Analytic Functions MCQ Quiz in తెలుగు - Objective Question with Answer for Analytic Functions - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

f(z) = ϕ + iψ

ϕ = Real part, ψ = Imaginary part

If f(z) is analytic function

$\begin{array}{c}\frac{\partial \varphi }{\partial x}=\frac{\partial \psi }{\partial y}\\ \frac{\partial \psi }{\partial x}=-\frac{\partial \varphi }{\partial y}\end{array}\}\to C.Requation$

$\therefore d\psi =\frac{\partial \psi }{\partial x}dx+\frac{\partial \psi }{\partial y}dy$

$d\psi =\frac{-\partial \varphi }{\partial y}dx+\frac{\partial \varphi }{\partial x}dy$

Calculation:

Given, f(z) = x² – y² + i ψ (x, y)

$\varphi =x^2-y^2$

∵ f(z) is analytic function

$d\psi =\frac{-\partial \varphi }{\partial y}dx+\frac{\partial \varphi }{\partial x}\cdot dy$

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

$d\psi =2ydx+2xdy$

$\int d\psi =2\int d\left(xy\right)$

ψ = 2 xy

Given, z = 1 + i

Comparing it with z = x + iy, we get:

∴ x = 1, y = 1

$\therefore {\left(\psi \right)}_{\left(1,1\right)}=2\times 1\times 1=2$

Given region |z|≤ 1

a) $\frac{z^2-1}{z}$

z = 0 |z| = 0 ≤ 1

The pole is lies inside the given region.

Hence, the function is not analytic.

b)$\frac{z^2-1}{z+2}$

z + 2 = 0 → z = -2 |z| = 2 ≥ 1

the pole is lies outside the given region.

Hence, the function is analytic.

c) $\frac{z^2-1}{z-0.5}$

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) $\frac{z^2-1}{z+j0.5}$

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.

Concept:

Cauchy-Riemann Equation (Condition for a function to be analytic):

If z = x + iy and w = f(z) = u(x,y) + iv(x,y) then

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}and\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$ i.e. u_x = v_y and u_y = -v_x

Calculation:

Given:

w = z² ⇒ (x + iy)²

∴ w = (x² – y²) + i(2xy)

Comparing with standard form i.e. w = f(z) = u(x,y) + iv(x,y)

u(x, y) = x² – y² and v(x, y) = 2xy

u_x = 2x, -v_x = -2y

u_y = -2y, v_y = 2x

C-R equations are satisfied every where in the finite part of complex plane and all first order partial derivatives u_x, u_y, v_x, v_y are continuous function of x and y

∴ f(z) = z² is analytic everywhere i.e., f(z) = z² is entire function.

Additional Information

Entire function:

A function f(z) which is analytic at every point of the finite complex plane is known as entire functions. 

Eg. polynomial and exponential functions are entire functions.

Concept:

Let the real part u(x, y) of an analytic function f(z) = u + iv be given

To find the function v(x, y),

The total derivative of V is given by

$dv=\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy$ 

By using CR equations, u_x = v_y and u_y = -v_x

$dv=\frac{-\partial u}{\partial y}dx+\frac{\partial u}{\partial x}dy$ 

Calculation:

Given that, u(x, y) = In (x² + y²)

$u_x=\frac{2x}{\left(x^2+y^2\right)},u_y=\frac{2y}{\left(x^2+y^2\right)}$ 

From CR equation,

$u_x=v_y=\frac{2x}{\left(x^2+y^2\right)}$ 

By integrating on both the sides,

$v=\int \frac{2x}{x^2+y^2}dy$ 

$\Rightarrow v=2{\tan }^{-1}\left(\frac{y}{x}\right)+g\left(x\right)$ 

By differentiating w.r.t. ‘x’

$\Rightarrow v_x=\frac{-2y}{x^2+y^2}+g^{\prime }\left(x\right)$ 

$\Rightarrow -u_y=\frac{-2y}{x^2+y^2}+g^{\prime }\left(x\right)$ 

$\Rightarrow \frac{-2y}{x^2+y^2}=\frac{-2y}{x^2+y^2}+g^{\prime }\left(x\right)$ 

⇒ g’(x) = 0

⇒ g(x) = C

Explanation:

$x^2+y^2=1$

$\Rightarrow \left(x+jy\right)\left(x-jy\right)=1$

If $z=x+jy$ then its conjugate is given as $x-jy=\frac{1}{z}$

$\Rightarrow \frac{1+x+jy}{1+x-jy}=\frac{1+z}{1+\frac{1}{z}}$

$\Rightarrow \frac{1+x+jy}{1+x-jy}=\frac{z\left(1+z\right)}{z+1}=z=x+jy$

Concept:

f(z) = u + iv

u = real part

v = imaginary part

If f(z) is an analytic function

$\begin{array}{c}\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\ \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}\end{array}\}\to C.Requation$

$\therefore dv=\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy$

$dv=\frac{-\partial u}{\partial y}dx+\frac{\partial u}{\partial x}dy$ (This is an exact differential equation)

Calculation:

Given,

u = x³ – 3xy²

$\frac{\partial u}{\partial x}=3x^2-3y^2$

$\frac{\partial u}{\partial y}=-6xy$

$dv=\frac{-\partial u}{\partial y}dx+\frac{\partial u}{\partial x}dy$

$dv=6xydx+\left(3x^2-3y^2\right)dy$

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

Concept:

Let f(z) = u + iv be the analytic function,

Cauchy-Riemann equations are 

vy = ux

vx = - uy

Calculation:

Given $f\left(z\right)=g\left(z\right)+h\left(z\right)$

Let $g\left(z\right)=gu\left(z\right)+jgv\left(z\right)$

$h\left(z\right)=hu\left(z\right)+jhv\left(z\right)$

If g and h are differentiable, then it will satisfy C-R equations. So, we have

$\begin{array}{cc}g{u}_x=g{v}_y,& g{u}_y=-g{v}_x\\ h{u}_x=h{v}_y,& huy=-h{v}_x\end{array}$

$f\left(z\right)=\left(gu\left(z\right)+hu\left(z\right)\right)+j(\left(gv\left(z\right)+hv\left(z\right)\right)$

By observing the above equations, we get

$\begin{array}{c}\left(g{u}_x+h{u}_x\right)=\left(g{v}_y+h{v}_y\right)\\ \left(g{v}_y+h{v}_y\right)=-\left(g{v}_x+h{v}_x\right)\end{array}$

Explanation:

The argument of a complex number z = x + iy

$\arg z={\tan }^{-1}\left(\frac{y}{x}\right)$

$Z_1=5+\left(5\sqrt{3}\right)i,Z_2=\frac{2}{\surd 3}+2i$

Then $\arg Z_1={\tan }^{-1}\left(\frac{5\sqrt{3}}{5}\right)={60}^{\circ }$

And $\arg Z_2={\tan }^{-1}\left(\frac{2}{\frac{2}{\sqrt{3}}}\right)={60}^{\circ }$

So, $\arg \left(\frac{Z_1}{Z_2}\right)=\arg \left(Z_1\right)-\mathrm{a}\mathrm{r}\mathrm{g}\left(Z_2\right)$

$z=\varphi \left(x,y\right)+i\psi \left(x,y\right)$

Given that z is an analytic function

It satisfies the Cauchy Riemann equations

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}and\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

${\varphi }_x={\psi }_y,{\varphi }_y=-{\psi }_x$

$\varphi \left(x,y\right)=3x^{2}y-y^3$

${\varphi }_x=6xy={\psi }_y$

${\psi }_y=6xy$

By Integrating with respect to y an both sides

$\int {\psi }_y=\int 6xy$

$\Rightarrow \psi =3x{y}^2+f\left(x\right)$       ----(1)

$\Rightarrow {\psi }_x=3y^2+f^{\prime }\left(x\right)$       ----(2)

$\varphi \left(x,y\right)=3x^{2}y-y^3$

${\psi }_x=-{\varphi }_y=-\left(3x^2-3y^2\right)=3y^2-3x^2$      ----(3)

From (2) and (3)

$\Rightarrow 3y^2+f^{\prime }\left(x\right)=3y^2-3x^2$

$\Rightarrow f^{\prime }\left(x\right)=-3x^2$

By integrating w.r.t. x on both sides

$f\left(x\right)=\int -3x^{2}dx=-x^3+c$

From equation (2)

$\psi \left(x,y\right)=3x{y}^2-x^3$

⇒ 0 = 0 – 0 + c ⇒ c = 0

$\psi \left(x,y\right)=3x{y}^2-x^3$

$\psi \left(2,3\right)=3\left(2\right){\left(3\right)}^2-2^3=46$

Concept:

From CR equations,

$\frac{\partial \varphi }{\partial \mathrm{x}}=\frac{\partial \psi }{\partial \mathrm{y}}$

$\frac{\partial \varphi }{\partial \mathrm{y}}=\frac{\partial \psi }{\partial \mathrm{x}}$

Calculation:

$\frac{\partial \varphi }{\partial \mathrm{x}}=\frac{\partial }{\partial \mathrm{y}}\left[{\mathrm{x}}^2-{\mathrm{y}}^2+\frac{\mathrm{x}}{{\mathrm{x}}^2+{\mathrm{y}}^2}\right]=-2\mathrm{y}-\frac{2\mathrm{x}\mathrm{y}}{{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^2}$

Integrating w.r.t ‘x’ by keeping y constant

$\Rightarrow \varphi =-2\mathrm{x}\mathrm{y}+\frac{\mathrm{y}}{{\mathrm{x}}^2+{\mathrm{y}}^2}+\mathrm{C}$

$\frac{\partial \psi }{\partial \mathrm{y}}=\frac{\partial }{\partial \mathrm{x}}\left[{\mathrm{x}}^2-{\mathrm{y}}^2+\frac{\mathrm{x}}{{\mathrm{x}}^2+{\mathrm{y}}^2}\right]=-2\mathrm{x}-\frac{{\mathrm{x}}^2-y^2}{{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^2}$

Integrating w.r.t ‘y’ by keeping x constant 

$\Rightarrow \varphi =-2\mathrm{x}\mathrm{y}+\frac{\mathrm{y}}{{\mathrm{x}}^2+{\mathrm{y}}^2}+\mathrm{C}$