Real And Complex Analysis
- 1.
Which is compact in Rn?
- A.
{(x1,x2,x3,……….xn ) ∶ |xi | ≤ 1, 1 ≤ i ≤n}.
- B.
{(x1,x2,x3,……….xn ) ∶ x1+x2+x3+......+xn=0}.
- C.
{(x1,x2,x3,……….xn ) ∶ xi > 0, 1 ≤ i ≤ n}.
- D.
{(x1,x2,x3,……….xn ) ∶ 1≤ |xi | < 2, 1 ≤ i ≤ n}.
Correct Answer
D. {(x1,x2,x3,……….xn ) ∶ 1≤ |xi | < 2, 1 ≤ i ≤ n}.Explanation
The correct answer is {(x1,x2,x3,...,xn) : 1 ≤ |xi| < 2, 1 ≤ i ≤ n}. This set is compact in Rn because it is a closed and bounded set. The condition 1 ≤ |xi| < 2 ensures that the set is bounded, as all the coordinates of the vectors in the set are within a certain range. Additionally, the set is closed because it includes its boundary points, i.e., the points where |xi| = 1 or |xi| = 2. Therefore, the set satisfies the necessary conditions for compactness in Rn.Rate this question:
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- 2.
If G is an open set and f : G → C is differentiable function then
- A.
F is analytic on G
- B.
F is non-analytic on G
- C.
F is constant on C
- D.
None of these
Correct Answer
A. F is analytic on GExplanation
If G is an open set and f: G → C is a differentiable function, then f is analytic on G. Analyticity is a stronger condition than differentiability. A function is said to be analytic if it can be locally represented as a power series. Since f is differentiable on G, it satisfies the necessary condition for analyticity. Therefore, f is analytic on G.Rate this question:
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- 3.
Listed below are four subsets of C2. Which is bounded in C2. ( R(z) denotes real part of z )
- A.
{ (z,w) ∈ C2 : z 2+w 2 = 1}
- B.
{ (z,w) ∈ C 2 : | R(z)| 2+ |R(w)| 2=1}
- C.
{ (z,w) ∈ C 2 : |z| 2+ |w| 2 = 1}
- D.
{ (z,w) ∈ C 2 : |z| 2 - |w| 2 =1}
Correct Answer
C. { (z,w) ∈ C 2 : |z| 2+ |w| 2 = 1}Explanation
The correct answer is { (z,w) ∈ C2 : |z|2 + |w|2 = 1}. This subset is bounded in C2 because it represents the set of points in the complex plane where the sum of the squares of the absolute values of z and w is equal to 1. This forms a circle with radius 1 centered at the origin, which is a bounded set.Rate this question:
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- 4.
The image of the region {z ∈ C: Re (z )>0, Im (z) >0 } under the mapping z → e z^2 is
- A.
{w ∈ C: Re (w )>0, Im (w) >0 }
- B.
{w ∈ C: Re (w )>0, Im (w) >0, |w| >1 }
- C.
{w ∈ C: |w|>1 }
- D.
{w ∈ C: Im (w) >0, |w| >1 }
Correct Answer
C. {w ∈ C: |w|>1 }Explanation
The image of the region {z ∈ C: Re (z )>0, Im (z) >0 } under the mapping z → e z^2 is {w ∈ C: |w|>1 }. This is because the function e z^2 maps the region to points in the complex plane where the magnitude of w is greater than 1. The condition Re (w )>0, Im (w) >0 is not necessary for the image, as the function can map points outside this region as well.Rate this question:
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- 5.
Consider the function f(z) = z2 (1-cosz), z∈ C (Choose all possible options)
- A.
The function f has zero of order 2 at 0
- B.
The function f has zero of order 1 at 2nπ, n = ±1, ±2, ….
- C.
The function f has zero of order 4 at 0
- D.
The function f has zero of order 2 at 2nπ, n = ±1, ±2, ….
Correct Answer(s)
C. The function f has zero of order 4 at 0
D. The function f has zero of order 2 at 2nπ, n = ±1, ±2, ….Explanation
The function f(z) = z^2(1-cosz) has a zero of order 4 at 0 because when z = 0, the term z^2 becomes 0 and the term (1-cosz) also becomes 0. Therefore, the function has a zero at 0 of order 4. Additionally, the function has zeros of order 2 at 2nπ, where n = ±1, ±2, ... because when z = 2nπ, the term (1-cosz) becomes 0, resulting in a zero of order 2.Rate this question:
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- 6.
Which is closed in C ?
- A.
{ z ∈ C: Re z= Im z}
- B.
{z∈ C: Re z ≠ 0}
- C.
{z∈ C: |z| ≠0}
- D.
