let's analyze the given function:

\[
f(z) = 
\begin{cases} 
\cos z & \text{if } z \neq \pm \frac{\pi}{2}, \\
-\frac{1}{\pi} & \text{if } z = \pm \frac{\pi}{2}.
\end{cases}
\]

### Step-by-step reasoning:
- The function \( f(z) \) is defined piecewise. At \( z \neq \pm \frac{\pi}{2} \), it behaves as \( \cos z \), which is a well-known entire function (holomorphic over the entire complex plane).
- At \( z = \pm \frac{\pi}{2} \), we have a special definition, setting \( f(z) = -\frac{1}{\pi} \), different from what \( \cos z \) would naturally take at these points (since \( \cos z \) has singularities at these points, where \( z = \pm \frac{\pi}{2} \)).
  
Let's consider the behavior at \( z = \pm \frac{\pi}{2} \):
- The function \( \cos z \) has simple poles at \( z = \pm \frac{\pi}{2} \), which suggests singularities.
- However, the function \( f(z) \) redefines the values at these points to \( -\frac{1}{\pi} \), potentially "removing" the singularities.

#### Removable singularities
We can check if these points \( z = \pm \frac{\pi}{2} \) are removable singularities. For removable singularities, if the limit of the function as \( z \to \pm \frac{\pi}{2} \) exists and matches the value of the function at that point, the singularity is removable.

Let's compute the limit as \( z \to \pm \frac{\pi}{2} \):

\[
\lim_{z \to \pm \frac{\pi}{2}} \left( \cos z \cdot \left( z^2 - \left( \frac{\pi}{2} \right)^2 \right)^2 \right) = 0
\]

Thus, the limit exists, and the function is made continuous by setting \( f(z) = -\frac{1}{\pi} \). Therefore, \( z = \pm \frac{\pi}{2} \) are removable singularities.

### Answer:
The correct statement is **C: \( z = \pm \frac{\pi}{2} \) are removable singularities**.

Would you like more details on the singularities or any further explanation? Here are 5 related questions:

1. What is the definition of a removable singularity?
2. How are singularities classified in complex analysis?
3. What are simple poles in the context of complex functions?
4. Can you explain the behavior of the function \( \cos z \) near its singularities?
5. How does the concept of "entire function" differ from other types of holomorphic functions?

**Tip:** Always check the behavior of limits around singularities to determine their type—whether they are poles, essential, or removable.

For the function f(z) = cos(z), if z ≠ ±π/2, and f(z) = -1/π, if z = ±π/2, which of the following is true? A. z = ±π/2 are singular points B. z = ±π/2 are simple poles C. z = ±π/2 are removable singularities D. f(z) is an entire function

Explore the classification of singularities for the function f(z) defined piecewise with cos(z) and constants. Learn why z = ±π/2 are removable singularities through detailed limit analysis. Ideal for students studying complex analysis at the university level.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation