1.

If 𝐹⃗ = (2𝑥 − 𝑦 + 4𝑧)𝑖̂ + (𝑥 + 𝑦 − 𝑧 2 )𝑗̂ + (3𝑥 − 2𝑦 + 4𝑧 3 )𝑘̂ then

evaluate ∫𝐶 𝐹⃗ ⋅ 𝑑𝑟⃗ over the circle 𝑥 2 + 𝑦 2 = 9, 𝑧 = 0.

Ans. 303

𝜋 𝜋 1 3
2. Show that ∫𝐶 𝐹⃗ × 𝑑𝑟⃗ = − (2 + ) 𝑖̂ + ( − ) 𝑗̂ + 𝑘̂ where 𝐹⃗ = 𝑦𝑖̂ +
4 2 2 2
̂
𝑧𝑗̂ + 𝑥𝑘 and the integral is taken along the curve 𝑥 = 𝑐𝑜𝑠𝑡, 𝑦 = 𝑠𝑖𝑛𝑡, 𝑧 =
𝜋
2𝑐𝑜𝑠𝑡 from 𝑡 = 0 𝑡𝑜 𝑡 = .
2

Ans. 𝐹⃗ × 𝑑𝑟⃗ =

= 𝑖̂(−4 𝑠𝑖𝑛𝑡 𝑐𝑜𝑠𝑡 − cos2 𝑡)𝑑𝑡 + 𝑗̂(−𝑠𝑖𝑛𝑡 𝑐𝑜𝑠𝑡 + 2 sin2 𝑡)𝑑𝑡

+ 𝑘̂ (sin 𝑡 𝑐𝑜𝑠𝑡 + 2sint cost)dt

3. Verify Green’s theorem for ∫𝐶 [(3𝑥 − 8𝑦 2 )𝑑𝑥 + (4𝑦 − 6𝑥𝑦)𝑑𝑦] where C

is the boundary of the region bounded by 𝑥 = 0, 𝑦 = 0 and 𝑥 + 𝑦 = 1.

Ans.
5
∮𝐶 (𝑓𝑑𝑥 + 𝑔𝑑𝑦) = 3
𝜕𝑔 𝜕𝑓 5
∬𝑅 ( 𝜕𝑥 − 𝜕𝑦
) 𝑑𝑥𝑑𝑦 = 3
Green’s theorem is verified

4. Apply Green’s theorem to show that the area enclosed by a simple closed
1
curve Γ in a plane is ∮Γ(𝑥𝑑𝑦 − 𝑦𝑑𝑥). Hence find the area of an ellipse
2
𝑥2 𝑦2
+ = 1, where 𝑎 and 𝑏 are the semi-major and semi-minor axes of the
𝑎2 𝑏2
ellipse.

Ans. 𝜋𝑎𝑏
 5. Evaluate by Green’s theorem
∮C{(𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦 − 𝑥𝑦)𝑑𝑥 + 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦𝑑𝑦}
where 𝐶 is the circle 𝑥 2 + 𝑦 2 = 1.
Ans. 0

6. Verify the divergence theorem for the vector function 𝐹⃗ = 4𝑥𝑧𝑖̂ − 𝑦 2 𝑗̂ +

𝑦𝑧𝑘̂ taken over the cube bounded by 𝑥 = 0, 𝑥 = 1; 𝑦 = 0, 𝑦 = 1; 𝑧 =
0, 𝑧 = 1.
Ans.

Divergence theorem states

∭ 𝛻⃗⃗ ⋅ 𝐹⃗ 𝑑𝑣 = ∬ 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠
𝑉 𝑆
Here, 𝛻⃗⃗ ⋅ 𝐹⃗ = 4𝑧 − 2𝑦 + 𝑦 = 4𝑧 − 𝑦
3
∴ ∭𝑉 𝛻⃗⃗ ⋅ 𝐹⃗ 𝑑𝑣 = 2----------(=1)
Now ∬𝑆 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 = ∬𝑆 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 + ∬𝑆 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 +
1 2
⋯ + ∬ 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠
𝑆6
where 𝑆1 is the face DEFG, 𝑆2 the face ABCO, 𝑆3
the face ABEF, 𝑆4 the face OGDC, 𝑆5 the face
BCDE and 𝑆6 the face AFGO
In the integral ∬𝑆 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠, the unit vector 𝑛̂ is
1
normal to the face 𝑆1 . So 𝑛̂ = 𝑖̂
∴ ∬𝑆 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 = 2, ∬𝑆 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 = 0
1 2
1
∬𝑆 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 = −1, ∬𝑆 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 = 0, ∬𝑆 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 = 2,
3 4 5
3
∬𝑆 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 = 0, ∬𝑆 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 = 2------------(2)
6

From (1) and (2) the divergence theorem is verified.

