Complex Analysis MCQ Quiz in தமிழ் - Objective Question with Answer for Complex Analysis - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Taylor series expansion of f(x) about x = 0 is

f(x) = 

Explanation:

f(x) = ln(cosh x) ⇒ f(0) = ln(cosh 0) = 0

f'(x) = tanh x ⇒ f'(0) = 0

f''(x) = sech²x ⇒ f''(0) = 1

f'''(x) = 2sech x (- sech x tanh x) = - 2sech²x tahn x ⇒ f'''(0) = 0

f⁽⁴⁾(x) = - 4sech x(- sech x tanh x)tanh x - 2sech²x sech2x ⇒ f(4)(0) = - 2 

Hence using Taylor series expansion at x = 0 we get

f(x) =  = 

(2) is correct

Given :-  is an analytic function of .

Concept used :- If w is an analytic function then 

Solution :- As we know that

 and 

subtracting both equation and putting  , we get

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

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

Explanation -

we have 

Let assume  here Re(z) = x and Im(z) = y

so we get -

Clearly 

we know that the C-R equations - 

So f(z) satisfy C-R equations at z = 0. but it is not differentiable at 0.

Hence option (iii) is correct.

Concept:

The radius of convergence of the series  is  

Explanation:

Here in the series ,  and k = 2

The radius of convergence of the series is

1/R =  

⇒ 1/R =  

⇒ 1/R =  

⇒ 

(2) is correct

Explanation:

f(z) =  = 1 -  

⇒ f(z) = 1 - ()

⇒ f(z) = 1 + 

Now for |z| ⇒  and 

⇒ f(z) = 1 + 

⇒ f(z) = 1 +  - 

⇒ f(z) = 1 +  -  ( for |x|  

(1) is correct