Theorems of Complex analysis MCQ Quiz in বাংলা - Objective Question with Answer for Theorems of Complex analysis - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

Liouville's theorem:  Every bounded entire function must be constant.

Explanation:

F = {f : ℂ → ℂ| f is an entire function, | f'(z) | ≤ | f(z) | for all z ∈ ℂ}

f(z) is entire so f'(z) is entire

Hence \(f'(z)\over f(z)\) is entire

Now,

|f'(z) | ≤ | f(z) | for all z ∈ ℂ.

⇒ \(|f'(z) | \over | f(z) |\) ≤ 1 for all z ∈ ℂ.

⇒ |\(f'(z)\over f(z)\)| ≤ 1 for all z ∈ ℂ.

So \(f'(z)\over f(z)\) is entire and bounded

Hence \(f'(z)\over f(z)\) is constant

Let \(f'(z)\over f(z)\) = α 

Integrating both sides

⇒ log f(z) = αz + log β 

⇒ f(z) = β e^αz  where α, β ∈ ℂ

Now, f'(z) = βα eαz 

So, |f'(z) | ≤ | f(z) |

⇒ |βα eαz| ≤  |β eαz| 

⇒ |α| ≤ 1

i.e., F = {βeαz :  β ∈ ℂ, |α| ≤ 1}

Hence option (4) is true and option (3) is false

Also F is infinite 

So, option (2) is true and option (1) is false

Concept

(i) Riemann mapping theorem: If U is a non-empty simply connected open subset of the complex number plane \(\mathbb {C}\)  which is not all of \(\mathbb {C}\), then there exists a biholomorphic mapping f (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from U onto the open unit disk D = {z ∈ ℂ ∶ |z| < 1}

(ii) Schwarz lemma: Let  f(z) is an analytic function on the unit disc such that f(z) ≤ 1 for all z and f(0) = 0 then |f(z)| ≤ |z| and |f’(0)| ≤ 1.

Solution- 

(a): D = {z ∈ ℂ ∶ |z| < 1}

So by direct theorem, f ∶ D → D be a holomorphic function. Suppose α, β are distinct complex numbers in D such that f(α) = α and f(β) = β. Then f(z) = z for all z ∈ D.

(a) is true

(b): By Reimann mapping theorem, There exist a bijective holomorphic function from D to the set of all complex numbers whose imaginary part is positive.

(b) false

(c): f(α) = α and f'(α) = 1, α ∈ D

If we take α = 0 then f(0) = 0 and f'(0) = 1 by Schwarz lemman 

|f(z)| ≤ |z| and |f’(0)| ≤ 1 which implies f(z) = z

(c) is true

Therefore option (2) and (4) are correct

Explanation:

 f is analytic on the closed unit disk {|z| ≤ 1},

there is some k > 0 such that |f(z)| ≤ k ∀ |z| ≤ 1, as the continuous image of compact set is compact.

Also given that \(\rm |f(z)| ≤ \frac{1}{|z|^2}\) ≤ 1 for all |z| ≥ 10

so we have |f(z)| ≤ max{1, k} ∀ z ∈ C

Thus by Liouville's theorem f is bounded entire function and hence is constant function.

As f(0) = 1 then we must have f(z) = 1 for all z, which is a contradiction. So there does not exist such function.

(1) is incorrect

By corollaries of Liouville’s Theorem, If f is a non constant entire function, then image if f is dense in complex plane C.

(2) correct

Also we can prove using Liouville theorem that (3) correct

Explanation:

Every analytic function in \(D \subset C\) is obviously meromorphic 

Consequently, sums and products of meromorphic functions are meromorphic 

The Quotient of meromorphic function is meromorphic provided that denominator term 

is not identically zero. 

Therefore, Correct option is option 1 and option 2.

