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The quadratic approximation of (x) = x3 - 3x2 - 5 at the point x = 0 is- a)3x2 − 6x − 5
- b)−3x2 − 5
- c)−3x2 + 6x − 5
- d)3x2 – 5
Correct answer is option 'B'. Can you explain this answer?
The quadratic approximation of (x) = x3 - 3x2 - 5 at the point x = 0 is
a)
3x2 − 6x − 5
b)
−3x2 − 5
c)
−3x2 + 6x − 5
d)
3x2 – 5
|
Sanya Agarwal answered |
Concept:
The Taylor's series expansion of f(x) about origin (i.e x = 0) is given by
f(x) = f(0) + x × f′(0) + x2/2! × f"(0)+.....
The Taylor's series expansion of f(x) about origin (i.e x = 0) is given by
f(x) = f(0) + x × f′(0) + x2/2! × f"(0)+.....
It is also called Maclaurin's series.
Calculation:
Calculation:
f(x) = x3 - 3x2-5
f(0) = 03 - 3 × 02 - 5 = - 5
f'(0) = 3x2 - 6x = 0
f"(0) = 6x - 6 = - 6
The quadratic approximation of f(x) at the point x = 0 is
If the principal part of the Laurent’s series vanishes, then the Laurent’s series reduces to- a)Cauchy’s series
- b)Maclaurin’s series
- c)Taylor’s series
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
If the principal part of the Laurent’s series vanishes, then the Laurent’s series reduces to
a)
Cauchy’s series
b)
Maclaurin’s series
c)
Taylor’s series
d)
None of the above
|
|
Sanya Agarwal answered |
Taylor Series:
If f(z) is analytic inside a circle 'C', centre at z = a, and radius 'r', then for all z inside 'C'; the Taylor series is given by-
Laurent Series:
If f(z) is analytic at every point inside and on the boundary of a ring shaped region 'R' bounded by two concentric circle C1 and C2 having centre at 'a' & respective radii r1 and r2 (r1 > r2).
The negative part of Laurent's series i.e
is called the singular part, and if that vanishes the terms that remain will be 
, which is nothing but Taylor series.
Laurent Series:
If f(z) is analytic at every point inside and on the boundary of a ring shaped region 'R' bounded by two concentric circle C1 and C2 having centre at 'a' & respective radii r1 and r2 (r1 > r2).
The negative part of Laurent's series i.e
is called the singular part, and if that vanishes the terms that remain will be , which is nothing but Taylor series.The residue at the singular point z = -2 of f (z) = 
- a)1/2
- b)1/3
- c)4/3
- d)3/2
Correct answer is option 'B'. Can you explain this answer?
The residue at the singular point z = -2 of f (z) =
a)
1/2
b)
1/3
c)
4/3
d)
3/2
|
Vertex Academy answered |
Residue Theorem:
If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then
f(z)dz = 2πi × [sum of residues at the singualr points with in C]Formula to find residue:
1. If f(z) has a simple pole at z = a, then
2. If f(z) has a pole of order n at z = a, then
Calculation:
1. If f(z) has a simple pole at z = a, then
2. If f(z) has a pole of order n at z = a, then
Calculation:
Let f (x) = 
. Then f(100)(54) is given by- a)Undefined
- b)100
- c)10
- d)0
Correct answer is option 'A'. Can you explain this answer?
Let f (x) = . Then f(100)(54) is given by
. Then f(100)(54) is given bya)
Undefined
b)
100
c)
10
d)
0
|
Gate Gurus answered |
Concept:
Taylor’s series method:
The Taylor series can be used to calculate the value of an entire function at every point, if the value of the function, and all of its derivatives, are known at a single point.
Taylor's series expansion for f (x + h) is
f(x+h) = f(x) + hf′(x) + h2/2!f″(x) + h3/3! + f‴(x)+…∞
f(x) = f(a) + (x−α)f′(x) +
(x) +........∞
f(x+h) = f(x) + hf′(x) + h2/2!f″(x) + h3/3! + f‴(x)+…∞
f(x) = f(a) + (x−α)f′(x) +
(x) +........∞Calculation:
Given:
f(100)(54) = ?
Using Taylor series expansion for Sin x at a = 54
f(100)(54) = ?
Using Taylor series expansion for Sin x at a = 54
Now the function transforms into:
After Observing carefully the first term in the above infinite series, the (x - 54) term is always in the denominator, which will become zero when we put x = 54.
Every derivative will also have the same term till infinite.
So, every term will have zero in its denominator after putting x = 54.
⇒ f(100)(54) is Undefined.
Which of the following is not true?- a)log(1 + z) = z −
…..about z = 0 - b)
+……about z = 2 - c)
=0 where C is the circle |z-1| = 2 - d)
has no singularity
Correct answer is option 'D'. Can you explain this answer?
Which of the following is not true?
a)
log(1 + z) = z −…..about z = 0
…..about z = 0b)
+……about z = 2c)
=0 where C is the circle |z-1| = 2d)
has no singularity
|
Engineers Adda answered |
Concept:
Taylor series expansion
Option 1:
The standard expansion of log(1 + z) is given as
Hence, Option 1 is true
Option 2:
Given complex function is
→ Let’s Resolve f(z) into partial fractions
For expanding about z = 2, let z – 2 = t ⇒ z = 2 + t
Option 3:
Cauchy’s Integral Formula:
Option 1:
The standard expansion of log(1 + z) is given as
Hence, Option 1 is true
Option 2:
Given complex function is
→ Let’s Resolve f(z) into partial fractions
For expanding about z = 2, let z – 2 = t ⇒ z = 2 + t
Option 3:
Cauchy’s Integral Formula:
If f(z) is an analytic function within a closed curve and if a is any point within C, then
Residue Theorem:
Residue Theorem:
If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then
f(z)dz = 2πi × [sum of residues at the singualr points with in C]Formula to find residue:
1. If f(z) has a simple pole at z = a, then
2. If f(z) has a pole of order n at z = a, then
Given complex integral is
where Cis the circle |z-1| = 2;
2. If f(z) has a pole of order n at z = a, then
Given complex integral is
where Cis the circle |z-1| = 2;Now for the given complex function, the pole is -4 with order 2;
The pole - 4 lies outside the given circle C;
Therefore, no residue inside the circle, hence integration will be zero.
Option 3 is also correct
Option 4:
The given complex function is f(z) = 
In this function, the singularities are z = 0, +i, -i;
In this function, the singularities are z = 0, +i, -i;
Therefore, the given function has 3 singularities...
Option 4 is incorrect
If f(z) is an analytic function whose real part is constant then f(z) is- a)function of z
- b)function of x only
- c)function of y only
- d)constant
Correct answer is option 'D'. Can you explain this answer?
If f(z) is an analytic function whose real part is constant then f(z) is
a)
function of z
b)
function of x only
c)
function of y only
d)
constant
|
Akanksha Tiwari answered |
Explanation:
Introduction:
In complex analysis, an analytic function is a function that is locally given by a convergent power series. It is differentiable at every point in its domain. The real part of an analytic function is the real-valued function obtained by taking the real part of the complex-valued function.
Given:
We are given that f(z) is an analytic function whose real part is constant.
Real Part of an Analytic Function:
The real part of a complex function f(z) is defined as Re(f(z)) and is denoted by u(x, y), where z = x + iy.
