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

\({\partial ^2 f\over \partial x^2}+{\partial ^2 f\over \partial y^2}\) = 0

Explanation:

(1): u = x2 + y2

\({\partial ^2 f\over \partial x^2}=2, {\partial ^2 f\over \partial y^2}= 2\)

So, \({\partial ^2 f\over \partial x^2}+{\partial ^2 f\over \partial y^2}\) ≠ 0

u = x2 + y2 is not harmonic.

(1) is correct

(2): u = x2 - y2

\({\partial ^2 f\over \partial x^2}=2, {\partial ^2 f\over \partial y^2}=- 2\)

So, \({\partial ^2 f\over \partial x^2}+{\partial ^2 f\over \partial y^2}\) = 0

u = x2 - y2 is harmonic.

(3): u = sin hx cos y

\({\partial ^2 f\over \partial x^2}\) = sin hx cos y and \( {\partial ^2 f\over \partial y^2}\) = - sin hx cos y

So, \({\partial ^2 f\over \partial x^2}+{\partial ^2 f\over \partial y^2}\) = 0

u = sin hx cos y is harmonic.

(4): Similarly we can show that

u = \(\frac{1}{2}\)log(x2 + y2) is harmonic.

Concept:

A function u = f(x, y) is called hamonic function if \(\frac{\partial^2u}{\partial^2x}+\frac{\partial^2u}{\partial^2y}=0\)

Solution:

Given,  \(\mathrm{u}=\frac{1}{2} \log \left(x^2+y^2\right)\)is harmonic

Now, \(\frac{\partial u}{\partial x}\) = \(\frac{1}{2}\times\frac{2x}{x^2+y^2}\) = \(\frac{x}{x^2+y^2}\) 

\(\frac{\partial u}{\partial y}\) = \(\frac{1}{2}\times\frac{2y}{x^2+y^2}\) = \(\frac{y}{x^2+y^2}\)

∴ dv =\(\left ( \frac{\partial v}{\partial x} \right )\)dx + \(\left ( \frac{\partial v}{\partial y} \right )\)dy 

=\(\left ( \frac{-\partial u}{\partial x} \right )\)dx + \(\left ( \frac{\partial u}{\partial y} \right )\)dy [Using CR equations]

=\(\left (\frac{-y}{x^2+y^2}\right )\)dx + \(\left (\frac{x}{x^2+y^2}\right )\)dy

=\(\frac{x dy-ydx}{x^2+y^2}\)

Integrating both sides,

v = \(\tan^{-1}\left ( \frac{y}{x} \right )\) + C

∴ The harmonic conjugate is \(\tan ^{-1}\left(\frac{y}{x}\right)+\mathrm{c}\)

The correct answer is option 1.

Additional Informationd(xy) = xdy + ydx

d(\(\frac{x}{y}\)) = \(\frac{ydx-xdy}{y^2}\)

d(\(\frac{x dy-ydx}{x^2+y^2}\)) = d(\(\tan^{-1}\left ( \frac{y}{x} \right )\))

Given -​

Let Ω be an open connected subset of ℂ containing 𝑈 = { 𝑧 ∈ ℂ ∶ |𝑧| ≤ \(\frac{1}{2}\)}.

Let 𝔍 = { 𝑓 ∶ Ω → ℂ ∶ 𝑓 is analytic and \(\rm \displaystyle sup_{z,w \in U}\) |𝑓(𝑧) − 𝑓(𝑤)| = 1 }.

Concept:

If f(z) is analytic within and on  \(\rm C\ and\ |f(z)| \le M\) . Let a be a point inside C. \(\rm |f^n(a)|\le \frac{M\cdot n!}{R^n}\ \ for\ |z-a|=R\)

Calculation:

Let \(\rm g(z)=f(z)-f(-z)\ \forall\ z \in U\)

\(\rm \implies|g(z)|\le 1\)

Now,

\(\rm g'(z)=f'(z)+f'(-z)\)

\(\rm \implies g'(0)=f'(0)+f'(0)=2f'(0)\)

\(\rm \implies |g'(a)| \le \frac{M}{1/2}=2\)

\(\rm \implies |g'(0)| \le 2\)

\(\rm \implies |2f'(0)| \le 2\)

\(\rm \implies |f'(0)| \le 1<2\)

Hence the statement P is incorrect.

Now,

\(\rm g''(z)=f''(z)-f''(-z)\)

\(\rm g'''(z)=f'''(z)+f'''(-z)\)

\(\rm g'''(0)=2f'''(0)\)

\(\rm |g'''(0)| \le \frac{1\cdot3!}{(1/2)^3}=48\)

\(\rm \implies |2f'''(0)| \le 48\)

\(\rm \implies |2f'''(0)| \le 48\)

\(\rm \implies |f'''(0)| \le 24<48\)

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)| = \(\rm\sqrt{(x+a)^2+(y+b)^2}\)

⇒ & |z − w| = |(x − a) + i(y − b)| = \(\rm\sqrt{(x−a)^2+(y−b)^2}\)

for |z + w| = |z−w| to be hold

\(\rm\sqrt{(x+a)^2+(y+b)^2}=\sqrt{(x−a)^2+(y−b)^2}\)

⇒(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

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

Polar form:

f(z) = u(r, θ) + f v(r, θ)