{z∈ C: Im z ≠ 0}
Correct Answer
A. { z ∈ C: Re z= Im z}Explanation
The correct answer is { z ∈ C: Re z = Im z}. In complex numbers, the real part (Re z) represents the horizontal component and the imaginary part (Im z) represents the vertical component. In this case, the question is asking for the set of complex numbers in which the real part is equal to the imaginary part. This means that the complex numbers lie on the line y = x in the complex plane.Rate this question:
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- 7.
F : D→ D be holomorphic with f(0)= 0, f(1/2)=0 where D = {z : |z| < 1}. Then (Choose all possible options)
- A.
| f'(1/2) | ≤ 4/3
- B.
|f(1/2)| ≤ 1
- C.
| f'(1/2) | ≤ 4/3 and |f'(0)| ≤1
- D.
F(z) = z, z∈ D
Correct Answer(s)
A. | f'(1/2) | ≤ 4/3
B. |f(1/2)| ≤ 1
C. | f'(1/2) | ≤ 4/3 and |f'(0)| ≤1Explanation
Given that f is a holomorphic function with f(0)=0 and f(1/2)=0 in the disk D, we can apply Cauchy's integral formula to obtain the following inequalities:
1. |f'(1/2)| = |(1/2πi) ∮γ f(z)/(z-1/2)^2 dz| ≤ (1/2π) ∮γ |f(z)|/|z-1/2|^2 |dz| ≤ (1/2π) ∮γ |f(z)|/(1/4) |dz| = 4/π ∮γ |f(z)| dz,
where γ is the unit circle centered at 1/2.
2. |f(1/2)| = 0.
3. |f'(0)| = |(1/2πi) ∮γ f(z)/z^2 dz| ≤ (1/2π) ∮γ |f(z)|/|z|^2 |dz| ≤ (1/2π) ∮γ |f(z)|/1 |dz| = 1/2π ∮γ |f(z)| dz.
Therefore, we have |f'(1/2)| ≤ 4/π ∮γ |f(z)| dz, |f(1/2)| = 0, and |f'(0)| ≤ 1/2π ∮γ |f(z)| dz. Hence, the given answer options are all correct.Rate this question:
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- 8.
Let f(z) = (z - 1) / (e2πi/z -1). Then (Choose all possible options)
- A.
F has isolated singularity at z = 0.
- B.
F has removable singularity at z=1.
- C.
F has infinitely many poles.
- D.
Each pole of f is of order 1.
Correct Answer(s)
A. F has isolated singularity at z = 0.
B. F has removable singularity at z=1.
C. F has infinitely many poles.
D. Each pole of f is of order 1.Explanation
The function f(z) has an isolated singularity at z = 0 because the denominator becomes zero at z = 0, causing a singularity. However, this singularity can be removed by defining f(0) separately, so f also has a removable singularity at z = 1. The function also has infinitely many poles because the denominator becomes zero at infinitely many values of z. Each pole of f is of order 1 because the numerator and denominator have a simple zero at each pole.Rate this question:
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- 9.
Suppose f : C → C is a holomorphic function such that the real part of f''(z) is strictly positive for all z∈ C. What is the possible number of solutions of the equation f(z)= az + b as a and b vary over C?
- A.
1
- B.
0
- C.
∞
- D.
2
Correct Answer
D. 2Explanation
The possible number of solutions of the equation f(z) = az + b as a and b vary over C is 2. This is because if the real part of f''(z) is strictly positive for all z in C, then f(z) is a convex function. A convex function can intersect a line at most twice, so there can be at most two solutions for the equation f(z) = az + b.Rate this question:
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- 10.
Let f : C → C be a non-constant entire function and let Im (f) = {w∈ C : ∃ z ∈ C such that f(z) = w}. Then
- A.
Then interior of Im (f) is empty
- B.
Im (f) intersects every line passing through the origin
- C.
There exists a disc in complex plane, which is disjoint from Im (f)
- D.
Im(f) contains all its limit points
Correct Answer
B. Im (f) intersects every line passing through the originExplanation
The given answer, "Im (f) intersects every line passing through the origin," is correct because an entire function is defined as a function that is analytic on the entire complex plane. Since the function is non-constant, it means that it takes on different values at different points. Therefore, for any line passing through the origin, there will always be at least one point on that line where the function takes on a value in Im (f). Hence, Im (f) intersects every line passing through the origin.Rate this question:
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- 11.
For z ∈ C of the form z = x+ iy, define H+ = {z∈ C : y > 0 } , H - = {z∈ C : y < 0 }, L+ = {z∈ C : x> 0 }, L- = {z∈ C : x < 0 }. The function f(z)= (2z+1)/(5z+3)
- A.
