Cauchy’s Integral Theorem - 2 - Free MCQ Practice Test with solutions,

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 

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

Calculation:

Given:

C:  |z - i| = 2

The above path is a circle with a radius of 2 and center as (0, 1). The graph of the circle is given below,


⇒ After putting the denominator as zero, the poles will be 'z = ± πi '. These pole points A and B are outside of the given circle as shown above. So, the complex function f(z) will be an analytic function for all points inside the circle.

By Cauchy’s Theorem (Using equation (1))

Concept:
Residue theorem: if f(z) is an analytic function in a closed curve C except at a finite number of singular points within C then 
∮_cf(z)dz = 2πi × (sum of the residues at the singular point within curve C)
Residue for simple pole z = a:
Res f(a) =  
Calculation:

Given:

pole z = i

Check for singularity at pole z = i

f(z) = 2z⁴ - 3z³ + 7z² - 3z + 5

f(i) = 2(i)⁴ - 3(i)³ + 7(i)² - 3i + 5 

f(i) = 2 ×1 - 3(-i) - 7 - 3i + 5 = 0

since, f(i) = 0 ⇒ z = i  is a singular point

From Residue theorem:
dz = 2πi (Residue at z = i )

at z = i , Res = 0/0 form, applying L'hospital rule

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)+.....

It is also called Maclaurin's series.
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

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

 Resf(α) = 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

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

Concept:
Residue Theorem:
If f(z) is analytic inside and on a closed curve ‘C’ except at a finite number of singularities inside ‘C’

Then:

∮f(z) dz = 2πi (sum of residues)

Singularity: A point where the function f(z) fails to be analytic

Analysis:

x = 0 pole of order 2
x = 2i, -2i
Given curve or region is |x - i| = 2

x = -2i lies outside |x - i| = 2
So the singular points are x = 0 & x = 2i
Residue at x = 0

 
Residue at x = 2i

Concept:
If f(z) is analytic within and on a closed curve, and if ‘a’ is any point within C then according to Cauchy Integral formula:

Application:
|z| = 3

Pole z = -2j, which lies Inside the given C i.e. |z| = 3
∴ Using the Cauchy Integral formula, we get:

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, or if the simple closed curve does not contain any singular point of f(z) then,
⁡f(z) dz = 0

Given,
The integral is where C is |z -i| = 1

Now, for point singularity consider Z² + 6 = 0

Therefore point of singularity is a = ± √6i

For |z -i| = 1 we have,

radius r = 1,  centre c = (0,1),

a = ± √6i is out of circle,

By Cauchy's Integral Formula We have,

Concept:

Cauchy’s Integral Formula  - If f(z) is analytic within and on the closed curve C and if zo is any point inside C then

Cauchy’s Residue Theorem – If f(z) is analytic in a closed curve C except at a finite number of singular point lies inside C then,

∫_c ⁡f(z)dz = 2πi × (sum of the residues at the singular point within curve C)

If f(z) has a simple pole at z = a then,

Calculation:

∴ Sum of residue must be equal to 2

∴ Z = 1 must lies inside C, Z = 2 lies outside C.