Explanation:

(1) f(z) = sin πz and z₀ = 0

when n = 1 then \(\rm\frac{1}{n}\) = 1

⇒ f(1) = 0

So we are getting only one point 1 in  \(\left\{\frac{1}{n}|n \right.\)∈ ℕ\(\left. \right\}\) for which f(z) = 0

Hence f can take the value z0 at finitely many points in \(\left\{\frac{1}{n}|n \right.\)∈ ℕ\(\left. \right\}\) 

(2) \(\rm\frac{1}{n}\) is a convergent sequence in ℂ so \(\rm f\left(\frac{1}{n}\right)\) is constant then by identity theorem f(z) = z₀.

(3) f(z) = sin πz and z₀ = 0

(4) f(r) = z₀, r ∈ Q ∩ [1, 2] has limit point in ℂ, hence f(z) must be constant.

The correct options are (1), (2) and (4).

Concept:

Liouville’s theorem: If a complex function is entire and bounded then it is constant.

Explanation:

|f(z)| ≥ 2025

⇒ \(\frac1{|f(z)|}\leq \frac1{2025}\)

Let g(z) = \(\frac1{f(z)}\)

|g(z)| = \(|\frac1{f(z)}|\) = \(\frac1{|f(z)|}\leq \frac1{2025}\)

So, g(z) is entire and bounded and hence constant.

g(z) = c

⇒ f(z) = 1/c

Since f(0) = 1 ⇒ c = 1

Hence f(z) = 1 for all z ∈ ℂ 

Concept:

Residue Theorem:
 

The integral of a function around a closed contour in the complex plane is \( 2\pi i \) times the sum of the residues

of the function inside the contour. Hence, \(I_r \) will depend on whether \(a\) and /or \(b\) lie inside the contour defined by \(\gamma_r\) .

Explanation: The contour \( \gamma_r \) is a circle of radius \( \gamma_r \) centered at the origin, and the integrand has singularities

(poles) at \(z = a\) and \(z = b \) .

Poles of the integrand:

The function \( \frac{z^2 + 1}{(z - a)(z - b)} \) has two poles

at \( z = a\) and at \(z = b \)

\(I_r = \frac{1}{2\pi i} \int_{\gamma_r} \frac{z^2 + 1}{(z - a)(z - b)} \, dz\), where \(\gamma_r(t) = re^{it} \) for \(t \in [0, 2\pi]\) and \(a, b \) are real numbers such that \(a < 0 < b\) . 

The poles of the function \(\frac{z^2 + 1}{(z - a)(z - b)} \) within the contour \(\gamma_r\), which is a circle of radius  \(r\) centered at the origin.

The poles of the function are at \(z = a \) and \( z = b\) . Since \(a < 0\) and \(b > 0\) , the two poles are located on opposite sides of the origin.

Depending on the radius r , the contour \(\gamma_r\) may or may not enclose one or both of the poles.

Conditions for \( I_r \):

If the radius r is smaller than |a| (the absolute value of a ), then the contour does not enclose

any poles, so by the Cauchy integral theorem,  \( I_r \) = 0 .

If the radius r is greater than \(\max\{|a|, b\}\) , the contour encloses both poles, and by the residue theorem, \( I_r \) will be non-zero.

If r is between |a| and b , the contour may enclose exactly one pole, which will also make \( I_r \) non-zero.

Option 3: Ir = 0 if r > max {|a|, b} and |a| = b

This condition is true because if r > \(\max \{|a|, b\} \), the contour encloses both poles, leading to a situation

where the integral sums the residues at both poles, potentially canceling each other out.

Additionally, if \(|a| = b \), both poles are symmetrically placed, reinforcing the cancellation.

Thus, Option 3) is the correct answer.

Concept -

Maximum Modulus Principle -

The Maximum Modulus Principle states that if a function f  is holomorphic and non-constant within a domain D, then the maximum of |f(z)| on D cannot occur inside D unless f is constant throughout D. The maximum must occur on the boundary of D.

