Test: Analytic Functions - 1 - Civil Engineering (CE) MCQ

Concept:
If a function f satisfies Laplace's equation ∇²f = 0, then f is said to be a harmonic function.
Calculation:
If f (z) = u + iv is an analytic function, then

Now 

Both u and v are satisfying Laplace’s equation (∇²f = 0).
∴ Both u and v are harmonic functions.

Given u = 2kxy & V = x² – y²
Concept:
For function to be analytic:

∴ Apply any of the above condition to get the answer.
By first condition:
2ky = -2y ⇒ k = -1

Concept:
f(z) = u + iv
u = real part
v = imaginary part

If f(z) is an analytic function

(This is an exact differential equation)

Calculation:
Given,
u = x³ – 3xy²
∂u/∂x = 3x² − 3y²
∂u/∂y = −6xy

It is an exact differential equation the solution is obtained by treating y as constant in the first term and in the second term only that part is integrated which is not containing x.

Integrating the above equation
v = 3x²y − y³ + constant

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 =
dV = e^x (sin y + cos y) dx + e^z(cosy – siny) dy
∴ dV = d[ex(siny + cosy)]

Now,
On integrating
V = ex (siny + cosy) + c1
F(z) = U + iV = ex(cosy - siny) + i e^x (siny + cosy) ic₁
F = ex(cosy + isiny) + iex (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

Concept:
Let f(z) = u + iv be the analytic function,
then According to Cauchy-Riemann Equations
u_x = v_y & u_y = - v_x
Calculation:
Given:
modulus is constant
= k ⇒ u² + v² = k
Now differentiating both sides w.r.t x and y, we get
2uu_x + 2vv_x = 0 ⇒ 2uu_x = -2vv_x      --- (1)
2uu_y + 2vv_y = 0 ⇒  2uu_y = -2vv_y     --- (2)
Now applying Cauchy-Riemann Equations u_x = v_y & u_y = - v_x
2uu_x – 2vu_y = 0 ⇒ 2uu_x = 2vu_y     --- (3)
2uu_y + 2vu_x = 0 ⇒ 2uu_y = -2vu_x     --- (4)
Now dividing equations 3 and 4, we get 

⇒ u_x = u_y = 0 ⇒ u is a constant function.
Similarly, we will get v is a constant function.
∴ f(z) is a constant function.

Concept:

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

u_x = v_y, u_y = -v_x

Polar form:
f(z) = u(r, θ) + f v(r, θ)

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:
Let f(Z) = u(x, y) + i(v(x, y)) be an analytical function.
If the real part u(x, y) of analytic function f(z) is given then to find imaginary part v(x, y) of f(z) we can use the following procedure:
1). Find u_x and u_y
2). Consider dv =

3). dv = v_x dx + v_y dy = -u_y dx + u_x dy.

​4). v = ∫(−u_y)dx + ∫(u_x)dy + c

Calculation:
Given:
u = 5x + 2xy
u_x = 5 + 2 × y, uy = 2 × x.
dv = v_x dx + v_y dy = -u_y dx + u_x dy = (-2 × x) dx + (5 + 2 × y) dy.

Now by integrating we get:

v = ∫(−u_y) dx + ∫(u_x)dy

v = ∫(−2x) dx + ∫(5+2y)dy + c

y² - x² + 5y + c

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.

are continuous function of x and y.
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.

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

u_x = v_y, u_y = -v_x

Polar form: