Theorems of Complex analysis MCQ Quiz - Objective Question with Answer for Theorems of Complex analysis - Download Free PDF

Concept:

• Holomorphic Function: A function that is complex differentiable at every point in its domain.
• Removable Singularity: A point where a function is not defined or not holomorphic, but can be redefined to make it holomorphic at that point.
• Pole: A singularity at which the modulus of the function tends to infinity.
• Essential Singularity: A point where the function exhibits wild behavior; near which, it takes on almost every complex value infinitely often.
• Picard’s Great Theorem: Near an essential singularity, a function takes on all possible complex values, with at most one exception, infinitely often.
• Liouville’s Theorem: Every bounded entire function is constant.

Calculation:

Given,

\( f:\mathbb{C} \setminus \{-1, 1\} \to \mathbb{C} \) is holomorphic and avoids the open disc \( \{z \in \mathbb{C} : |z - 1| < 1 \} \)

Step 1: Analyze singularities at \( -1 \) and \( 1 \)

⇒ If \( f \) has an essential singularity at either point, Picard’s theorem implies it takes all complex values (except maybe one)

⇒ But here, \( f \) avoids an entire open disc

⇒ Contradiction

⇒ Essential singularity not possible

Step 2: Could \( f \) have a pole?

⇒ Near a pole, \( f \) tends to infinity

⇒ So in every neighborhood, it takes arbitrarily large values

⇒ Then it must take values in the avoided disc (which is bounded)

⇒ Contradiction

⇒ Poles not possible

Step 3: So the singularities at \( -1 \) and \( 1 \) must be removable

⇒ \( f \) can be extended to an entire function over \( \mathbb{C} \)

⇒ Statement 2 is TRUE

Step 4: Is \( f \) bounded?

⇒ Yes, since it avoids an open disc of values

⇒ Bounded entire

⇒ Liouville’s theorem

⇒ constant

⇒ Statement 3 is TRUE

⇒ Statement 1 is TRUE

Step 5: Type of singularities?

⇒ Already shown they are removable

⇒ Not poles or essential singularities

⇒ Statement 4 is FALSE

∴ Final Answer: Statements 1, 2, and 3 are TRUE. Statement 4 is FALSE.

Concept:

Entire Functions and Functional Equations:

• An entire function is a complex function that is holomorphic (analytic) over the entire complex plane.
• The condition f(z) = f(iz) implies rotational symmetry by 90°, suggesting periodicity or structure involving powers of z⁴.
• If f(z) = f(iz), then f(z) = f(i⁴z) = f(z) ⇒ this leads to f(z) = g(z⁴) for some entire function g.
• The Taylor series of f(z) must contain only powers of z⁴ due to the symmetry f(iz) = f(z).

Calculation:

Given,

f is entire and f(z) = f(iz) ∀ z ∈ ℂ

⇒ Apply the condition iteratively:

⇒ f(z) = f(iz)

⇒ f(iz) = f(i²z) = f(−z)

⇒ f(−z) = f(i³z) = f(−iz)

⇒ f(−iz) = f(i⁴z) = f(z)

⇒ So, f(z) = f(iz) = f(−z) = f(−iz)

⇒ Therefore, f is symmetric under rotations by π/2

⇒ Implies f(z) depends only on z⁴

⇒ Let f(z) = g(z⁴), where g is an entire function

⇒ Then f is even: f(−z) = g((−z)⁴) = g(z⁴) = f(z)

⇒ f'(0) = f''(0) = f'''(0) = 0

⇒ All derivatives up to 3 vanish at z = 0 because f(z) = a₀ + a₄z⁴ + a₈z⁸ + ...

∴ The correct statements are: 1 , 2 and 3 .

Statement 4 is False . 

since f can be non-constant, e.g., f(z) = z⁴ is entire and satisfies f(z) = f(iz).

Concept:

Identity theorem: Let f(z) and g(z) be analytic functions in a domain D and S be any subset of D having a limit point in D. If f(z) = g(z) ∀ z ∈ S then f(z) = g(z) for all z ∈ D.

Explanation:

(1): f:ℂ → ℂ be analytic

z = (1 + k/n) + i 

Limit point is i which lies inside ℂ.

So, by identity theorem, f(z) = i for all z ∈ ℂ.

(1) is false.

(2): If f(z) ≠ 0 anywhere on S, f cannot be constant because a constant function g(z) = c has g′(z) = 0 for all z.

Since Im(f′(z)) > 0, f'(z) ≠ 0, so f is not constant. 

(2) is true.

(3): The integers ℤ form a countable subset of ℂ and are not dense in ℂ.

By the identity theorem, an analytic function constant on a non-dense subset of ℂ is not necessarily constant everywhere.

(3) & (4) is true.

Concept:

Identity theorem: Let f and g be two analytic functions in a domain D and let S = {z ∈ D: f(z) = g(z)} has a limit point in D. Then f(z) = g(z) ∀ z ∈ D. 