Constant Real Part:
Since the real part of f(z) is constant, it means that Re(f(z)) = c, where c is a constant.
Implication:
If the real part of a function is constant, it implies that the function is constant.
Proof:
To prove that f(z) is constant, we can consider the Cauchy-Riemann equations. The Cauchy-Riemann equations are a necessary and sufficient condition for a complex function to be analytic.
The Cauchy-Riemann equations are given by:
∂u/∂x = ∂v/∂y ...(1)
∂u/∂y = -∂v/∂x ...(2)
where u(x, y) is the real part of f(z) and v(x, y) is the imaginary part of f(z).
Since the real part of f(z) is constant, the partial derivatives of u(x, y) with respect to x and y are both zero. This implies that the partial derivative of v(x, y) with respect to x and y are also zero.
Therefore, the Cauchy-Riemann equations are satisfied for all points in the domain of f(z), which means that f(z) is analytic.
Since f(z) is analytic and its real part is constant, it implies that f(z) is constant.
Conclusion:
Hence, the correct answer is option D: f(z) is constant.
Introduction:
In complex analysis, an analytic function is a function that is locally given by a convergent power series. It is differentiable at every point in its domain. The real part of an analytic function is the real-valued function obtained by taking the real part of the complex-valued function.
Given:
We are given that f(z) is an analytic function whose real part is constant.
Real Part of an Analytic Function:
The real part of a complex function f(z) is defined as Re(f(z)) and is denoted by u(x, y), where z = x + iy.
Constant Real Part:
Since the real part of f(z) is constant, it means that Re(f(z)) = c, where c is a constant.
Implication:
If the real part of a function is constant, it implies that the function is constant.
Proof:
To prove that f(z) is constant, we can consider the Cauchy-Riemann equations. The Cauchy-Riemann equations are a necessary and sufficient condition for a complex function to be analytic.
The Cauchy-Riemann equations are given by:
∂u/∂x = ∂v/∂y ...(1)
∂u/∂y = -∂v/∂x ...(2)
where u(x, y) is the real part of f(z) and v(x, y) is the imaginary part of f(z).
Since the real part of f(z) is constant, the partial derivatives of u(x, y) with respect to x and y are both zero. This implies that the partial derivative of v(x, y) with respect to x and y are also zero.
Therefore, the Cauchy-Riemann equations are satisfied for all points in the domain of f(z), which means that f(z) is analytic.
Since f(z) is analytic and its real part is constant, it implies that f(z) is constant.
Conclusion:
Hence, the correct answer is option D: f(z) is constant.
Which of the following function f(z), of the complex variable z, is NOT analytic at all the points of the complex plane?- a)f(z) = z2
- b)f(z) = ez
- c)f(z) = sin z
- d)f(z) = log z
Correct answer is option 'D'. Can you explain this answer?
Which of the following function f(z), of the complex variable z, is NOT analytic at all the points of the complex plane?
a)
f(z) = z2
b)
f(z) = ez
c)
f(z) = sin z
d)
f(z) = log z
|
Sneha Nair answered |
Analytic Functions in Complex Analysis
An analytic function is a complex function that is differentiable at every point in its domain.
If a function is analytic at all the points in the complex plane, then it is called an entire function.
If a function is not analytic at any point in its domain, then it is called a non-analytic function.
Out of the given options, the function f(z) = log z is not analytic at all the points in the complex plane.
Explanation
The function f(z) = log z is not analytic at z = 0 and any other point where z is negative or zero.
The reason for this is that the complex logarithm is a multivalued function. For any non-zero complex number z, there are infinitely many complex numbers w such that ez = w. So, we define the complex logarithm as follows:
log z = ln |z| + i arg(z)
where arg(z) is any angle whose tangent is the imaginary part divided by the real part of z.
However, when z is negative or zero, arg(z) is not well-defined, and so log z is not analytic at these points.
Hence, the function f(z) = log z is not analytic at all the points in the complex plane.
An analytic function is a complex function that is differentiable at every point in its domain.
If a function is analytic at all the points in the complex plane, then it is called an entire function.
If a function is not analytic at any point in its domain, then it is called a non-analytic function.
Out of the given options, the function f(z) = log z is not analytic at all the points in the complex plane.
Explanation
The function f(z) = log z is not analytic at z = 0 and any other point where z is negative or zero.
The reason for this is that the complex logarithm is a multivalued function. For any non-zero complex number z, there are infinitely many complex numbers w such that ez = w. So, we define the complex logarithm as follows:
log z = ln |z| + i arg(z)
where arg(z) is any angle whose tangent is the imaginary part divided by the real part of z.
However, when z is negative or zero, arg(z) is not well-defined, and so log z is not analytic at these points.
Hence, the function f(z) = log z is not analytic at all the points in the complex plane.
Evaluate 
along the straight line joining the points (0, 0) and (3, 1)- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Evaluate along the straight line joining the points (0, 0) and (3, 1)
along the straight line joining the points (0, 0) and (3, 1)a)
b)
c)
d)
|
|
Gate Gurus answered |
Concept:
Integral of a complex function f(z) is given
∫ f(z) dz = ∫ (udx -vdy) + i ∫ (vdx + udy)
Noting f(z) = u(x, y) + i v(x, y) and dz = dx + i dy;
Calculation:
Given Along the straight line joining the points (0, 0) and (3, 1);
The equation of straight line will be x = 3y
⇒ dx = 3 dy ⇒ dz = (3 + i) dy;
Along the line x = 3y, the complex number z will be
z = x + iy = 3y + iy = (3 + i) y
Substituting both in the integral,
Substituting both in the integral,
Find the Laurent expansion of f(z) = 
in the region 1 < z + 1 < 3- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Find the Laurent expansion of f(z) = in the region 1 < z + 1 < 3
in the region 1 < z + 1 < 3a)
b)
c)
d)
|
|
Sanya Agarwal answered |
Calculation:
Let z +1 = u ⇒ z = u – 1
Given 1 < z + 1 < 3
⇒ 1 < u < 3
Let z +1 = u ⇒ z = u – 1
Given 1 < z + 1 < 3
⇒ 1 < u < 3
A harmonic function is analytic if it satisfies the Laplace equation. If u(x, y) = 2x2 − 2y2 + 4xy is a harmonic function, then its conjugate harmonic function v(x, y) is- a)4xy − 2x2 + 2y2 + constant
- b)4y2 − 4xy + constant
- c)2x2 − 2y2 + xy + constant
- d)−4xy + 2y2 − 2x2 + constant
Correct answer is option 'A'. Can you explain this answer?
A harmonic function is analytic if it satisfies the Laplace equation. If u(x, y) = 2x2 − 2y2 + 4xy is a harmonic function, then its conjugate harmonic function v(x, y) is
a)
4xy − 2x2 + 2y2 + constant
b)
4y2 − 4xy + constant
c)
2x2 − 2y2 + xy + constant
d)
−4xy + 2y2 − 2x2 + constant
|
Sharmila Gupta answered |
To determine if the function u(x, y) = 2x^2 satisfies the Laplace equation and is therefore analytic, we need to calculate its Laplacian.
The Laplacian of a function u(x, y) is given by the second partial derivatives with respect to x and y:
∇²u = ∂²u/∂x² + ∂²u/∂y²
Let's calculate the partial derivatives of u(x, y):
∂u/∂x = 4x
∂²u/∂x² = 4
∂u/∂y = 0
∂²u/∂y² = 0
Now, let's calculate the Laplacian:
∇²u = ∂²u/∂x² + ∂²u/∂y² = 4 + 0 = 4
Since the Laplacian of u(x, y) = 2x^2 is equal to 4 and not zero, it does not satisfy the Laplace equation. Therefore, u(x, y) = 2x^2 is not analytic.
The Laplacian of a function u(x, y) is given by the second partial derivatives with respect to x and y:
∇²u = ∂²u/∂x² + ∂²u/∂y²
Let's calculate the partial derivatives of u(x, y):
∂u/∂x = 4x
∂²u/∂x² = 4
∂u/∂y = 0
∂²u/∂y² = 0
Now, let's calculate the Laplacian:
∇²u = ∂²u/∂x² + ∂²u/∂y² = 4 + 0 = 4
Since the Laplacian of u(x, y) = 2x^2 is equal to 4 and not zero, it does not satisfy the Laplace equation. Therefore, u(x, y) = 2x^2 is not analytic.
The value of the following complex integral, with C representing the unit circle centered at origin in the counterclockwise sense, is: 
- a)8πi
- b)-8πi
- c)-πi
- d)πi
Correct answer is option 'C'. Can you explain this answer?
The value of the following complex integral, with C representing the unit circle centered at origin in the counterclockwise sense, is:
a)
8πi
b)
-8πi
c)
-πi
d)
πi
|
Pioneer Academy answered |
Concept:
Cauchy’s Theorem:
If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then
Cauchy’s Integral Formula:
If f(z) is an analytic function within a closed curve and if a is any point within C, then
Residue Theorem:
If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then
f(z)dz = 2πi × [sum of residues at the singualr points with in C]
Cauchy’s Integral Formula:
If f(z) is an analytic function within a closed curve and if a is any point within C, then
Residue Theorem:
If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then
f(z)dz = 2πi × [sum of residues at the singualr points with in C]Formula to find residue:
1. If f(z) has a simple pole at z = a, then
2. If f(z) has a pole of order n at z = a, then
Application:
The simple poles are: z = 0, 2
The given region is a unit circle.
The residue at z = 2 is zero as it lies outside the given region.
The reside at z = 0, is given by
The value of the given integral =
The value of the given integral =
Given two complex numbers Z1 = 5 + (5√3)i, and Z2 = 2/√3 + 2i the argument of Z1/Z2 in degrees is- a)0
- b)30
- c)60
- d)90
Correct answer is option 'A'. Can you explain this answer?
Given two complex numbers Z1 = 5 + (5√3)i, and Z2 = 2/√3 + 2i the argument of Z1/Z2 in degrees is
a)
0
b)
30
c)
60
d)
90
|
|
Naina Das answered |
Given Complex Numbers:
Z1 = 5 + (5√3)i
Z2 = 2/√3 + 2i
Calculating Z1/Z2:
To find the argument of Z1/Z2, we first need to calculate the quotient Z1/Z2.
Z1/Z2 = (5 + (5√3)i) / (2/√3 + 2i)
= [(5 + (5√3)i) * (√3 - i)] / [(2/√3 + 2i) * (√3 - i)]
= [(5√3 + 5i) - 5√3 - 5] / (2 + 2√3i + 2√3 - 2i)
= (5i - 5) / (2 + 2√3i + 2√3 - 2i)
= (5i - 5) / (2√3 - 2 + 2√3i + 2i)
= [5(i - 1)] / [2(√3 - 1) + 2(i + √3)]
= [5(i - 1)] / [2(i + √3) + 2(√3 - 1)i]
= [5(i - 1)] / [2i + 2√3 + 2√3i - 2]
= [5(i - 1)] / [-2 + 2√3 + 2√3i + 2i]
= [5(i - 1)] / [-2 + 2√3(1 + i)]
Calculating Argument:
To find the argument of Z1/Z2, we need to find the angle that the complex number makes with the positive real axis in the complex plane.
Since the real part of Z1/Z2 is 0, the argument is 0 degrees.
Therefore, the correct answer is option 'A' (0).
Z1 = 5 + (5√3)i
Z2 = 2/√3 + 2i
Calculating Z1/Z2:
To find the argument of Z1/Z2, we first need to calculate the quotient Z1/Z2.
Z1/Z2 = (5 + (5√3)i) / (2/√3 + 2i)
= [(5 + (5√3)i) * (√3 - i)] / [(2/√3 + 2i) * (√3 - i)]
= [(5√3 + 5i) - 5√3 - 5] / (2 + 2√3i + 2√3 - 2i)
= (5i - 5) / (2 + 2√3i + 2√3 - 2i)
= (5i - 5) / (2√3 - 2 + 2√3i + 2i)
= [5(i - 1)] / [2(√3 - 1) + 2(i + √3)]
= [5(i - 1)] / [2(i + √3) + 2(√3 - 1)i]
= [5(i - 1)] / [2i + 2√3 + 2√3i - 2]
= [5(i - 1)] / [-2 + 2√3 + 2√3i + 2i]
= [5(i - 1)] / [-2 + 2√3(1 + i)]
Calculating Argument:
To find the argument of Z1/Z2, we need to find the angle that the complex number makes with the positive real axis in the complex plane.
Since the real part of Z1/Z2 is 0, the argument is 0 degrees.
Therefore, the correct answer is option 'A' (0).
If f(z) has a pole of order n at z = a, then residue of function f(z) at a is- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
If f(z) has a pole of order n at z = a, then residue of function f(z) at a is
a)
b)
c)
d)
|
|
Vertex Academy answered |
Residue Theorem:
If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then
f(z) dz = 2πi × [sum of residues at the singualr points with in C]
f(z) dz = 2πi × [sum of residues at the singualr points with in C]Formula to find residue:
1. If f(z) has a simple pole at z = a, then
2. If f(z) has a pole of order n at z = a, then
Important Points:
1. If f(z) has a simple pole at z = a, then
2. If f(z) has a pole of order n at z = a, then
Important Points:
Cauchy’s Theorem:
If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then
Cauchy’s Integral Formula:
If f(z) is an analytic function within a closed curve and if a is any point within C, then
Cauchy’s Integral Formula:
If f(z) is an analytic function within a closed curve and if a is any point within C, then
If f(z) = u + iv is an analytic function of z = x + iy and u – v = ex (cosy - siny), then f(z) in terms of z is- a)e−z2 + (1 + i)c
- b)e- z + (1 + i)c
- c)ez + (1 + i)c
- d)e- 2z + (1 + i)c
Correct answer is option 'C'. Can you explain this answer?
If f(z) = u + iv is an analytic function of z = x + iy and u – v = ex (cosy - siny), then f(z) in terms of z is
a)
e−z2 + (1 + i)c
b)
e- z + (1 + i)c
c)
ez + (1 + i)c
d)
e- 2z + (1 + i)c
|
Pranavi Gupta answered |
And v are real-valued functions, then the Cauchy-Riemann equations must hold:
∂u/∂x = ∂v/∂y
and
∂u/∂y = -∂v/∂x
Conversely, if these equations hold for a given function f(z), then it is analytic.
∂u/∂x = ∂v/∂y
and
∂u/∂y = -∂v/∂x
Conversely, if these equations hold for a given function f(z), then it is analytic.
If f (z) is an analytic function whose modulus is constant, then f (z) is a- a)Function of z
- b)Constant
- c)Function whose only imaginary part is constant
- d)Function whose only real part is constant
Correct answer is option 'B'. Can you explain this answer?
If f (z) is an analytic function whose modulus is constant, then f (z) is a
a)
Function of z
b)
Constant
c)
Function whose only imaginary part is constant
d)
Function whose only real part is constant
|
|
Sanvi Kapoor answered |
Concept:
Let f(z) = u + iv be the analytic function,
then According to Cauchy-Riemann Equations
ux = vy & uy = - vx
Calculation:
Given:
modulus is constant
= k ⇒ u2 + v2 = k
Now differentiating both sides w.r.t x and y, we get
2uux + 2vvx = 0 ⇒ 2uux = -2vvx --- (1)
2uuy + 2vvy = 0 ⇒ 2uuy = -2vvy --- (2)
Now applying Cauchy-Riemann Equations ux = vy & uy = - vx
2uux – 2vuy = 0 ⇒ 2uux = 2vuy --- (3)
2uuy + 2vux = 0 ⇒ 2uuy = -2vux --- (4)
Now dividing equations 3 and 4, we get
⇒ ux = uy = 0 ⇒ u is a constant function.
Similarly, we will get v is a constant function.
∴ f(z) is a constant function.
Let f(z) = u + iv be the analytic function,
then According to Cauchy-Riemann Equations
ux = vy & uy = - vx
Calculation:
Given:
modulus is constant
= k ⇒ u2 + v2 = kNow differentiating both sides w.r.t x and y, we get
2uux + 2vvx = 0 ⇒ 2uux = -2vvx --- (1)
2uuy + 2vvy = 0 ⇒ 2uuy = -2vvy --- (2)
Now applying Cauchy-Riemann Equations ux = vy & uy = - vx
2uux – 2vuy = 0 ⇒ 2uux = 2vuy --- (3)
2uuy + 2vux = 0 ⇒ 2uuy = -2vux --- (4)
Now dividing equations 3 and 4, we get
⇒ ux = uy = 0 ⇒ u is a constant function.
Similarly, we will get v is a constant function.
∴ f(z) is a constant function.
If f (z) = u + iv is an analytic function, then- a)u is harmonic function
- b)v is harmonic function
- c)Both u and v are harmonic functions
- d)Both u and v are not harmonic functions
Correct answer is option 'C'. Can you explain this answer?
If f (z) = u + iv is an analytic function, then
a)
u is harmonic function
b)
v is harmonic function
c)
Both u and v are harmonic functions
d)
Both u and v are not harmonic functions
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|
Rishika Sen answered |
Explanation:
To understand why option C is the correct answer, let's first define what it means for a function to be analytic.
An analytic function is a complex-valued function that is differentiable at every point within its domain. In other words, the function has a derivative at each point within its domain.
Given that f(z) = u + iv is an analytic function, we can conclude the following:
Harmonic Functions:
A harmonic function is a real-valued function that satisfies Laplace's equation, which states that the sum of the second-order partial derivatives of the function is equal to zero. In other words, if u is a harmonic function, then it satisfies ∇²u = 0, where ∇² is the Laplacian operator.
Similarly, if v is a harmonic function, then it satisfies ∇²v = 0.
Derivatives of Analytic Functions:
Since f(z) is an analytic function, it means that both u and v have continuous first-order partial derivatives. This allows us to use the Cauchy-Riemann equations.
The Cauchy-Riemann equations relate the partial derivatives of u and v. They state that if f(z) = u + iv is an analytic function, then the following conditions must hold:
1. ∂u/∂x = ∂v/∂y
2. ∂u/∂y = -∂v/∂x
From these equations, we can see that the real part (u) and the imaginary part (v) of an analytic function are related to each other.
Conclusion:
Based on the above information, we can conclude that both u and v are harmonic functions. This is because if u and v are related to each other through the Cauchy-Riemann equations, and f(z) is an analytic function, then both u and v must satisfy Laplace's equation.
Therefore, the correct answer is option C: Both u and v are harmonic functions.
To understand why option C is the correct answer, let's first define what it means for a function to be analytic.
An analytic function is a complex-valued function that is differentiable at every point within its domain. In other words, the function has a derivative at each point within its domain.
Given that f(z) = u + iv is an analytic function, we can conclude the following:
Harmonic Functions:
A harmonic function is a real-valued function that satisfies Laplace's equation, which states that the sum of the second-order partial derivatives of the function is equal to zero. In other words, if u is a harmonic function, then it satisfies ∇²u = 0, where ∇² is the Laplacian operator.
Similarly, if v is a harmonic function, then it satisfies ∇²v = 0.
Derivatives of Analytic Functions:
Since f(z) is an analytic function, it means that both u and v have continuous first-order partial derivatives. This allows us to use the Cauchy-Riemann equations.
The Cauchy-Riemann equations relate the partial derivatives of u and v. They state that if f(z) = u + iv is an analytic function, then the following conditions must hold:
1. ∂u/∂x = ∂v/∂y
2. ∂u/∂y = -∂v/∂x
From these equations, we can see that the real part (u) and the imaginary part (v) of an analytic function are related to each other.
Conclusion:
Based on the above information, we can conclude that both u and v are harmonic functions. This is because if u and v are related to each other through the Cauchy-Riemann equations, and f(z) is an analytic function, then both u and v must satisfy Laplace's equation.
Therefore, the correct answer is option C: Both u and v are harmonic functions.
Evaluate the line integral 
(x + 4iy2)dz where c is the line x = 2y and x varies from 0 to 1 and z = x + iy- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Evaluate the line integral (x + 4iy2)dz where c is the line x = 2y and x varies from 0 to 1 and z = x + iy
(x + 4iy2)dz where c is the line x = 2y and x varies from 0 to 1 and z = x + iya)
b)
c)
d)
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|
Gate Gurus answered |
Calculation:
z = x + iy ⇒ dz = dx + i dy
given line is x = 2y ⇒ dy/dx = 1/2
lets substitute y in terms of x
I = ∫ x + i x2 (dx + i/2 dx)
I = ∫ x dx + i/2 x dx + i x2 dx - x2/2 dx
where c is the upper half of the circle |z| = 1.- a)-2/3
- b)2/3
- c)3/2
- d)-3/2
Correct answer is option 'B'. Can you explain this answer?
where c is the upper half of the circle |z| = 1.a)
-2/3
b)
2/3
c)
3/2
d)
-3/2
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|
Sanya Agarwal answered |
Given counter c is the circle, |z| = 1
⇒ z = eiθ ⇒ dz = ieiθdθ
Now, for upper half of the circle, 0 ≤ θ ≤ π
Integration of the complex function f (z) = 
in the counterclockwise direction, around |z – 1| = 1, is- a)-πi
- b)0
- c)πi
- d)2πi
Correct answer is option 'C'. Can you explain this answer?