7. Evaluate ∬𝑆(𝑦 2 𝑧 2 𝑖̂ + 𝑧 2 𝑥 2 𝑗̂ + 𝑧 2 𝑦 2 𝑘̂ ). 𝑛̂ 𝑑𝑠 where 𝑆 is the part of the

sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 1 above the 𝑥𝑦- plane and bounded by this plane.
Ans.
Divergence theorem states

∬ 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 = ∭ 𝛻⃗⃗ ⋅ 𝐹⃗ 𝑑𝑣
𝑆 𝑉
𝜋
∬𝑆(𝑦 2 𝑧 2 𝑖̂ + 𝑧 2 𝑥 2 𝑗̂ + 𝑧 2 𝑦 2 𝑘̂). 𝑛̂ 𝑑𝑠 = 2 ∭𝑉 𝑧𝑦 2 𝑑𝑣=12

𝑜𝑛 𝑆, 𝑥 2 + 𝑦 2 + 𝑧 2 = 1 𝑎𝑛𝑑 𝑠𝑜 𝑏𝑦 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑙𝑎𝑟 𝑐𝑜 − 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠

[ ]
𝑥 = 𝑟𝑠𝑖𝑛𝜃𝑐𝑜𝑠∅, 𝑦 = 𝑟𝑠𝑖𝑛𝜃𝑠𝑖𝑛∅, 𝑧 = 𝑟𝑐𝑜𝑠𝜃, 𝑑𝑣 = 𝑟 2 𝑠𝑖𝑛𝜃𝑑𝑟 𝑑𝜃 𝑑∅

8. Verify the Stoke’s theorem for the vector function 𝐹⃗ = (𝑥 2 + 𝑦 2 )𝑖̂ − 2𝑥𝑦𝑗̂
taken round the rectangle bounded by the lines 𝑥 = ±𝑎, 𝑦 = 0, 𝑦 = 𝑏.
Here

𝑐𝑢𝑟𝑙 𝐹⃗ = −4𝑦𝑘̂
For the surface 𝑆, 𝑛̂ = 𝑘̂ , since 𝑆 lies 𝑋 − 𝑌 plane

∴ 𝑐𝑢𝑟𝑙 𝐹⃗ . 𝑛̂ = −4𝑦𝑘̂ . 𝑘̂ = −4𝑦

∴ ∬ 𝐶𝑢𝑟𝑙 𝐹⃗ ⋅ 𝑛̂ 𝑑𝑠 = −4𝑎
𝑆 𝑏2
1
∮ 𝐹⃗ . 𝑑𝑟⃗ == −4𝑎𝑏 ------(2)
2
𝐶

Hence Stoke’s theorem is verified.

9. Apply Stoke’s theorem to evaluate where ∮𝐶 (𝑦𝑑𝑥 + 𝑧𝑑𝑦 + 𝑥𝑑𝑧) where 𝐶

is the curve of interaction of 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 and 𝑥 + 𝑧 = 𝑎.
𝜋𝑎2
Ans. −
√2
1 1 𝑥−𝑦 1 1 𝑥−𝑦
10. Show that ∫0 𝑑𝑥 ∫0 (𝑥+𝑦)3
𝑑𝑦 ≠ ∫0 𝑑𝑦 ∫0 (𝑥+𝑦)3
𝑑𝑥

1 1 𝑥−𝑦 1 1 1 𝑥−𝑦 1
Ans. ∫0 𝑑𝑥 ∫0 (𝑥+𝑦)3
𝑑𝑦 = , ∫0 𝑑𝑦 ∫0 (𝑥+𝑦)3
𝑑𝑥 = −
2 2

11. Evaluate∬𝑅 √4𝑥 2 − 𝑦 2 𝑑𝑥𝑑𝑦, where R is the triangular region bounded

by the lines 𝑦 = 0, 𝑥 = 1 𝑎𝑛𝑑 𝑦 = 𝑥.
1 √3 𝜋
Ans. ( + )
3 2 3
12. Evaluate∬𝑅 𝑑𝑥𝑑𝑦, where the field of integration is a circular ring R
between two circles 𝑥 2 + 𝑦 2 = 1 and 𝑥 2 + 𝑦 2 = 4.

