All questions of Complex variables for Electrical Engineering (EE) Exam

To determine if the function u(x, y) = 2x^2 satisfies the Laplace equation and is therefore analytic, we need to calculate its Laplacian.

The Laplacian of a function u(x, y) is given by the second partial derivatives with respect to x and y:

∇²u = ∂²u/∂x² + ∂²u/∂y²

Let's calculate the partial derivatives of u(x, y):

∂u/∂x = 4x
∂²u/∂x² = 4

∂u/∂y = 0
∂²u/∂y² = 0

Now, let's calculate the Laplacian:

∇²u = ∂²u/∂x² + ∂²u/∂y² = 4 + 0 = 4

Since the Laplacian of u(x, y) = 2x^2 is equal to 4 and not zero, it does not satisfy the Laplace equation. Therefore, u(x, y) = 2x^2 is not analytic.

To find the function f(z), we need to find the expressions for both u(x, y) and v(x, y).

Given that u(x, y) = 2kxy, we can find the expression for u(x, y) in terms of z:

u(x, y) = 2kxy = 2k(x)(iy) = 2kxiy

Similarly, given that v(x, y) = x^2, we can find the expression for v(x, y) in terms of z:

v(x, y) = x^2 = (x)(x) = (x)(-i)(-i) = (x)(-iy)(-iy) = (-ix)(-iy) = -ixy

Therefore, the function f(z) is given by:

f(z) = u(x, y) + iv(x, y) = 2kxiy - ixy = (2kx - i)x(iy - 1)

So, the function f(z) is f(z) = (2kx - i)x(iy - 1).

And v are real-valued functions, then the Cauchy-Riemann equations must hold:

∂u/∂x = ∂v/∂y

and

∂u/∂y = -∂v/∂x

Conversely, if these equations hold for a given function f(z), then it is analytic.

The real part of an analytic function f(z) is given by e^(-y) * cos(x). We are required to find the imaginary part of f(z).

To find the imaginary part of f(z), we can make use of the Cauchy-Riemann equations. According to these equations, if f(z) is an analytic function, then it satisfies the following conditions:

∂u/∂x = ∂v/∂y (1)
∂u/∂y = -∂v/∂x (2)

where u(x, y) is the real part of f(z) and v(x, y) is the imaginary part of f(z).

Let's differentiate the given real part of f(z) with respect to x and y:

∂u/∂x = -e^(-y) * sin(x) (3)
∂u/∂y = -e^(-y) * cos(x) (4)

Comparing equations (1) and (3), we can see that:

∂v/∂y = -e^(-y) * sin(x) (5)

Comparing equations (2) and (4), we can see that:

∂v/∂x = e^(-y) * cos(x) (6)

Now, integrating equation (5) with respect to y, we get:

v(x, y) = -e^(-y) * sin(x) + g(x) (7)

where g(x) is an arbitrary function of x.

Next, substituting equation (7) into equation (6), we can solve for g(x):

∂v/∂x = e^(-y) * cos(x) (6)
e^(-y) * cos(x) = e^(-y) * cos(x) + g'(x) (8)
g'(x) = 0 (9)

Since g'(x) = 0, it implies that g(x) is a constant.

Therefore, the imaginary part of f(z) is given by:

v(x, y) = -e^(-y) * sin(x) + C (10)

where C is a constant.

Comparing equation (10) with the given options, we can see that the correct answer is option B, i.e., e^(-y) * sin(x).

Analytic Functions in Complex Analysis

An analytic function is a complex function that is differentiable at every point in its domain.

If a function is analytic at all the points in the complex plane, then it is called an entire function.

If a function is not analytic at any point in its domain, then it is called a non-analytic function.

Out of the given options, the function f(z) = log z is not analytic at all the points in the complex plane.

Explanation

The function f(z) = log z is not analytic at z = 0 and any other point where z is negative or zero.

The reason for this is that the complex logarithm is a multivalued function. For any non-zero complex number z, there are infinitely many complex numbers w such that ez = w. So, we define the complex logarithm as follows:

log z = ln |z| + i arg(z)

where arg(z) is any angle whose tangent is the imaginary part divided by the real part of z.

However, when z is negative or zero, arg(z) is not well-defined, and so log z is not analytic at these points.

Hence, the function f(z) = log z is not analytic at all the points in the complex plane.

To evaluate the integral of 1/z^2 dz, we can use the Cauchy's Integral Formula, which states that for a function f(z) that is analytic within and on a simple closed contour C, and a point a inside C,

f(a) = (1/2πi) ∮C [f(z)/(z-a)] dz.

In this case, f(z) = 1 and a = 0. The contour C is the unit circle traversed clockwise.

Applying the formula, we have:

1 = (1/2πi) ∮C [1/(z-0)] dz
= (1/2πi) ∮C 1/z dz.

The integral of 1/z with respect to z along the unit circle traversed clockwise is equal to -2πi. Therefore,

1 = (1/2πi) (-2πi)
= -1.

So, the value of 1/z^2 dz, where the contour is the unit circle traversed clockwise, is -1.

Explanation:

To understand why option C is the correct answer, let's first define what it means for a function to be analytic.

An analytic function is a complex-valued function that is differentiable at every point within its domain. In other words, the function has a derivative at each point within its domain.

Given that f(z) = u + iv is an analytic function, we can conclude the following:

Harmonic Functions:

A harmonic function is a real-valued function that satisfies Laplace's equation, which states that the sum of the second-order partial derivatives of the function is equal to zero. In other words, if u is a harmonic function, then it satisfies ∇²u = 0, where ∇² is the Laplacian operator.

Similarly, if v is a harmonic function, then it satisfies ∇²v = 0.

Derivatives of Analytic Functions:

Since f(z) is an analytic function, it means that both u and v have continuous first-order partial derivatives. This allows us to use the Cauchy-Riemann equations.

The Cauchy-Riemann equations relate the partial derivatives of u and v. They state that if f(z) = u + iv is an analytic function, then the following conditions must hold:

1. ∂u/∂x = ∂v/∂y
2. ∂u/∂y = -∂v/∂x

From these equations, we can see that the real part (u) and the imaginary part (v) of an analytic function are related to each other.

Conclusion:

Based on the above information, we can conclude that both u and v are harmonic functions. This is because if u and v are related to each other through the Cauchy-Riemann equations, and f(z) is an analytic function, then both u and v must satisfy Laplace's equation.

Therefore, the correct answer is option C: Both u and v are harmonic functions.