Problems based on cauchy's integral formula

(b) PROBLEMS BASED ON CAUCHY'S INTEGRAL FORMULA

Example 4.1.b.1 Evaluate  where C is [z]=3.[A.U M/J 2014]

Solution: We know that, Cauchy's integral formula is

Given: f(z) =  z, a = 2, C is | z | = 3

a = 2 lies inside | z | = 3

f (a) = f(2) = 2

Example 4.1.b.2 Evaluate   where C is | 1 | = 3

Solution: We know that, Cauchy's integral formula is

Here, f (z) = e^-z

a = 0 is lies inside | z | = 1

f (a) = f (0) = e^-0 = 1

= 2 πi (1) = 2 πi

Example 4.1.b.3 Evaluate   over the entire complex plane. [A.U N/D 2019 (R17)]

Solution: We know that, Cauchy's integral formula is

Example 4.1.b.4. Evaluate  where C is | z|= 2, using Cauchy's integral formula. [Anna, May 1998]

Solution: We know that, Cauchy's integral formula is

Example 4.1.b.5. Evaluate  where C is | z | = 1 [Anna, May 2001]

Solution: Given:

z = 0, we get e^-1/0 = e^-∞ = 0

z = 0 lies inside | z | = 1

Cauchy's integral formula is

Example 4.1.b.6 Using Cauchy's integral formula, evaluate  [Anna, Oct. 1997 & May 2001]

Solution : Cauchy's integral formula is

Example 4.1.b.7. Evaluate where C is |z❘ = 3 [A.U A/M 2015 R-13]

Solution : z = 1 lies inside | z❘ = 3

z = -2 lies inside ❘z❘ = 3

z² /(z - 1)² (z + 2) =A/z – 1 + B/(z - 1)² + C/ z + 2 ......(1)

z² = A (z − 1) (z + 2)+ B (z + 2) + C(z - 1)²

= (5/9)2π if (1) + 1/3 2π if' (1) + 4/9 2π if (-2) by Cauchy's integral formula

= (5/9)2π (1) + 1/3 2π (0) + 4/9 2π (1)  [f(z) = 1]

= 2πi [ 5/9 + 4/9 ] = 2πi

Example 4.1.b.8 Evaluate  where C is ellipse x²+4y² = 4.

Solution : Cauchy's integral formula is

Example 4.1.b.9. Using Cauchy's integral formula, evaluate  where C is the circle| z - 1|= 1 [A.U N/D 2016 R-13]

Solution: Given | z - 1|= 1

Here, centre 1, radius 1

z = 1 lies inside C : | z -1 |=1

z = -2 lies outside C:| z - 1| = 1

Example 4.1.b.10.Evaluate where C is the circle |z –i |= 1 by using Cauchy's integral formula. [A.U A/M 2018, R-17]

Solution :

Cauchy's integral formula is

Given: | z – i |= 1 is a circle, whose centre is (0, 1) and radius is 1.

Example 4.1.b.11. Evaluate  where C is | z | = 3.

Solution : Cauchy's integral formula is

Here, f (z) = cos л z²

z = 1 lies inside | z | = 3, f (1) = cos л = -1

z = 2 lies inside | z | = 3, ƒ (2) = cos 4 π = 1

Consider, 1/ (z - 1) (z - 2) = 1/(z - 2)(z - 1) = [ 1/z – 2 + 1/ z – 1]

Example 4.1.b.12 Evaluate  dz, where C is | z | = 3.

[Anna, May, 2000] [AU CBT J/J 2009] [A.U M/J 2013]

Solution:

We know that, Cauchy's integral formula is

Example 4.1.b.13. Evaluate using Cauchy's integral formula  where C is the circle | z | = 2. [A.U M/J 2016 R-13]

Solution:

Given: | z | = 2

⇒ Centre 0, radius = 2

z = 3 lies outside | z | = 2; z = 1 lies inside | z | = 2

Example 4.1.b.14. Evaluate  dz where C is | z – 1 | = 1, using Cauchy’s integral formula.

Solution : Cauchy's integral formula is

Example 4.1.b.15. Evaluate  where C is |z – 2| = by using Cauchy's integral formula.

[AU M/J 2006, N/D 2009, M/J 2012]

Solution : Cauchy's integral formula is

Example 4.1.b.16. Using Cauchy's integral formula, evaluate

 where C is the circle | z + 1 - i | = 2

[A.U, Nov. 2001, N/D 2007]

Solution: Given:

| z + 1 - i | = 2 i.e., | z-(-1+ i) | = 2

is a circle whose centre is -1+ i and radius is 2.

i.e., centre is (-1, 1) and radius is 2.

Consider, z² + 2z + 5 = 0

Hence, by Cauchy's integral formula

Example 4.1.b.18. Evaluate  dz where C is | z + 1+ i | = 2

using Cauchy's integral formula. [A.U. A/M 2011] [A.U N/D 2013 R-8]

Solution : z² + 2z + 4 = 0

Example 4.1.b.17. Evaluate  where C is | z + 1 + I | = 2

[Anna, Nov. 1996] [A.U N/D 2007, 2010, 2011, 2012]

Solution: Given: | z + 1 + i | = 2

i.e., | z - [ - (1 + i)] | = 2 is a circle

whose centre is - (1 + i) and radius is 2.

i.e., centre is (-1, -1) and radius is 2.

Example 4.1.b.18. Evaluate  dz where C is ❘z + 1 + i | = 2 using Cauchy's integral formula. [A.U. A/M 2011] [A.U N/D 2013 R-8]

Solution: z² + 2z + 4 = 0

Example 4.1.b.19. Evaluate  where C is the circle | z + 1+ i | = 2 using Cauchy's integral formula. [A.U N/D 2016 R-8] [A.U N/D 2014 R-13]

Solution :

Given:

| z + 1 + i | = 2

| z - (-1 - i)| = 2 is a circle,

where centre is -1- i and radius is 2.

i.e., centre (-1, -1) and radius is 2.

z²+ 2z + 4 = 0

Example 4.1.b.20. If 'c' is the circle | z | = 3 and if

 then g(2)

[A.U A/M 2018, R-17]

Solution:

We know that, Cauchy's integral formula is