Ans. 4(2√4 − 𝑥 2 − √1 − 𝑥 2 ).
13. Evaluate ∬𝑅 𝑦𝑑𝑥𝑑𝑦 over the over the domain R bounded by the parabolas
𝑥 2 = 4𝑦 𝑎𝑛𝑑 𝑦 2 = 4𝑥.

48
Ans.
5
1 𝑒 𝑑𝑥𝑑𝑦
14. Change the order of integration and hence evaluate ∫0 ∫𝑒 𝑥 2 .
𝑦 𝑙𝑜𝑔𝑦
1
Ans. 1 −
𝑒
𝑎 √𝑎2 −𝑦 2
15.Evaluate ∫0 ∫0 (𝑥 2 + 𝑦 2 )𝑑𝑥𝑑𝑦 by changing to polar coordinates.
𝜋𝑎4
Ans.
8
16. Use the transformation 𝑢 = 𝑥 + 𝑦 and 𝑢𝑣 = 𝑦, evaluate the double
𝑦
1 1−𝑥
integration ∫0 𝑑𝑥 ∫0 𝑒 𝑥+𝑦 𝑑𝑦.
1
Ans. (𝑒 − 1)
2

17.Evaluate ∭(𝑥 + 𝑦 + 𝑧 + 1)4 𝑑𝑥𝑑𝑦𝑑𝑧, over the region bounded 𝑥 ≥ 0,

𝑦 ≥ 0, 𝑧 ≥ 0, 𝑥 + 𝑦 + 𝑧 ≤ 1.

117
Ans.
70

18. Evaluate ∭ 𝑧 2 𝑑𝑥𝑑𝑦𝑑𝑧, extended over the hemisphere 𝑧 ≥ 0, 𝑥 2 + 𝑦 2 +

𝑧 2 ≤ 𝑎2 .

2
Ans. 𝜋𝑎5
15

1st order ODE

1. Show that (3𝑥 + 4𝑦 + 5)𝑑𝑥 + (4𝑥 − 3𝑦 + 3)𝑑𝑦 = 0 is an exact

equation and hence solve it.
Ans. 𝑥𝑦 + sin 𝑦 = 𝑐, 𝑐 is any arbitrary constant

2. Solve: (𝑥 2 𝑦 − 2𝑥𝑦 2 )𝑑𝑥 − (𝑥 3 − 3𝑥 2 𝑦)𝑑𝑦 = 0

 𝑦3 𝑥
Ans. 𝑙𝑜𝑔 + = 𝑐, 𝑐 is any arbitrary constant
𝑥2 𝑦

3. Solve:
(𝑥𝑦 𝑠𝑖𝑛 𝑥𝑦 + cos 𝑥𝑦)𝑦𝑑𝑥 + (𝑥𝑦 sin 𝑥𝑦 − cos 𝑥𝑦)𝑥𝑑𝑦 = 0
Ans. ky = xsec xy, 𝑘 is any arbitrary constant

1
4. Solve: (𝑥𝑦 2 − 𝑒 𝑥3 ) 𝑑𝑥 − 𝑥 2 𝑦𝑑𝑦 = 0
1
Ans. 2x 2 ex3 − 3y 2 = cx 2 , 𝑐 is any arbitrary constant

5. Solve: 𝑥𝑦𝑑𝑥 + (2𝑥 2 + 3𝑦 2 − 20)𝑑𝑦 = 0

Ans. x 2 y 4 + y 6 − 10y 4 = 5c = k, 𝑘 is any arbitrary constant

−4
6. Prove that (𝑥 + 𝑦 + 1) is an integrating factor of the differential
equation
(2𝑥𝑦 − 𝑦 2 − 𝑦)𝑑𝑥 + (2𝑥𝑦 − 𝑥 2 − 𝑥)𝑑𝑦 = 0
and hence solve it.