\({u_r} = \frac{1}{r}{v_\theta }\) and uθ = -rvr

​​\(\dfrac{\partial u}{\partial r} = \dfrac{1}{r} \dfrac{\partial v}{\partial \theta} \ \text{and} \ \dfrac{\partial v}{\partial r}=-\dfrac{1}{r} \dfrac{\partial u}{\partial\theta }\)

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

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

Polar form:

f(z) = u(r, θ) + f v(r, θ)

\({u_r} = \frac{1}{r}{v_\theta }\) and uθ = -rvr

​​\(\dfrac{\partial u}{\partial r} = \dfrac{1}{r} \dfrac{\partial v}{\partial \theta} \ \text{and} \ \dfrac{\partial v}{\partial r}=-\dfrac{1}{r} \dfrac{\partial u}{\partial\theta }\)

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

Calculation:

Given:

v = xy​

\(\frac{{\partial v}}{{\partial y}} = x \Rightarrow \frac{{\partial u}}{{\partial x}} = x\)

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

If u = f(x, y)

\(du = \frac{{\partial u}}{{\partial x}}dx + \frac{{\partial u}}{{\partial y}}dy\)

du = xdx - ydy

Integrating both sides

\(\smallint du = \smallint \left( { x} \right)dx - \smallint ydy\)

\(u = \frac{1}{2}\left( {{x^2} - {y^2}} \right)\)

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.

• \(\frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}}\;\;and\;\frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}}\)
• \(\frac{{\partial u}}{{\partial x}},\frac{{\partial u}}{{\partial y}},\frac{{\partial v}}{{\partial x}},\frac{{\partial v}}{{\partial y}}\;{\rm{are\;continuous\;function\;of\;x\;and\;y}}.\)

f(z) = log z

\(\frac{{\partial }}{{\partial z}}f(z) = \frac{1}{z}\)

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

Concept:

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

\({{\nabla }^{2}}~f\left( x,~y \right)=0=\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}\)

Calculation:

Given function: f = 2x – x² + my²

So, for harmonic it should satisfy Laplace’s equation

\({{\nabla }^{2}}f=0=\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}\)

\(\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}=\frac{\partial \left( \frac{\partial f}{\partial x} \right)}{\partial x}=\frac{\partial }{\partial x}~\left( 2-2x \right)=-2\)

\(\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}=\frac{\partial \left( \frac{\partial f}{\partial y} \right)}{\partial y}=\frac{\partial \left( 2my \right)}{\partial y}=2m\)

\(\Rightarrow \frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}=-2+2m=0\)

Concept:

Let w = u + iν be a function of complex variable.

Function of a complex variable is analytic, if it satisfies Cauchy-reimann equation;

\(i.e.\frac{{\partial u}}{{\partial x}} = \frac{{\partial \nu }}{{\partial y}}\;and\;\frac{{\partial u}}{{\partial y}} = - \frac{{\partial \nu }}{{\partial x}}\)

Calculation:

Given:

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

\(\frac{{\partial u}}{{\partial x}} = \frac{{\partial \nu }}{{\partial y}}\)

\(\frac{{\partial u }}{{\partial x}}=4x + 4y = \frac{{\partial \nu }}{{\partial y}}\)

Integrating w.r.t y keeping x constant

ν(x, y) = 4xy + 2y² + f(x)

\(\frac{\partial v}{\partial x}=4y\;+\;f'(x)\)

\(\frac{{\partial u}}{{\partial y}} = - \frac{{\partial \nu }}{{\partial x}}\)

\(\frac{\partial u}{\partial y}= - 4y + 4x \)

4y – 4x = 4y + f’(x)

\(f\left( x \right) = - \frac{{4{x^2}}}{2} + C= - 2{x^2} + C\)

Concept:

Let f(z) = u + iv

if f(z) is analytic

\(\begin{array}{l} \frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}}\\ \frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}} \end{array}\)

Calculation:

Given:

u = constant

since u is constant

\(\begin{array}{l} \frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}} = 0 \Rightarrow v = f(x)\\ \frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}} = 0 \Rightarrow v = f(y) \end{array}\)

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:

\(\frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{y}\;\;and\;\frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}}\)

Calculation:

Given: u = e^-y cos x

\(\frac{{\partial u}}{{\partial x}} = - {e^{ - y}}\sin x\)

\(\frac{{\partial u}}{{\partial y}} = - {e^{ - y}}\cos x\)

\(\frac{{\partial v}}{{\partial y}} = - {e^{ - y}}\sin x\)    ---(1)

\(\frac{{\partial v}}{{\partial x}} = {e^{ - y}}\cos x\)      ---(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

\(dV\; = \;\frac{{ - \partial U}}{{\partial x}}dy\)

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

\(\begin{array}{l} \Rightarrow \;f\left( z \right)\; = \;{e^z} + \frac{i}{{1 + i}}{c_1}\; = \;{e^z} + \frac{{i\left( {1 - i} \right)}}{{\left( {1 + i} \right)\left( {1 - i} \right)}}{c_1}\\ = \;{e^z} + \frac{{\left( {i + 1} \right)}}{2}{c_1} \end{array}\)

∴ f(z) = e^z + (1 + i) c

Explanation-

Given that,

The function f(x, y) satisfies the Laplace equation \(\rm \nabla ^2 f(x, y) = 0\)

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.