Explanation:

Given  f ∶ ℂ → ℂ be an entire function with \(f\left(\frac{1}{n} \right)=\frac{1}{n^4}\) for all n ∈ ℕ. 

So let \(\frac1n\) = z

So f(z) = z⁴ = g(z)

Hence \(f\left(\frac{1}{n} \right)=g\left(\frac{1}{n}\right)\) for all n ∈ ℕ

Now, the limit point of {\(\frac1n\)} is 0 which belongs to ℂ.

Therefore by Identity theorem, f(z) = z4 for all z ∈ ℂ

Option (3) is correct.

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:

A function f(z) is said to be holomorphic in a domain D if f(z) has no singularities in D.

Explanation:

g'(z) =\(f(z)=\frac{\sin z}{z^p}\) z ∈ \(\mathbb{C}\) \{0}

⇒ g'(z) = \(\frac{1}{z^p}(z-\frac{z^3}{3!}+\frac{z^5}{5!}-...)\)

⇒ g'(z) = \((\frac{z^{1-p}}{1!}-\frac{z^{3-p}}{3!}+\frac{z^{5-p}}{5!}-...)\)

Integrating both sides we get

g(z) = \((\frac{z^{2-p}}{1!(2-p)}-\frac{z^{4-p}}{3!(4-p)}+\frac{z^{6-p}}{5!(6-p)}-...)\)

So g(z) can not be holomorphic if p is a multiple of 2, 3 and 4.

∴ Options (1), (3) and (4) are not correct.

Hence option (2) is correct

Concept:

By Liouville's theorem, any entire function (a function that is holomorphic across the entire

complex plane) that is bounded must be constant. 

Explanation:

Option 1:

If both the real and imaginary parts of \( f\) are bounded, the entire function \( f\) itself is bounded.

By Liouville’s theorem, \( f\) must be constant.

This statement is true.

Option 2:

If \( e^{|\text{Re}(f)| + |\text{Im}(f)|}\)  is bounded, then f is constant.

The exponential function grows very fast. If the exponential of the sum of the absolute values of

the real and imaginary parts is bounded, then both real and imaginary parts of \( f\) must be very

restricted (in fact, constant). Therefore, \( f\) must be constant.

This statement is true.

Option 3:

If the sum \(\text{Re}(f) + \text{Im}(f)\)  and the product \(\text{Re}(f)\text{Im}(f) \) are bounded, then f is constant.

Boundedness of both the sum and product of \(\text{Re}(f) \) and \(\text{Im}(f) \) imposes significant

restrictions on the behavior of \(f\). If both these quantities are bounded, then \( f\) must be constant.

This statement is true.

 Option 4:

If \( \sin(\text{Re}(f) + \text{Im}(f)) \) is bounded, then f is constant.

The sine function is inherently bounded (since \(\sin(x)\) \(\in \) [-1, 1]  for any real \(x \), so the boundedness

of \(\sin(\text{Re}(f) + \text{Im}(f))\) does not necessarily imply that \(f\) is constant. The function \(f\) could still vary without violating this condition.

This statement is false.

Hence, our answer is option 4).

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:

A function f(z) is said to be holomorphic in a domain D if f(z) has no singularities in D.

Explanation:

g'(z) =\(f(z)=\frac{\sin z}{z^p}\) z ∈ \(\mathbb{C}\) \{0}

⇒ g'(z) = \(\frac{1}{z^p}(z-\frac{z^3}{3!}+\frac{z^5}{5!}-...)\)

⇒ g'(z) = \((\frac{z^{1-p}}{1!}-\frac{z^{3-p}}{3!}+\frac{z^{5-p}}{5!}-...)\)

Integrating both sides we get

g(z) = \((\frac{z^{2-p}}{1!(2-p)}-\frac{z^{4-p}}{3!(4-p)}+\frac{z^{6-p}}{5!(6-p)}-...)\)

So g(z) can not be holomorphic if p is a multiple of 2, 3 and 4.

∴ Options (1), (3) and (4) are not correct.

Hence option (2) is correct

Concept:

By Liouville's theorem, any entire function (a function that is holomorphic across the entire

complex plane) that is bounded must be constant. 

Explanation:

Option 1:

If both the real and imaginary parts of \( f\) are bounded, the entire function \( f\) itself is bounded.

By Liouville’s theorem, \( f\) must be constant.

This statement is true.

Option 2:

If \( e^{|\text{Re}(f)| + |\text{Im}(f)|}\)  is bounded, then f is constant.

The exponential function grows very fast. If the exponential of the sum of the absolute values of

the real and imaginary parts is bounded, then both real and imaginary parts of \( f\) must be very

restricted (in fact, constant). Therefore, \( f\) must be constant.

This statement is true.

Option 3:

If the sum \(\text{Re}(f) + \text{Im}(f)\)  and the product \(\text{Re}(f)\text{Im}(f) \) are bounded, then f is constant.

Boundedness of both the sum and product of \(\text{Re}(f) \) and \(\text{Im}(f) \) imposes significant

restrictions on the behavior of \(f\). If both these quantities are bounded, then \( f\) must be constant.