Integration of the complex function f (z) = in the counterclockwise direction, around |z – 1| = 1, is
in the counterclockwise direction, around |z – 1| = 1, isa)
-πi
b)
0
c)
πi
d)
2πi
|
|
Vertex Academy answered |
Concept:
Cauchy’s Theorem:
If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then
Cauchy’s Integral Formula:
If f(z) is an analytic function within a closed curve and if a is any point within C, then
Residue Theorem:
If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then
= 2πi × [sum of residues at the singualr points with in C]
Formula to find residue:
Residue Theorem:
If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then
= 2πi × [sum of residues at the singualr points with in C]Formula to find residue:
1. If f(z) has a simple pole at z = a, then
2. If f(z) has a pole of order n at z = a, then
Application:
Given function is
2. If f(z) has a pole of order n at z = a, then
Application:
Given function is
Poles: z = 1, -1
|z – 1| = 1
⇒ |x – 1 + iy| = 1
The given region is a circle with the centre at (1, 0) and the radius is 1.
|z – 1| = 1
⇒ |x – 1 + iy| = 1
The given region is a circle with the centre at (1, 0) and the radius is 1.
Only pole z = 1, lies within the given region.
Residue at z = 1 is, 
The value of the integral = 2πi × 0.5 = πi
The value of the integral = 2πi × 0.5 = πi
The Laurent series expansion of the function 
valid in the region 0 < |z| < 2, is given by- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The Laurent series expansion of the function valid in the region 0 < |z| < 2, is given by
valid in the region 0 < |z| < 2, is given bya)
b)
c)
d)
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|
Sanya Agarwal answered |
ex – 1 has a zero at ‘0’ of multiplicity one and hence f(z) has pole at 0 of order 1. So, the Laurent series f(z) is given by
Since (ez – 1) f(z) = 1
By comparing both the sides,
a-1 = 1
If f(z) is analytic in a simply connected domain D, then for every closed path C and D- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If f(z) is analytic in a simply connected domain D, then for every closed path C and D
a)
b)
c)
d)
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|
Sanvi Kapoor answered |
Cauchy's Theorem:
If f(z) is single-valued and an analytic function of z and f'(z) is continuous at each point within and on the closed curve c, then according to the theorem, 
Cauchy's Integral Formula:
For Simple Pole:
If f(z) is analytic within and on a closed curve c and if a (simple pole) is any point within c, then
For Multiple Pole:
Cauchy's Integral Formula:
For Simple Pole:
If f(z) is analytic within and on a closed curve c and if a (simple pole) is any point within c, then
For Multiple Pole:
If f(z) is analytic within and on a closed curve c, and if a (multiple poles) are points within c, then
Let 
defined in the complex plane. The integral ∮c f(z)dz over the contour of a circle c with center at the origin and unit radius is ______.
Correct answer is '0'. Can you explain this answer?
Let defined in the complex plane. The integral ∮c f(z)dz over the contour of a circle c with center at the origin and unit radius is ______.
defined in the complex plane. The integral ∮c f(z)dz over the contour of a circle c with center at the origin and unit radius is ______.
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Gate Funda answered |
Concept:
Cauchy integral formula:
If f(z) is analytic within a closed curve and if a is any point within C (contour), then
Calculation:
Given
Contour is a unit radius circle with center is the origin
Pole of f(z) is -3, -3 and both poles are outside of the given unit circle (it is shown in below fig.)
Here all-poles of f(z) outside the circle so
Contour is a unit radius circle with center is the origin
Pole of f(z) is -3, -3 and both poles are outside of the given unit circle (it is shown in below fig.)
Here all-poles of f(z) outside the circle so
Given f(z) = 
Then- a)z = ia is a simple pole and ia/2 is a residue at z = ia of f(z)
- b)z = ia is a simple pole ia is a residue at z = ia of f(z)
- c)z = ia is a simple pole and −ia/2 is a residue at z = ia of f(z)
- d)none of the above
Correct answer is option 'A'. Can you explain this answer?
Given f(z) = Then
Thena)
z = ia is a simple pole and ia/2 is a residue at z = ia of f(z)
b)
z = ia is a simple pole ia is a residue at z = ia of f(z)
c)
z = ia is a simple pole and −ia/2 is a residue at z = ia of f(z)
d)
none of the above
|
|
Sanya Agarwal answered |
Concept:
Pole:
The value for which f(z) fails to exists i.e. the value at which the denominator of the function f(z) = 0.
When the order of a pole is 1, it is known as a simple pole.
Residue:
If f(z) has a simple pole at z = a, then
If f(z) has a pole of order n at z = a, then
Calculate:
Given:
For calculating pole:
If f(z) has a pole of order n at z = a, then
Calculate:
Given:
For calculating pole:
z2 + a2 = 0
∴ (z + ia)(z - ai) = 0
∴ z = ai, -ai.
∴ z has simple pole at z = ai and -ai.
Residue:
If f(z) has a simple pole at z = a, then
For pole at z = ai
For pole at z = -ai
∴ z has a simple pole at z = ai and
is a residue at z = ia of f(z)
For pole at z = ai
For pole at z = -ai
∴ z has a simple pole at z = ai and
is a residue at z = ia of f(z)Given z = x +iy, i = √-1 C is a circle of radius 2 with the centre at the origin. If the contour C is traversed anticlockwise, then the value of the integral 
is ________ (round off to one decimal place.)
Correct answer is '0.2'. Can you explain this answer?
Given z = x +iy, i = √-1 C is a circle of radius 2 with the centre at the origin. If the contour C is traversed anticlockwise, then the value of the integral is ________ (round off to one decimal place.)
is ________ (round off to one decimal place.)|
|
Sanya Agarwal answered |
Concept:
if f(z) is analytic in closed curve C except at a finite number of singular points within C then
f(Z)dz = 2πi × [sum of residues at singular points within C]Calculation:
Given:
Singular points: z = i, -4i
C is a circle of radius 2, only z = i will lie inside the circle
Let (-1 - j), (3 - j), (3 + j) and (-1 + j) be the vertices of a rectangle C in the complex plane. Assuming that C is traversed in counter-clockwise direction, the value of the countour integral 
is- a)0
- b)jπ/16
- c)jπ/2
- d)-jπ/8
Correct answer is option 'D'. Can you explain this answer?
Let (-1 - j), (3 - j), (3 + j) and (-1 + j) be the vertices of a rectangle C in the complex plane. Assuming that C is traversed in counter-clockwise direction, the value of the countour integral is
isa)
0
b)
jπ/16
c)
jπ/2
d)
-jπ/8
|
|
Sanya Agarwal answered |
Concept:
Residue Theorem:
If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then
∫cf(z) dz = 2πj × [sum of residues at the singular points within C]
Formula to find residue:
1. If f(z) has a simple pole at z = a, then
2. If f(z) has a pole of order n at z = a, then
Application:
Given (-1 - j), (3 - j), (3 + j) and (-1 + j) are the vertices of a rectangle C in the complex plane
f(z) from the given data is,
2. If f(z) has a pole of order n at z = a, then
Application:
Given (-1 - j), (3 - j), (3 + j) and (-1 + j) are the vertices of a rectangle C in the complex plane
f(z) from the given data is,
Poleas of f(z) is
z = 0 of order n = 2, lies in side the closed curve.
z = 4 of order n = 1, lies outside the closed curve.
The function f(x, y) satisfies the Laplace equation
∇2f(x,y)=0
on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. The value of this function on the circular boundary of this domain is equal to 3.
The numerical value of f(0, 0) is:- a)0
- b)2
- c)3
- d)1
Correct answer is option 'C'. Can you explain this answer?
The function f(x, y) satisfies the Laplace equation
∇2f(x,y)=0
on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. The value of this function on the circular boundary of this domain is equal to 3.
The numerical value of f(0, 0) is:
∇2f(x,y)=0
on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. The value of this function on the circular boundary of this domain is equal to 3.