Maps H+ onto H+ and H- onto H-
- B.
Maps H+ onto H- and H- onto H +
- C.
Maps H+ onto L+ and H- onto L-
- D.
Maps H+ onto L- and H- onto L+
Correct Answer
A. Maps H+ onto H+ and H- onto H-Explanation
The given function f(z) = (2z + 1)/(5z + 3) maps H+ onto H+ and H- onto H-. This means that for any complex number z with a positive imaginary part, the function f(z) will also have a positive imaginary part. Similarly, for any complex number z with a negative imaginary part, the function f(z) will also have a negative imaginary part. Therefore, the function preserves the orientation of the upper half-plane (H+) and the lower half-plane (H-), mapping them onto themselves.Rate this question:
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- 12.
Let A be a closed subset of R, A ≠ ∅ , A ≠ R. Then A is
- A.
The closure of the interior of A
- B.
A countable set
- C.
A compact set
- D.
Not open
Correct Answer
D. Not openExplanation
A closed subset of R is a set that contains all of its limit points. If A is not open, it means that there exists a point in A that does not have an open neighborhood entirely contained within A. In other words, there is a point in A that is not an interior point of A. Therefore, A cannot be the closure of the interior of A because the closure of the interior of A would include all interior points of A. Hence, the correct answer is "Not open".Rate this question:
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- 13.
Let I = { 1 } ∪ { 2 } for x ∈ R, Ï• (x )= dist (x , I) = Inf { |x - y| : y ∈ I }, then Ï• is
- A.
Discontinuous somewhere.
- B.
Continuous on R but not differentiable only at x=1.
- C.
Continuous on R but not differentiable only at x=1, 2.
- D.
Continuous on R but not differentiable only at x=1, 3 /2 ,2.
Correct Answer
D. Continuous on R but not differentiable only at x=1, 3 /2 ,2.Explanation
The function Ï•(x) is defined as the distance between x and the set I, which consists of the numbers 1 and 2. Since the set I only contains two distinct points, the function Ï•(x) will be continuous on the entire real number line R. However, it will not be differentiable at the points where x is equal to 1, 3/2, and 2. This is because the distance between x and the set I will have a sharp change at these points, resulting in a non-differentiable point. Therefore, the correct answer is "Continuous on R but not differentiable only at x=1, 3/2, 2."Rate this question:
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- 14.
Let X ⊂ R be an infinite countable bounded subset of R then which of the statements is true
- A.
X cannot be compact.
- B.
X contains an interior point.
- C.
X may be closed.
- D.
Closure of X is countable.
Correct Answer
A. X cannot be compact.Explanation
An infinite countable bounded subset of R cannot be compact because compact sets must be closed and bounded, and an infinite countable set cannot be bounded. Since X is infinite and countable, it cannot be bounded, and therefore cannot be compact.Rate this question:
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- 15.
Let A be a connected open subset of R2.The number of continuous surjective functions from the closure of A in R2 to Q is
- A.
1
- B.
0
- C.
2
- D.
Not finite
Correct Answer
B. 0Explanation
Since A is a connected open subset of R2, its closure in R2 will also be connected. However, the set of rational numbers Q is not connected, as it contains gaps between each pair of rational numbers. Therefore, it is not possible to have a continuous surjective function from the closure of A in R2 to Q. Thus, the number of such functions is 0.Rate this question:
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- 16.
Let f : X→X such that f [ f ( x ) ] = x, for all x ∈ X then
- A.
F is one-to-one and onto.
- B.
F is one-to-one but not onto
- C.
F is onto but not one-to-one
- D.
F need not be either one-to-one or onto
Correct Answer
A. F is one-to-one and onto.Explanation
The given function f satisfies the equation f[f(x)] = x for all x in X. This means that for any input x, applying the function f twice will result in the original input x. This implies that f is an invertible function, meaning it has an inverse function that undoes its operation. Since f has an inverse, it must be both one-to-one and onto. One-to-one means that each input has a unique output, and onto means that every element in the codomain is mapped to by at least one element in the domain. Therefore, the correct answer is that f is one-to-one and onto.Rate this question:
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- 17.
Suppose f : R → R is a function that satisfies | f(x) - f(y) | ≤ | x - y | β , β > 0 (Choose all possible options)
- A.
If β =1 then f is differentiable.
- B.
If β >0 then f is uniformly continuous .
- C.
If β >1 then f is constant function.
- D.
F must be a polynomial.
Correct Answer(s)
B. If β >0 then f is uniformly continuous .
C. If β >1 then f is constant function.Explanation
If β > 0, then f is uniformly continuous. This is because the condition |f(x) - f(y)| ≤ |x - y|β implies that for any ε > 0, there exists a δ > 0 such that |x - y| < δ implies |f(x) - f(y)| < ε. This is the definition of uniform continuity.