Explanation -

We know that |f(z)| achieves its maximum on the boundary, but there is an interesting point at z = i where |f(i)| = |f(1+i)|, and neither of these points is the maximum.

This does not necessarily indicate a violation of the principle but hints at a specific characteristic of |f(z)| around z = i.

For Option (1) - This is Incorrect.

Although the principle states the maximum of |f(z)|  occurs on the boundary, the fact that |f(i)| = |f(1+i)| does not imply that |f(z)| is constant unless |f(i)| was the global maximum, which it is not stated to be.

For Option (2) - This is Correct.

This is a direct statement of the Maximum Modulus Principle. It correctly states that the maximum of |f(z)| cannot occur inside D unless f  is constant.

This choice adheres to the principle given that  f is non-constant and holomorphic.

For Option (3) - This is Incorrect.

The equality |f(i)| = |f(1+i)| does not imply a singularity at  z = i ; it merely points to a peculiar but permissible behavior under the continuity and

holomorphic conditions of f.

For Option (4) - This is Incorrect.

There is no basis provided in the question or in complex analysis theory to assume that f  is not holomorphic at z = i  merely based on the equality of

modulus values with another boundary point.

Hence Option (2) is correct.

Concept -

Application of Identity Theorem -

Given ( f ) and ( g ) are holomorphic and agree on a sequence {z_n} converging to z₀ , by the Identity Theorem, ( f ) and ( g ) are identical on ( D ).

This fundamental point confirms that any properties of ( f ) at ( z₀ ) also apply to ( g ) at the same point.

Understanding Zeros of Holomorphic Functions -

The fact that ( f ) has a zero of order 3 at ( z₀ ) means that near ( z₀ ), f(z) can be expressed as (z - z₀)³h(z), where h(z) is holomorphic and h(z₀) ≠ 0.

Since ( f ) and ( g ) are identical, ( g(z) ) must also have this form.

Explanation -

For Option (1) - This is Correct.

Since ( f ) and ( g ) are identical and ( f ) has a zero of order 3 at ( z₀ ), ( g ) also has a zero of order 3 at ( z₀ ). 

For Option (2) - This is Incorrect

Because it suggests uncertainty about the order of the zero, which is explicitly defined by the identity of ( f ) and ( g ).

For Option (3) - This is Incorrect

As it suggests the possibility of ( g ) having a zero of order higher than 3, which conflicts with the exact identity established between ( f ) and ( g ).

For Option (4) - This is Incorrect

Because, given the conditions and the application of the Identity Theorem, we can conclusively determine the nature of ( g ) is zero at z₀.

Hence Option (1) is correct.

Concept:

Identity theorem: Let D ⊂ C be a domain and f : D → C is analytic. If there exists an infinite sequence {z_k } ⊂ D, such that f (z_k) = 0, ∀ k ∈ \(\mathbb N\) and limit point of the sequence lies inside D then f (z) = 0 for all z ∈ D

Explanation:

f(x)=|x^k| for all x∈ (−1, 1)

So f(x) = −x^k for x ∈ (−1,0]

and f(x) = x^k for x ∈ [0, 1)

let g(z) = f(z) − z^k for all z ∈ (−1,1)

g(z) is zero on {− \(1\over2n\)} whose limit point is lies inside (−1, 0]

So, By identity theorem we have f(z) − zk = 0 for all z ∈ (−1,1)

i.e., f(z) = zk  for all z ∈ (−1,1)

Similarly, g(z) = f(z) + zk for all z ∈ (−1,1)

we can have the sequence of zeros {\(1\over2n\)} whose limit point is lies inside [0, 1]

So, By identity theorem we have f(z) + zk = 0 for all z ∈ (−1,1)

i.e., f(z) = - zk  for all z ∈ (−1,1)

So, f(z) = z^k = − z^k for all z∈(−1,1)

, which is a contradiction.

So the cardinality of the set is 0.

Option (1) is true