The numerical value of f(0, 0) is:
a)
0
b)
2
c)
3
d)
1
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|
Manasa Bose answered |
Understanding the Laplace Equation
The Laplace equation, ∇²f(x, y) = 0, describes a function f that is harmonic within a given domain. In this case, the domain is a circular area with a radius of 1 centered at the origin (0, 0).
Boundary Conditions
- The function f(x, y) is known to take a constant value on the circular boundary defined by the equation x² + y² = 1.
- It is given that f(x, y) = 3 for all points on this boundary.
Properties of Harmonic Functions
- A key property of harmonic functions is that the average value of the function over any closed boundary is equal to its value at the center of that domain.
- Therefore, if the function is constant on the boundary, it must also be constant throughout the entire domain.
Finding the Value at the Center
- Since f(x, y) = 3 on the boundary and the function is harmonic, by the mean value property of harmonic functions, we conclude that the value of the function at the center (0, 0) must also be equal to the boundary value.
Conclusion
- Thus, f(0, 0) = 3, which corresponds to option 'C'.
This demonstrates that within a circular domain, if a harmonic function is constant on the boundary, it will also take that same value at the center.
The Laplace equation, ∇²f(x, y) = 0, describes a function f that is harmonic within a given domain. In this case, the domain is a circular area with a radius of 1 centered at the origin (0, 0).
Boundary Conditions
- The function f(x, y) is known to take a constant value on the circular boundary defined by the equation x² + y² = 1.
- It is given that f(x, y) = 3 for all points on this boundary.
Properties of Harmonic Functions
- A key property of harmonic functions is that the average value of the function over any closed boundary is equal to its value at the center of that domain.
- Therefore, if the function is constant on the boundary, it must also be constant throughout the entire domain.
Finding the Value at the Center
- Since f(x, y) = 3 on the boundary and the function is harmonic, by the mean value property of harmonic functions, we conclude that the value of the function at the center (0, 0) must also be equal to the boundary value.
Conclusion
- Thus, f(0, 0) = 3, which corresponds to option 'C'.
This demonstrates that within a circular domain, if a harmonic function is constant on the boundary, it will also take that same value at the center.
The value of y at x = 0.1 to five places of decimals, by Taylor's series method, given that dy/dx = x2y−1, y(0) = 1, is- a)0.68281
- b)0.81122
- c)0.90033
- d)0.70127
Correct answer is option 'C'. Can you explain this answer?
The value of y at x = 0.1 to five places of decimals, by Taylor's series method, given that dy/dx = x2y−1, y(0) = 1, is
a)
0.68281
b)
0.81122
c)
0.90033
d)
0.70127
|
Lekshmi Kaur answered |
To find the value of y at x = 0.1 using Taylor series method, we need to find the coefficients of the Taylor series expansion of y.
The given differential equation is dy/dx = x^2y.
Let's differentiate both sides of the equation with respect to x:
d^2y/dx^2 = 2xy + x^2(dy/dx).
Now, let's substitute x = 0 into the differential equation:
d^2y/dx^2 = 0^2 * y = 0.
This means that the second derivative of y with respect to x is zero at x = 0.
We can use this information to write the Taylor series expansion of y as follows:
y = y(0) + x(dy/dx)(0) + (x^2/2!)(d^2y/dx^2)(0) + ...
Since the second derivative is zero at x = 0, the Taylor series expansion simplifies to:
y = y(0) + x(dy/dx)(0) + ...
Now, let's find the values of y(0) and (dy/dx)(0) using the given differential equation.
At x = 0, dy/dx = 0^2 * y = 0.
So, (dy/dx)(0) = 0.
Now, let's differentiate the differential equation with respect to x again:
d^2y/dx^2 = 2xy + x^2(dy/dx).
Substituting x = 0, we get:
d^2y/dx^2 = 0.
Therefore, the second derivative is also zero at x = 0, which means (d^2y/dx^2)(0) = 0.
Now, we can write the Taylor series expansion as:
y = y(0) + 0 + ...
Since we only need the value of y at x = 0.1 to five decimal places, we can stop here.
Therefore, the value of y at x = 0.1 is equal to y(0).
However, without specific initial conditions or more information, we cannot determine the exact value of y(0) or the value of y at x = 0.1 using the Taylor series method.
The given differential equation is dy/dx = x^2y.
Let's differentiate both sides of the equation with respect to x:
d^2y/dx^2 = 2xy + x^2(dy/dx).
Now, let's substitute x = 0 into the differential equation:
d^2y/dx^2 = 0^2 * y = 0.
This means that the second derivative of y with respect to x is zero at x = 0.
We can use this information to write the Taylor series expansion of y as follows:
y = y(0) + x(dy/dx)(0) + (x^2/2!)(d^2y/dx^2)(0) + ...
Since the second derivative is zero at x = 0, the Taylor series expansion simplifies to:
y = y(0) + x(dy/dx)(0) + ...
Now, let's find the values of y(0) and (dy/dx)(0) using the given differential equation.
At x = 0, dy/dx = 0^2 * y = 0.
So, (dy/dx)(0) = 0.
Now, let's differentiate the differential equation with respect to x again:
d^2y/dx^2 = 2xy + x^2(dy/dx).
Substituting x = 0, we get:
d^2y/dx^2 = 0.
Therefore, the second derivative is also zero at x = 0, which means (d^2y/dx^2)(0) = 0.
Now, we can write the Taylor series expansion as:
y = y(0) + 0 + ...
Since we only need the value of y at x = 0.1 to five decimal places, we can stop here.
Therefore, the value of y at x = 0.1 is equal to y(0).
However, without specific initial conditions or more information, we cannot determine the exact value of y(0) or the value of y at x = 0.1 using the Taylor series method.
is
The function f(z) of complex variable z = x + iy, where i = √−1, is given as f(z) = (x3 – 3xy2) + i v(x,y). For this function to be analytic, v(x,y) should be- a)(3xy2 – y3) + constant
- b)(3x2y2 – y3) + constant
- c)(x3 – 3x2y) + constant
- d)(3x2y – y3) + constant
Correct answer is option 'D'. Can you explain this answer?
The function f(z) of complex variable z = x + iy, where i = √−1, is given as f(z) = (x3 – 3xy2) + i v(x,y). For this function to be analytic, v(x,y) should be
a)
(3xy2 – y3) + constant
b)
(3x2y2 – y3) + constant
c)
(x3 – 3x2y) + constant
d)
(3x2y – y3) + constant
|
Partho Jain answered |
The function f(z) of complex variable z = x + iy, where i = √(-1), can be written as f(z) = u(x, y) + iv(x, y), where u(x, y) and v(x, y) are real-valued functions of x and y.
For example, if f(z) = z^2, then we can write it as f(z) = (x + iy)^2 = x^2 + 2ixy - y^2. In this case, u(x, y) = x^2 - y^2 and v(x, y) = 2xy.
In general, the real part u(x, y) represents the real-valued function associated with the real component of f(z), and the imaginary part v(x, y) represents the real-valued function associated with the imaginary component of f(z).
For example, if f(z) = z^2, then we can write it as f(z) = (x + iy)^2 = x^2 + 2ixy - y^2. In this case, u(x, y) = x^2 - y^2 and v(x, y) = 2xy.
In general, the real part u(x, y) represents the real-valued function associated with the real component of f(z), and the imaginary part v(x, y) represents the real-valued function associated with the imaginary component of f(z).
