Test: Analytic Functions - 2 - Civil Engineering (CE) MCQ

Concept:

Conjugate of a complex number:

For any complex number z = x + iy  the conjugate z̅ is given by  z̅ = x - iy

We can expand the expression

A∠θ as Acosθ + jAsinθ ----(1)

Calculation: 

Given:

Z = 10∠45°

By using expression from equation (1)

Z  = 10cos45° + jsin45° 

Now the conjugate of above expression will be:

Z̅ = 10cos45° - jsin45° 

Z̅ = 10∠-45°

Given that,

The function f(x, y) satisfies the Laplace equation ∇²f(x,y)=0

on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. 

The value of this function on the circular boundary of this domain is equal to 3.

Here it is given that the value of the function is 3 for its domain, which signifies that it is a constant function whose value is 3.

So the value of the function at (0, 0) is 3.

Given region |z|≤ 1

a) 
z = 0 |z| = 0 ≤ 1

The pole is lies inside the given region.

Hence, the function is not analytic.

b) 
 z + 2 = 0 → z = -2 |z| = 2 ≥ 1

the pole is lies outside the given region.

Hence, the function is analytic.

c) 
z – 0.5 = 0 z = 0.5 |z| = 0.5 ≤ 1

The pole is lies inside the given region.

Hence, the function is not analytic.

d) 
 z + j 0.5 = 0 ⇒ z = -j 0.5 ⇒ |z| = 0.5 ≤ 1

The pole is lies inside the given region.

Hence, the function is not analytic.

i) f(z) = Re(z) = Re(x + iy) = x

here u = x, v = 0

u_x = 1, v_y = 0

f(z) is not satisfying CR equations Hence f(z) is not analytic.

ii) f(z) = z̅ = x - fy

u = x, v = -y

u_x = 1, v_y = -1

u_x ≠ v_y

f(x) is not satisfying CR equations.

Hence f(z) is not analytic

iii) f(z) = Im (z) = Im (x + iy) = y

u = 0, v = y

u_x = 0, v_y = 1

f(x) is not satisfying CR equations.

Hence f(z) is not analytic

iv) f(z) = sin z = (x + iy)

= sin x cos iy + cos x sin iy

= sin x cos hy + i cos x sin hy

u = sin x cos hy, v = cos x sin hy

u_x = cos x cos hy

v_y = cos x cos hy

u_y = sin x sin hy

v_x = - sin x sin hy

u_x = vy and u_y = - v_x

f(z) is satisfying CR equation.

Hence f(z) is analytic.

The argument of a complex number z = x + iy
arg⁡ z = tan⁻¹ (y/x)

Concept:
If f(z) = u + iv is an analytic function, then it satisfies the following:
Calculation:
Given: u = e^-y cos x
∂u/∂x = −e^−y sin⁡x
∂u/∂y = −e^−y cos⁡x
∂v/∂y = −e^−y sin⁡x    ---(1)
∂v/∂x = e^−y cos⁡x      ---(2)
Integrate equation (1) w.r.t. y, taking x as constant, we get:
v = e^-y sin x

Concept:
Let w = u + iν be a function of complex variable.
Function of a complex variable is analytic, if it satisfies Cauchy-reimann equation;

Calculation:
Given:

u(x, y) = 2x² – 2y² + 4xy, ν(x, y) = ?

∂u/∂x = ∂ν/∂y
∂u/∂x = 4x + 4y = ∂ν/∂y

Integrating w.r.t y keeping x constant

ν(x, y) = 4xy + 2y² + f(x)
∂v/∂x = 4y + f′(x)
∂u/∂y =−∂ν*∂x
∂u/∂y = −4y+4x
4y – 4x = 4y + f’(x)
f(x) = −4x²*2 + C = −2x² + C
∴ ν(x, y) = 4xy + 2y² – 2x² + C

Concept:

If f(x, y) is harmonic then it must satisfy Laplace’s equation 

Calculation:
Given function: f = 2x – x² + my²
So, for harmonic it should satisfy Laplace’s equation

⇒ m = 1

Concept:

if f(z) = u(x, y) + iv(x, y) is an analytic function then Cauchy-Riemann condition will be satisfied.

Calculation:
Given:
v = xy​

du = xdx - ydy
Integrating both sides

∫du = ∫ (x)dx − ∫ ydy
u = 1/2(x²−y²)

Concept:

If two functions u and v satisfy Cauchy-Riemann equations, then they are said to be harmonic conjugates with respect to each other.

Cauchy-Riemann equations are 
v_y = u_x
v_x = - u_y

Calculation:

Given u = x² – y², let v be the harmonic conjugate.

By Cauchy-Riemann equations,

v_y = u_x = 2x; v_x = - u_y = - (-2y) = 2y;

We have dv = v_x dx + v_y dy 

⇒ dv = 2y dx + 2x dy = d(2xy)

⇒ v = 2xy + k or v = 2xy
∴ The conjugate harmonic function is 2xy