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

$\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}$ = 0

Explanation:

(1): u = x2 + y2

$\frac{{\partial }^{2}f}{\partial x^2}=2,\frac{{\partial }^{2}f}{\partial y^2}=2$

So, $\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}$ ≠ 0

u = x2 + y2 is not harmonic.

(1) is correct

(2): u = x2 - y2

$\frac{{\partial }^{2}f}{\partial x^2}=2,\frac{{\partial }^{2}f}{\partial y^2}=-2$

So, $\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}$ = 0

u = x2 - y2 is harmonic.

(3): u = sin hx cos y

$\frac{{\partial }^{2}f}{\partial x^2}$ = sin hx cos y and $\frac{{\partial }^{2}f}{\partial y^2}$ = - sin hx cos y

So, $\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}$ = 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 $\frac{{\partial }^{2}u}{{\partial }^{2}x}+\frac{{\partial }^{2}u}{{\partial }^{2}y}=0$

Solution:

Given,  $\mathrm{u}=\frac{1}{2}\log (x^2+y^2)$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

=$\frac{xdy-ydx}{x^2+y^2}$

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($\frac{xdy-ydx}{x^2+y^2}$) = 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 ${\displaystyle \mathrm{s}\mathrm{u}{\mathrm{p}}_{\mathrm{z},\mathrm{w}\in \mathrm{U}}}$ |𝑓(𝑧) − 𝑓(𝑤)| = 1 }.

Concept:

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

Calculation:

Let $\mathrm{g}(\mathrm{z})=\mathrm{f}(\mathrm{z})-\mathrm{f}(-\mathrm{z})\mathrm{\forall }\mathrm{z}\in \mathrm{U}$

$⟹|\mathrm{g}(\mathrm{z})|\le 1$

Now,

${\mathrm{g}}^{\prime }(\mathrm{z})={\mathrm{f}}^{\prime }(\mathrm{z})+{\mathrm{f}}^{\prime }(-\mathrm{z})$

$⟹{\mathrm{g}}^{\prime }(0)={\mathrm{f}}^{\prime }(0)+{\mathrm{f}}^{\prime }(0)=2{\mathrm{f}}^{\prime }(0)$

$⟹|{\mathrm{g}}^{\prime }(\mathrm{a})|\le \frac{\mathrm{M}}{1/2}=2$

$⟹|{\mathrm{g}}^{\prime }(0)|\le 2$

$⟹|2{\mathrm{f}}^{\prime }(0)|\le 2$

$⟹|{\mathrm{f}}^{\prime }(0)|\le 1<2$

Hence the statement P is incorrect.

Now,

${\mathrm{g}}^{″}(\mathrm{z})={\mathrm{f}}^{″}(\mathrm{z})-{\mathrm{f}}^{″}(-\mathrm{z})$

${\mathrm{g}}^{‴}(\mathrm{z})={\mathrm{f}}^{‴}(\mathrm{z})+{\mathrm{f}}^{‴}(-\mathrm{z})$

${\mathrm{g}}^{‴}(0)=2{\mathrm{f}}^{‴}(0)$

$|{\mathrm{g}}^{‴}(0)|\le \frac{1\cdot 3!}{(1/2{)}^3}=48$

$⟹|2{\mathrm{f}}^{‴}(0)|\le 48$

$⟹|2{\mathrm{f}}^{‴}(0)|\le 48$

$⟹|{\mathrm{f}}^{‴}(0)|\le 24<48$

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)| = $\sqrt{(\mathrm{x}+\mathrm{a}{)}^2+(\mathrm{y}+\mathrm{b}{)}^2}$

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

for |z + w| = |z−w| to be hold

$\sqrt{(\mathrm{x}+\mathrm{a}{)}^2+(\mathrm{y}+\mathrm{b}{)}^2}=\sqrt{(\mathrm{x}-\mathrm{a}{)}^2+(\mathrm{y}-\mathrm{b}{)}^2}$

⇒(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

​​${\displaystyle \frac{\partial u}{\partial r}}={\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial v}{\partial \theta }}\text{and}{\displaystyle \frac{\partial v}{\partial r}}=-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial u}{\partial \theta }}$

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

​​${\displaystyle \frac{\partial u}{\partial r}}={\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial v}{\partial \theta }}\text{and}{\displaystyle \frac{\partial v}{\partial r}}=-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial u}{\partial \theta }}$

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

$\int du=\int \left(x\right)dx-\int ydy$

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

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.

• $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}and\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$
• $\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{o}\mathrm{u}\mathrm{s}\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{x}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{y}.$

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

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

Calculation:

Given function: f = 2x – x² + my²

So, for harmonic it should satisfy Laplace’s equation

${\mathrm{\nabla }}^{2}f=0=\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}$

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

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

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

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)

$\frac{\partial v}{\partial x}=4y+f^{\prime }(x)$

$\frac{\partial u}{\partial y}=-\frac{\partial \nu }{\partial x}$

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

4y – 4x = 4y + f’(x)

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

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 ${\mathrm{\nabla }}^2\mathrm{f}(\mathrm{x},\mathrm{y})=0$

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.