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

Concept:

(i) A complex function f(z) is entire function if it is analytic in whole complex plane.

(ii) If a complex function f(z) = u + iv is entire then it satisfy C-R equation i.e., u_x = v_y, u_y = - v_x

Explanation:

 f : ℂ → ℂ is a real-differentiable function.

 u, v : ℝ2 → ℝ by u(x, y) = Re f(x + i y) and v(x, y) = Im f(x + iy), x, y ∈ ℝ. 

Also, ∇u = (ux, uy)

(1): Then "For c1, c2 ∈ ℂ, the level curves u = c1 and v = c2 are orthogonal wherever they intersect" this statement will satisfy only if f(z) is analytic function.

(1) is false

(3): f(z) is entire function so ux = vy, uy = - vx

then ∇u . ∇v = (ux, uy) . (vx, vy) = uxvx + uyvy = uxvx - vxux = 0 at every point.  

(3) is true and (2) is false

(4): ∇u . ∇v = 0

⇒ (ux, uy) . (vx, vy) = 0

⇒ uxvx + uyvy = 0

⇒ uxvx = - uyvy 

which does not imply ux = vy, uy = - vx

f is not an entire function. 

(4) is false

Concept:

A complex function f(z) is differentiable then  = 0

Explanation:

f(z) = |z|2, z ∈ ℂ

so f(z) = x² + y²

then f(z) is continuous everywhere

Now, f(z) = |z|2 = 

So,  = z 

Therefore  = 0 at z = 0 only

 i.e., f(z) is differentiable at the origin only

Hence f(z) is continuous everywhere but nowhere differentiable except at the origin.

(3) is correct

Given -

Concept - 

If singular point  z = c of f(z) lies in | z - a | = r then 

If singular point  z = c of f(z) does not lie in | z - a | = r then \(\displaystyle \int_{|z-a|=r} F(z) dz = 2\pi i \times Rez(f(z))_{ z = c}=0\)

Explanation -

Where 

For Singularity - 

⇒ 

F(z) has singularity at z = 2 and z = -2 But the singularity z = 2 does not lie in  Hence the integral should be zero for z = 2.

Now  singularity z = -2  lies in \(|z+1|=2\) Hence we have to calculate the integral using the above concept -

So,   .........(i)

⇒ 

= 

Put this value in the above equation we get -

⇒ 

Hence the option (iii) is correct.

Explanation:

f be an entire function that satisfies |f(z)| ≤ ey for all z = x + iy ∈ ℂ

(1): f(z) = ce−iz 

So |f(z)| = |ce−iz| = |ce^{-i(x + iy)}| = |ce-ix ey| ≤ |c|ey ≤ ey for some c ∈ ℂ with |c| ≤ 1 (as |e-ix| ≤ 1 for all x ∈ )

Option (1) is correct

(2): f(z) = ceiz 

So |f(z)| = |ceiz| = |cei(x + iy)| = |ceix e-y| ≤ e-y for some c ∈ ℂ with |c| ≤ 1 (as |e-ix| ≤ 1 for all x ∈ )

Option (2) is false

(3): f(z) = e−ciz 

So |f(z)| = |e−ciz | = |e-ci(x + iy)| = |e-cix ecy| ≤ ecy ≤ ey for c = 1 only (as |e-ix| ≤ 1 for all x ∈ )

Option (3) is false

(4): f(z) = eciz 

So |f(z)| = |eciz | = |eci(x + iy)| = |ecix e-cy| ≤ e-cy ≤ ey for c = - 1 only (as |e-ix| ≤ 1 for all x ∈ )

Option (4) is false

Explanation: 

If R is the radius of convergence of  then the series converges when |x|

i.e., when - R

So interval of convergence is (- R, R)

(1) is correct

Concept:

Rouche’s Theorem: If f(z) and g(z) are two analytic functions within and on a simple closed curve C such that |f(z)| roots inside C.

Explanation:

z100 - 50z30 + 40z10 + 6z + 1 and the open disc {z ∈ ℂ : |z|

Let f(z) = z100 +  40z10 + 6z + 1 and g(z) = - 50z30 

Then |f(z)| = |z100 +  40z10 + 6z + 1| ≤ |z100|+  40|z10| + 6|z| + 1

and |g(z)| = | - 50z30| = 50|z30|

Hnece |f(z)|

Then By Rouche's theorem, 

f(z)+g(z) and g(z) has same roots inside {z ∈ ℂ : |z|

Now, g(z) = - 50z30 has 30 roots inside {z ∈ ℂ : |z|

Therefore z100 - 50z30 + 40z10 + 6z + 1 has 30 roots inside {z ∈ ℂ : |z|

(3) is correct

Given -

Let R denote the radius of convergence of power series . 

Concept -

If the power series is given by 

 then Radius of convergence R of the power series is 

 And The interval of convergence for the power series is  

Explanation -

⇒ R = 1

Hence option (iv) is false.

Now the interval of convergence for the power series is 

Now we check the convergence of the power series at end points -

Now at x = 1

put this value in the given series we get the series -

Clearly the series is divergent.

Now at x = -1

put this value in the given series we get the series -

Clearly the series is also divergent.

Hence the power series is convergent at (-1,1) or (-R,R)

Hence option (iii) is true.

Concept:

Argument theorem: Let f be meromorphic function and C be simple closed contour such that no zero or pole of f lies inside C. Let a₁, a₂,…,a_k be zeros of f of order n₁, n₂,…,n_k respectively and f(z) has no pole in C. Then
  = 

Explanation:

g(z) = z3 and and f(z) = z3 - z - 1.

Let a, b, c are the zeros of f(z) lies in C then

a + b + c = 0

a² + b² + c² = -1 and

abc = 1

Now, 

 = a³ + b³ + c³

                           = (a + b + c)(a² + b² + c² - ab - bc - ca) + 3abc

                           = 0(-1) + 3 = 3

Option (1) is true.

Concept:

Cauchy Residue Theorem:

Let  be a function that is analytic inside and on a simple closed contour C, except for a finite number of

isolated singularities (poles) inside C. If  has isolated singularities at   inside C, then the integral of  around  C is given by

Explanation:

, where the contour  is piecewise defined as:

We will apply the Cauchy Residue Theorem to solve this integral. The integrand is

This function has two singularities at   and  . If the contour  encloses these singularities, the value of the integral is determined by the residues of the function at these points.

Here winding number of z = 0 is 2 and winding number of z = 2 is 1.

Residue at  :

The residue at  is found by multiplying  by  and taking the limit as  :

Residue at :

The residue at  is found similarly,

Applying the Residue Theorem:

Since the contour  encloses both singularities at  and  , the value of the integral is

I = 

  = 

  =  = -πi

Hence, the correct option is (3).

Explanation:

The radius of convergent of ∑ anxn is 7 

So, the radius of convergent of ∑an(x + 3)n 

= max{0, R - |a|} = max{0, 7 - |3|} = max{0, 4} = 4

(4) is correct