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prove that the following function is analytic and where they are analytic

The given function is \( f_2(z) = -\frac{2j}{z} \), where \( j \) likely represents the imaginary unit (denoted by \( j \) in engineering, corresponding to \( i \) in mathematics), and \( z \) is a complex variable.

### To prove that this function is analytic:

A complex function \( f(z) \) is analytic at a point \( z_0 \) if it is differentiable in some neighborhood of \( z_0 \) and if its derivative exists at every point in that neighborhood.

#### 1. **Determine the domain of the function**:
   The function \( f_2(z) = -\frac{2j}{z} \) is undefined at \( z = 0 \) because division by zero is not allowed. Therefore, the function is not defined at \( z = 0 \). For \( z \neq 0 \), the function is defined and continuous.

#### 2. **Cauchy-Riemann Equations**:
   To check whether \( f_2(z) \) is analytic, we will test whether it satisfies the **Cauchy-Riemann equations** in the domain where it is defined (i.e., \( z \neq 0 \)).

   Let \( z = x + iy \), where \( x \) and \( y \) are real numbers, and \( i \) (or \( j \)) is the imaginary unit. Express \( f_2(z) \) in terms of \( x \) and \( y \).

   \[
   f_2(z) = -\frac{2j}{z} = -\frac{2j}{x + iy}
   \]
   Multiply numerator and denominator by the conjugate \( x - iy \):
   \[
   f_2(z) = -\frac{2j(x - iy)}{(x + iy)(x - iy)} = -\frac{2j(x - iy)}{x^2 + y^2}
   \]
   Simplify:
   \[
   f_2(z) = \frac{2(-y - ix)}{x^2 + y^2}
   \]
   Thus, we can write \( f_2(z) = u(x, y) + iv(x, y) \), where:
   \[
   u(x, y) = \frac{-2y}{x^2 + y^2}, \quad v(x, y) = \frac{-2x}{x^2 + y^2}
   \]

   Now, check the Cauchy-Riemann equations:
   \[
   \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
   \]

   Compute the partial derivatives:

   \[
   \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \left( \frac{-2y}{x^2 + y^2} \right) = \frac{-2y \cdot (-2x)}{(x^2 + y^2)^2} = \frac{4xy}{(x^2 + y^2)^2}
   \]

   \[
   \frac{\partial v}{\partial y} = \frac{\partial}{\partial y} \left( \frac{-2x}{x^2 + y^2} \right) = \frac{-2x \cdot (-2y)}{(x^2 + y^2)^2} = \frac{4xy}{(x^2 + y^2)^2}
   \]

   Hence, \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \).

   Next, compute the other derivatives:

   \[
   \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} \left( \frac{-2y}{x^2 + y^2} \right) = \frac{(x^2 + y^2)(-2) - (-2y)(2y)}{(x^2 + y^2)^2} = \frac{-2(x^2 - y^2)}{(x^2 + y^2)^2}
   \]

   \[
   \frac{\partial v}{\partial x} = \frac{\partial}{\partial x} \left( \frac{-2x}{x^2 + y^2} \right) = \frac{(x^2 + y^2)(-2) - (-2x)(2x)}{(x^2 + y^2)^2} = \frac{-2(x^2 - y^2)}{(x^2 + y^2)^2}
   \]

   Hence, \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \).

   Since the Cauchy-Riemann equations are satisfied, the function is analytic wherever it is defined, i.e., for \( z \neq 0 \).

#### 3. **Conclusion**:
The function \( f_2(z) = -\frac{2j}{z} \) is analytic in the complex plane except at \( z = 0 \), where it has a singularity.

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Do you want further clarification or additional details?

Here are 5 follow-up questions to explore:
1. How does analyticity relate to differentiability in the complex plane?
2. What types of singularities can a function have, and which one applies to this case?
3. Can you find the Laurent series expansion of \( f_2(z) \) around \( z = 0 \)?
4. How do the Cauchy-Riemann equations ensure the differentiability of a complex function?
5. What is the physical significance of this function in an applied context?

**Tip:** For complex functions, always check the points where the function might be undefined (e.g., division by zero), as they often lead to singularities.

Prove that the following function is analytic and where they are analytic.

Explore how to prove the analyticity of the complex function f(z) = -2j/z using the Cauchy-Riemann equations. Learn about the conditions under which the function is analytic and identify its domain of definition. Suitable for undergraduate mathematics students.

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