Analytic Functions MCQ Quiz - Objective Question with Answer for Analytic Functions - Download Free PDF

Concept:

A function f(x, y) is said to be harmonic if it satisfy

 = 0

Explanation:

(1): u = x2 + y2

So,  ≠ 0

u = x2 + y2 is not harmonic.

(1) is correct

(2): u = x2 - y2

So,  = 0

u = x2 - y2 is harmonic.

(3): u = sin hx cos y

 = sin hx cos y and  = - sin hx cos y

So,  = 0

u = sin hx cos y is harmonic.

(4): Similarly we can show that

u = log(x2 + y2) is harmonic.

Concept:

A function u = f(x, y) is called hamonic function if 

Solution:

Given,  is harmonic

Now,  =  =  

 =  = 

∴ dv =dx + dy 

=dx + dy [Using CR equations]

=dx + dy

=

Integrating both sides,

v =  + C

∴ The harmonic conjugate is 

The correct answer is option 1.

Additional Informationd(xy) = xdy + ydx

d() = 

d() = d()

Given -​

Let Ω be an open connected subset of ℂ containing 𝑈 = { 𝑧 ∈ ℂ ∶ |𝑧| ≤ }.

Let 𝔍 = { 𝑓 ∶ Ω → ℂ ∶ 𝑓 is analytic and  |𝑓(𝑧) − 𝑓(𝑤)| = 1 }.

Concept:

If f(z) is analytic within and on   . Let a be a point inside C. 

Calculation:

Let 

Now,

Hence the statement P is incorrect.

Now,

Hence the statement Q is incorrect.

Hence the option (2) and (3) are correct.

Calculation:

Let z = x + iy & w = a + ib

⇒ |z + w| = |(x + a) + i(y + b)| = 

⇒ & |z − w| = |(x − a) + i(y − b)| = 

for |z + w| = |z−w| to be hold

⇒(x + a)² + (y + b)² = (x − a)² + (y − b)²

⇒ x² + a² + 2ax + y² + b² + 2by = x² + a² − 2ax + y² + b² − 2by

⇒ 4(ax + by) = 0 ⇒ ax + by = 0

Now, z ⋅ w̅ = (x + iy) (a − ib) = ax − ibx + iay + by = (ax + by) − i(bx − ay) = − i(bx − ay),   {∵ ax + by = 0)

⇒ z ⋅ w̅ = − i(bx − ay) is purely imaginary.

The correct answer is option "4"

Cauchy-Riemann equations:

Rectangular form:

f(z) = u(x, y) + f v(x, y)

f(z) to be analytic it needs to satisfy Cauchy Riemann equations

ux = vy, uy = -vx

Polar form:

f(z) = u(r, θ) + f v(r, θ)

 and uθ = -rvr

​​

Cauchy-Riemann equations:

Rectangular form:

f(z) = u(x, y) + f v(x, y)

f(z) to be analytic it needs to satisfy Cauchy Riemann equations

ux = vy, uy = -vx

Polar form:

f(z) = u(r, θ) + f v(r, θ)

 and uθ = -rvr

​​

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 

vy = ux

vx = - uy

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

Concept:

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

i.e., 

Calculation:

Given:

v = xy​

If u = f(x, y)

du = xdx - ydy

Integrating both sides

Concept:

The complex function f(z) = u (x, y) + iv (x, y) is to be analytic if it satisfy the following two conditions of Cauchy-Reiman theorem.

f(z) = log z

Here, the function f(z) is analytic except z = 0. Since the function is not defined for these two values.

But in question it is asked at all points hence f(z) = log z is not analytic at all points.

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:

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) = ?

Integrating w.r.t y keeping x constant

ν(x, y) = 4xy + 2y² + f(x)

4y – 4x = 4y + f’(x)

∴ ν(x, y) = 4xy + 2y² – 2x² + C

Concept:

Let f(z) = u + iv

if f(z) is analytic

Calculation:

Given:

u = constant

since u is constant

this is possible only if v = constant

Hence, f(z) = constant (Both real part (u) and imaginary part (v) are constant)

Concept:

If f(z) = u + iv is an analytic function, then it satisfies the following:

Calculation:

Given: u = e^-y cos x

    ---(1)

      ---(2)

Integrate equation (1) w.r.t. y, taking x as constant, we get:

v = e^-y sin x

Explanation:

f(z) = u + iv

⇒ i f(z) = - v + i u

⇒ (1 + i) f(z) = (u - v) + i(u + v)

⇒ F(z) = U + iv, where F(z) = (1 + i) f(z)

U = u – v, V = u + v

Now,

Let F(z) be an analytic function

dV = e^x (sin y + cos y) dx + e^z(cosy – siny) dy

∴ dV = d[e^x(siny + cosy)]

Now,

On integrating

V = e^x (siny + cosy) + c₁

F(z) = U + iV = e^x(cosy - siny) + i e^x (siny + cosy) + ic₁

F = e^x(cosy + isiny) + ie^x (cosy + isiny) + ic₁

F(z) = (1 + i) e^{x + iy} + ic₁ = (1 + i)e^z + ic₁

⇒ (1 + i) F(z) = (1 + i) e^z + ic₁

∴ f(z) = e^z + (1 + i) c

Explanation-

Given that,

The function f(x, y) satisfies the Laplace equation 

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.