This statement is true.

 Option 4:

If \( \sin(\text{Re}(f) + \text{Im}(f)) \) is bounded, then f is constant.

The sine function is inherently bounded (since \(\sin(x)\) \(\in \) [-1, 1]  for any real \(x \), so the boundedness

of \(\sin(\text{Re}(f) + \text{Im}(f))\) does not necessarily imply that \(f\) is constant. The function \(f\) could still vary without violating this condition.

This statement is false.

Hence, our answer is option 4).

Concept:

If z₀ is a removable singular point of a function f, then f is
analytic and bounded in some deleted neighborhood 0 < |z − z₀| < ε of z₀.

Explanation: 

From the direct result we can say that option (1) is true.

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)| < |g(z)| at each point on C, then both f(z) + g(z) and g(z) have the same number of roots inside C.

Explanation:

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

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

Then |f(z)| = |z100 +  40z10 + 6z + 1| ≤ |z100|+  40|z10| + 6|z| + 1 < 1 + 40 + 6 + 1 = 48

and |g(z)| = | - 50z30| = 50|z30| < 50

Hnece |f(z)| < |g(z)|  inside {z ∈ ℂ : |z| < 1}

Then By Rouche's theorem, 

f(z)+g(z) and g(z) has same roots inside {z ∈ ℂ : |z| < 1}

Now, g(z) = - 50z30 has 30 roots inside {z ∈ ℂ : |z| < 1}

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

(3) is correct

Concept:

Liouville’s Theorem: A bounded entire function is a constant function.

Corollaries of Liouville’s Theorem:
(i) A non-constant entire function is not bounded.

(ii) If f is a non-constant entire function, then w-image if f is dense in complex plane ℂ.

Open mapping theorem: If U is a domain of the complex plane C and f: U → ℂ is a non-constant holomorphic function, then f is an open map i.e. it sends open subsets of U to open subsets of ℂ.

Minimum Modulus Theorem: If f is holomorphic and non-constant on a bounded domain D, then |f| attains its minimum either at a zero of f or on the boundary.

Explanation:

Given, f is a non-constant analytic function defined over ℂ

i.e., f is a non-constant entire function, then by corollary (i), f is unbounded and by corollary (ii), the image of f is dense in ℂ

(1), (4) are correct 

By open mapping theorem, (2) is correct

By Maximum Modulo theorem, (3) is false

Concept:

Extension of Liouville's theorem: If f(z) is entire function and |f(z)| ≤ λ|z|^α ∀ z ∈ \(\mathbb C\) then f(z) = c_[α]z^[α]    

Explanation:

Let g(z) = z f(z) - 1 + ez

So |g(z)| = |z f(z) - 1 + ez| ≤ 1 + |z| ≤ |z| for all z ∈ ℂ

Also since f(z) is entire function so g(z) is also entire function.

So by extension of Liouville's theorem, g(z) is a polynomial of at most one degree.

Let g(z) = a + bz

⇒ zf(z) - 1 + ez = a + bz

⇒ zf(z) - 1 + ez - a - bz = 0

Differentiating

f(z) + zf'(z) + ez - b = 0....(i)

Again differentiating

f'(z) + f'(z) + zf''(z) + ez = 0

⇒ 2f'(z) + zf''(z) + ez = 0....(ii)

Differentiating again

2f''(z) + f''(z) + zf'''(z) + ez = 0

⇒ 3f''(z) + zf'''(z) + ez = 0 .....(iii)

Putting z = 0 in (ii)

2f'(0) + 0 + 1 = 0 ⇒ 2f'(0) = -1 ⇒ f'(0) = -1/2

Option (2) is correct

Putting z = 0 in (iii)

3f''(0) + 0 + 1 = 0 ⇒ 3f''(0) = -1 ⇒ f''(0) = -1/3

Option (3) 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)| < |g(z)| at each point on C, then both f(z) + g(z) and g(z) have the same number of roots inside C.

Explanation:

z90 + 80z40 - 40z20 + 8z10 + 10 and the open disc {z ∈ ℂ : |z| < 1}

Let f(z) = z90 - 40z20 + 8z10 + 10 and g(z) = 80z40

Then |f(z)| = |z90 - 40z20 + 8z10 + 10 | ≤ |z90| + 40|z20| + 8|z¹⁰| + 10 < 1 + 40 + 8 + 10 = 59

and |g(z)| = | 80z⁴0| = 80|z40| < 80

Hnece |f(z)| < |g(z)|  inside {z ∈ ℂ : |z| < 1}

Then By Rouche's theorem, 

f(z)+g(z) and g(z) has same roots inside {z ∈ ℂ : |z| < 1}

Now, g(z) = 80z40 has 40 roots inside {z ∈ ℂ : |z| < 1}

Therefore z90 + 80z40 - 40z20 + 8z10 + 10 has 40 roots inside {z ∈ ℂ : |z| < 1}

(4) is correct