Which of the following statements is FALSE?- a)If a sequence X of real numbers converges to a real number and has two convergent subsequences X' and X" whose limits are not equal, then X is divergent.
- b)A Cauchy sequence of real numbers is unbounded.
- c)If a sequence (xn) of real number converges to a real number x, then any subsequence (xnK) of (xn) also converges to x.
- d)A bounded sequence of real numbers has a convergent subsequence.
Correct answer is option 'B'. Can you explain this answer?
Which of the following statements is FALSE?
a)
If a sequence X of real numbers converges to a real number and has two convergent subsequences X' and X" whose limits are not equal, then X is divergent.
b)
A Cauchy sequence of real numbers is unbounded.
c)
If a sequence (xn) of real number converges to a real number x, then any subsequence (xnK) of (xn) also converges to x.
d)
A bounded sequence of real numbers has a convergent subsequence.
|
|
Rahul Chauhan answered |
The statement (c) is FALSE.
In a sequence, if two subsequences converge to the same limit, it does not necessarily mean that the original sequence itself converges. It is possible for a sequence to have convergent subsequences without converging itself.
In a sequence, if two subsequences converge to the same limit, it does not necessarily mean that the original sequence itself converges. It is possible for a sequence to have convergent subsequences without converging itself.
What is the expansion of y = sin-1 x?- a)
- b)
- c)
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
What is the expansion of y = sin-1 x?
a)
b)
c)
d)
None of the above
|
|
Engineers Adda answered |
Concept:
Binomial expansion of (1 - x)-n is given by
(1 - x)-n = 1 + nx +
Formula used:
Calculation:
Given,
y = sin-1 x
....(1)
Using the above binomial expansion formula
Integrating both sides with respect to x,
Hence, option c is the correct answer.
(1 - x)-n = 1 + nx +
Formula used:
Calculation:
Given,
y = sin-1 x
....(1)Using the above binomial expansion formula
Integrating both sides with respect to x,
Hence, option c is the correct answer.
If C is a circle |z| = 4 and f(z) = 
then ∮ f (z) dz is- a)1
- b)0
- c)-1
- d)-2
Correct answer is option 'B'. Can you explain this answer?
If C is a circle |z| = 4 and f(z) = then ∮ f (z) dz is
then ∮ f (z) dz isa)
1
b)
0
c)
-1
d)
-2
|
|
Sanya Agarwal answered |
Poles of f(z) are z = 1, 2
Both Z = 1 and 2 are inside the curve |z| = 4
= 2πi [sum of residues at z = 1 and z = 2]If f has a pole of order n at z = a, then Residue of f(z) at z = a is
Residue at z = 1,
Residue at z = 2,
Residue at z = 1,
Residue at z = 2,
In the Laurent expansion of f(z) = 
valid in the region 1 < |z| < 2, the co-efficient of 1/z2 is- a)0
- b)1/2
- c)1
- d)-1
Correct answer is option 'D'. Can you explain this answer?
In the Laurent expansion of f(z) = valid in the region 1 < |z| < 2, the co-efficient of 1/z2 is
valid in the region 1 < |z| < 2, the co-efficient of 1/z2 isa)
0
b)
1/2
c)
1
d)
-1
|
|
Gate Gurus answered |
Concept:
Laurent series of f(z) is given by:
Calculation:
Given:
The given region is 1 < |z| < 2
Co-efficient of 1/z2 is -1
Calculation:
Given:
The given region is 1 < |z| < 2
Co-efficient of 1/z2 is -1
Expand the function 
in Laurent’s series for 1 < |z| < 2- a)
- b)
- c)
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
Expand the function in Laurent’s series for 1 < |z| < 2
in Laurent’s series for 1 < |z| < 2a)
b)
c)
d)
None of the above
|
|
Sanya Agarwal answered |
Concept:
Laurent Series:
If f(z) is analytic at every point inside and on the boundary of a ring-shaped region 'R' bounded by two concentric circle C1 and C2 having centre at 'a' & respective radii r1 and r2 (r1 > r2).
Calculation:
Laurent Series:
If f(z) is analytic at every point inside and on the boundary of a ring-shaped region 'R' bounded by two concentric circle C1 and C2 having centre at 'a' & respective radii r1 and r2 (r1 > r2).
Calculation:
Given:
and 1 < |z| < 2Here region of convergence is 1 < |z| and |z| < 2
Let C represent the unit circle centered at origin in the complex plane, and complex variable, z = x + iy. The value of the contour integral 
(where integration is taken counter clockwise) is- a)2πi
- b)πi
- c)0
- d)2
Correct answer is option 'B'. Can you explain this answer?
Let C represent the unit circle centered at origin in the complex plane, and complex variable, z = x + iy. The value of the contour integral (where integration is taken counter clockwise) is
(where integration is taken counter clockwise) isa)
2πi
b)
πi
c)
0
d)
2
|
|
Pioneer Academy answered |
Concept:
Cauchy's Integral Formula:
For Simple Pole:
If f(z) is analytic within and on a closed curve c and if a (simple pole) is any point within c, then
Calculation:
Given:
where C represents unit circle i.e. radius is unity.
Calculation:
Given:
where C represents unit circle i.e. radius is unity.The above equation can be written in standard form i.e. 
Therefore f(z) =
and a = 0.
Therefore f(z) =
and a = 0.The pole of the given function is at z = 0, and lie inside the circle.
Cauchy's Integral Formula:
The Taylor series expansion of 3 sin x + 2 cos x is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The Taylor series expansion of 3 sin x + 2 cos x is
a)
b)
c)
d)
|
|
Sanya Agarwal answered |
Concept:
Taylor series expansion for sin x and cos x are respectively:
Calculation:
Similarly,
Adding equation (1) and equation (2), we get:
3sinx + cosx = 2 + 3x −
Taylor series expansion for sin x and cos x are respectively:
Calculation:
Similarly,
Adding equation (1) and equation (2), we get:
3sinx + cosx = 2 + 3x −
Let z = x + iy be a complex variable. Consider that contour integration is performed along the unit circle in anticlockwise direction. Which one of the following statements is Not True?- a)The residue of z/z2 - 1 at z = 1 is 1/2
- b)
- c)
- d)
(Complex conjugate of z) is an analytical function
Correct answer is option 'D'. Can you explain this answer?
Let z = x + iy be a complex variable. Consider that contour integration is performed along the unit circle in anticlockwise direction. Which one of the following statements is Not True?
a)
The residue of z/z2 - 1 at z = 1 is 1/2
b)
c)
d)
(Complex conjugate of z) is an analytical function|
|
Sanya Agarwal answered |
Residue of
z = x + iy and
= x − iy i.e.u = x and v = − y
ux = 1 and vy = −1
uX ≠ vy →
is not Analytic Analytic
z = x + iy and
= x − iy i.e.u = x and v = − yux = 1 and vy = −1
uX ≠ vy →
is not Analytic AnalyticThe Residue of
for z = ia is- a)1/4α3
- b)−i/4α3
- c)−1/4α3
- d)i/4α3
Correct answer is option 'B'. Can you explain this answer?