If β > 1, then f is a constant function. This is because for any x and y, |f(x) - f(y)| ≤ |x - y|β. If β > 1, then |x - y|β will be smaller than |x - y|, which means that the difference between f(x) and f(y) must be zero. Therefore, f is constant.
Note: The statement "If β = 1 then f is differentiable" is not necessarily true. The given condition does not provide enough information to determine the differentiability of f.Rate this question:
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- 18.
Which of the following subsets of R2 is /are convex? (Choose all possible options)
- A.
{(x,y) : x ≤ 5 , y ≤ 10 }.
- B.
{(x , y) : x 2 + y 2 =1}.
- C.
{(x , y) : y ≥ x 2 }.
- D.
{(x , y) : y ≤ x 2 }.
Correct Answer(s)
A. {(x,y) : x ≤ 5 , y ≤ 10 }.
C. {(x , y) : y ≥ x 2 }.Explanation
The subset {(x,y) : x ≤ 5 , y ≤ 10} is convex because it includes all points within or on the boundary of the rectangle defined by x ≤ 5 and y ≤ 10. The subset {(x , y) : y ≥ x^2} is also convex because it includes all points above or on the parabola defined by y = x^2.Rate this question:
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- 19.
Let fn (x) : R→ R be defined by fn(x) = x / (1+n x2) , n∈ N. Then (Choose all possible options)
- A.
The sequence {fn(x)} converges uniformly on R.
- B.
The sequence {fn(x)} converges uniformly on [1, b] for any b > 1.
- C.
The sequence {fn'(x)} converges uniformly on R.
- D.
The sequence {fn'(x)} converges uniformly on [1,b] for any b > 1.
Correct Answer(s)
A. The sequence {fn(x)} converges uniformly on R.
B. The sequence {fn(x)} converges uniformly on [1, b] for any b > 1.
D. The sequence {fn'(x)} converges uniformly on [1,b] for any b > 1.Explanation
The given function fn(x) = x / (1+n x^2) is a sequence of functions defined on the real numbers. To determine if the sequence converges uniformly on R, we need to check if the sequence converges uniformly on any closed interval [a, b] within R.
Since the function fn(x) is continuous on R for all values of n, and the denominator (1+n x^2) is always positive, the function fn(x) is bounded on any closed interval [a, b].
By applying the Weierstrass M-test, we can show that the sequence {fn(x)} converges uniformly on R. The M-test states that if there exists a sequence of positive numbers {Mn} such that |fn(x)| ≤ Mn for all x in R, and the series ∑Mn converges, then the sequence {fn(x)} converges uniformly on R.
In this case, we can choose Mn = 1/(1+n), which satisfies the conditions of the M-test. The series ∑(1/(1+n)) converges (as it is a convergent p-series), so the sequence {fn(x)} converges uniformly on R.
Similarly, by choosing Mn = 1/(1+n) and applying the M-test, we can show that the sequence {fn(x)} converges uniformly on [1, b] for any b > 1.
However, the sequence {fn'(x)} does not converge uniformly on R or on [1, b] for any b > 1. This is because the derivative of fn(x) is fn'(x) = (1-nx^2)/(1+n x^2)^2, which does not converge uniformly to a limit function on R or on [1, b] for any b > 1.
Therefore, the correct options are: The sequence {fn(x)} converges uniformly on R, The sequence {fn(x)} converges uniformly on [1, b] for any b > 1, and The sequence {fn'(x)} does not converge uniformly on R or on [1, b] for any b > 1.Rate this question:
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- 20.
Let G1 and G2 be two subsets of R 2 and f: R 2→ R 2 be a function, then (Choose all possible options)
- A.
f -1( G1 ∪ G 2 ) = f -1(G1) ∪ f -1( G2 ) .
- B.
f -1(G1 )c = ( f -1( G1 ) ) c .
- C.
f (G 1 ∩ G 2) = f (G 1 ) ∩ f ( G 2 ).
- D.
If G1 is open and G2 is closed then G1 + G 2 = { x+y : x∈ G 1 , y∈ G 2 } is neither open nor closed.
Correct Answer(s)
A. f -1( G1 ∪ G 2 ) = f -1(G1) ∪ f -1( G2 ) .
B. f -1(G1 )c = ( f -1( G1 ) ) c .Explanation
The answer states that for a function f and subsets G1 and G2 of R2, the inverse image of the union of G1 and G2 is equal to the union of the inverse images of G1 and G2. Additionally, the complement of the inverse image of G1 is equal to the inverse image of the complement of G1. This is a property of inverse images under set operations.Rate this question:
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