The Residue of for z = ia is
for z = ia isa)
1/4α3
b)
−i/4α3
c)
−1/4α3
d)
i/4α3
|
|
Sanya Agarwal answered |
Cauchy's Residue Theorem:
Residue of f(z):
Residue of f(z) is denoted as Res[f(z) : z = z0]
z0 is a simple pole of the function f(z)
If f(z) = p(z) / q(z)
Where, p(z), q(z) are polynomials
Then residue is,
Res[f(z) : z = z0] = 
If f(z) has a pole of order 'm' at z = z0 then
If f(z) has a pole of order 'm' at z = z0 then
Res [f(z) : z = z0] =
Calculation:
Calculation:
Given,
Pole of f(z) has order "2"
Pole of f(z) has order "2"
The series expansion of sinx/x near origin is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The series expansion of sinx/x near origin is
a)
b)
c)
d)
|
|
Engineers Adda answered |
Concept:
Taylor series:
The Taylor series of a real or complex-valued function f (x) that is infinitely differentiable at a real or complex number ‘a’ is the power series.
Expression of Taylor series is:
Calculation:
Calculation:
Given:
We have to find the series expansion of sinx/x near origin, or a = 0.
Let f(x) = sin x
f(0) = sin (0) = 0,
f'(0) = cos (0) = 1,
f''(0) = -sin(0) = 0,
f'''(0) = -1 .... so on
Putting all the values in Taylor series expansion, we get:
Series expansion of sin x will be:
sinx = x − x3/3!+…
Therefore the series expansion of sin x/x near origin will be:
sinx = x − x3/3!+…
Therefore the series expansion of sin x/x near origin will be:
The value of the integral 
where z is a complex number and C is a unit circle with centre at 1 + 0j in the complex plane is __________ .
Correct answer is '1'. Can you explain this answer?
The value of the integral where z is a complex number and C is a unit circle with centre at 1 + 0j in the complex plane is __________ .
where z is a complex number and C is a unit circle with centre at 1 + 0j in the complex plane is __________ .|
|
Sanya Agarwal answered |
Simple poles, z = ±1
According to Cauchy’s residue theorem,
f(z)dz = 2πi [sum of residues]
At, z = +1, residue is
At, z = -1 residue is zero as z = -1 lies outside the curve C.
The residues of a function f(z) = 
are:- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The residues of a function f(z) = are:
are:a)
b)
c)
d)
|
|
Vertex Academy answered |
Concept:
For simple poles at z = a, b, c…
Residue of
For simple poles at z = a, b, c…
Residue of
For multiple poles at z = a, a, a … n times
{Residue of
Calculation:
Given, f (z) =
For a simple pole at z = 4
Residue of
For multiple pole (n = 3) at z = -1
Residue will be
{Residue of
Calculation:
Given, f (z) =
For a simple pole at z = 4
Residue of
For multiple pole (n = 3) at z = -1
Residue will be
The number of integral solutions of 
is- a)4
- b)5
- c)3
- d)6
Correct answer is option 'B'. Can you explain this answer?
The number of integral solutions of is
isa)
4
b)
5
c)
3
d)
6
|
|
Sanya Agarwal answered |
4x + 8 ≥ x2 + 8
∴ x2 – 4x ≤ 0
x(x – 4) ≤ 0 → (1)
Clearly the integral solution of (1) are 0, 1, 2, 3 and 4
∴ Total 5 values of x satisfies (1)
Residue at z = 2 of 
is- a)5/3
- b)1/3
- c)3/5
- d)2/3
Correct answer is option 'A'. Can you explain this answer?
Residue at z = 2 of is
isa)
5/3
b)
1/3
c)
3/5
d)
2/3
|
|
Vertex Academy answered |
Concept:
Pole – a point on which functional value is infinite.
If z = a is a simple pole of f(z) then the residue of f(z) at z = a is given by,
Calculation:
Calculation:
Given:
We have to find residue at z = 2.
Value of f(z) is infinite at x = 2. hence z = 2 is a simple pole.
We have to find residue at z = 2.
Value of f(z) is infinite at x = 2. hence z = 2 is a simple pole.
Residue of f(z) at z = 2,
The value of 
- a)π
- b)2π
- c)π2/32
- d)π/16
Correct answer is option 'D'. Can you explain this answer?
The value of
a)
π
b)
2π
c)
π2/32
d)
π/16
|
|
Sanya Agarwal answered |
Now it is known that
Applying this result to the second term
( n = 6) of the expression of I we get,
Again, applying this result to the term
of the expression of I we get,
The closed loop line integral 
evaluated counter-clockwise, is- a)+8jπ
- b)-8jπ
- c)-4jπ
- d)+4jπ
Correct answer is option 'A'. Can you explain this answer?
The closed loop line integral evaluated counter-clockwise, is
evaluated counter-clockwise, isa)
+8jπ
b)
-8jπ
c)
-4jπ
d)
+4jπ
|
|
Sanya Agarwal answered |
Concept:
Residue Theorem:
If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then
Residue Theorem:
If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then
∫cf(z) dz = 2πj × [sum of residues at the singular points within C]
Formula to find residue:
1. If f(z) has a simple pole at z = a, then
Resf(α) = 
a[(z−α)f(z)]
a[(z−α)f(z)]2. If f(z) has a pole of order n at z = a, then
Calculation:
z + 2 = 0 z = -2 |z| = 2 < 5
f(x) is not analytic at z = -2
By Cauchy’s residue theorem
f(x) dz = 2πi × (sum of residues)At z = -2
Residue of f(x) = 
= -8 + 4 + 8 = 4
in the region 1 ≤ |z| ≤ 2 and around z = 1 is 
The value of ∮1/z2 dz, where the contour is the unit circle traversed clockwise, is- a)-2πi
- b)0
- c)2πi
- d)4πi
Correct answer is option 'B'. Can you explain this answer?
The value of ∮1/z2 dz, where the contour is the unit circle traversed clockwise, is
a)
-2πi
b)
0
c)
2πi
d)
4πi
|
Swati Patel answered |
To evaluate the integral of 1/z^2 dz, we can use the Cauchy's Integral Formula, which states that for a function f(z) that is analytic within and on a simple closed contour C, and a point a inside C,
f(a) = (1/2πi) ∮C [f(z)/(z-a)] dz.
In this case, f(z) = 1 and a = 0. The contour C is the unit circle traversed clockwise.
Applying the formula, we have:
1 = (1/2πi) ∮C [1/(z-0)] dz
= (1/2πi) ∮C 1/z dz.
The integral of 1/z with respect to z along the unit circle traversed clockwise is equal to -2πi. Therefore,
1 = (1/2πi) (-2πi)
= -1.
So, the value of 1/z^2 dz, where the contour is the unit circle traversed clockwise, is -1.
f(a) = (1/2πi) ∮C [f(z)/(z-a)] dz.
In this case, f(z) = 1 and a = 0. The contour C is the unit circle traversed clockwise.
Applying the formula, we have:
1 = (1/2πi) ∮C [1/(z-0)] dz
= (1/2πi) ∮C 1/z dz.
The integral of 1/z with respect to z along the unit circle traversed clockwise is equal to -2πi. Therefore,
1 = (1/2πi) (-2πi)
= -1.
So, the value of 1/z^2 dz, where the contour is the unit circle traversed clockwise, is -1.
The Maclaurin's series expansion of esin x is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The Maclaurin's series expansion of esin x is
a)
b)
c)
d)
|
|
Sanya Agarwal answered |
f''(0) = 1
fiv(0) = -2 -1 -1 + 1 = -3
Substitue in Maclaurin Series
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