−4
Ans. 𝑥𝑦 = 𝑐(𝑥 + 𝑦 + 1) , 𝑐 is any arbitrary constant

7. Solve: 3𝑦𝑑𝑥 − 2𝑥𝑑𝑦 + 𝑥 2 𝑦 −1 (10𝑦𝑑𝑥 − 6𝑥𝑑𝑦) = 0

Ans. x 3 y −2 + 2x 5 y −3 = c, 𝑐 is any arbitrary constant
𝑑𝑦
8. Solve: (𝑥 + 𝑦 + 1) ( ) = 1.
𝑑𝑥

Ans. 𝑥 + 𝑦 + 2 = 𝑐𝑒 𝑦 , 𝑐 is any arbitrary constant

𝑑𝑦 𝑡𝑎𝑛𝑦
9. Solve: − = (1 + 𝑥)𝑒 𝑥 𝑠𝑒𝑐𝑦.
𝑑𝑥 1+𝑥
 Ans. 𝑠𝑖𝑛𝑦 = (1 + 𝑥)(𝑒 𝑥 + 𝑐), c is an arbitrary constant

𝑑𝑦
10. Solve: + 𝑦 = 𝑦 3 (𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥).
𝑑𝑥

1 2
Ans. = (𝑐𝑜𝑠𝑥 − 3𝑠𝑖𝑛𝑥) + 𝑐𝑒 2𝑥 , c is an arbitrary constant
𝑦2 5

11. Find the general solution and singular solution of

𝑝 = cos(𝑦 − 𝑝𝑥)
Ans.
General solution is: y = cx + cos −1 c, c is an arbitrary constant

Singular solution is:

12. Obtain the general solution and the singular solution of the equation

𝑦 = 𝑝𝑥 + √𝑎2 𝑝2 + 𝑏 2 .
Ans.

General solution is: 𝑦 = 𝑐𝑥 + √𝑎2 𝑐 2 + 𝑏 2 , c is an arbitrary constant.

𝑥2 𝑦2
Singular solution is: 2
+ =1
𝑎 𝑏2

𝑑𝑦 𝑑𝑥 𝑥 𝑦
13. Solve: − = − .
𝑑𝑥 𝑑𝑦 𝑦 𝑥

Ans. (𝑥𝑦 − 𝑐)(𝑥 2 − 𝑦 2 − 𝑐) = 0, where c is arbitrary constant.

14. Solve: 𝑝2 + 2𝑝𝑦𝑐𝑜𝑡𝑥 = 𝑦 2 .
Ans.
{𝑦(1 + 𝑐𝑜𝑠𝑥) − 𝑐}{𝑦(1 − 𝑐𝑜𝑠𝑥) − 𝑐} = 0 , where c is arbitrary constant.

15. Solve: 𝑦 = 2𝑝𝑥 + 𝑦𝑝2 .

Ans.
c c
𝑦 = 2x√ + y( ), where c is arbitrary constant.
c−x c−x
 16. Solve: 𝑦 + 𝑝𝑥 = 𝑝2 𝑥 4 .

Ans. 𝑥𝑦 − 𝑐𝑥 2 + 𝑐 = 0, where c is arbitrary constant.

𝑝
17. Solve: 𝑡𝑎𝑛−1 𝑝 = 𝑥 − .
1+𝑝2

Ans.
𝑝
𝑥 = tan−1 𝑝 +
1 + 𝑝2
1
𝑦=𝑐−
1 + 𝑝2
where 𝑝 is the parameter and c is arbitrary constant.
18. Solve: 𝑦 = 2𝑝𝑥 + 𝑦 2 𝑝3 .
Ans. 𝑦 2 = 2𝑐𝑥 − 𝑐 3 , where c is arbitrary constant.

ODE (2nd and higher order)

1. Show that when 𝑛 is a positive integer J−n (x) = (−1)n Jn (x)
Ans.
𝑥 −𝑛+2𝑟 1
We have 𝐽−𝑛 (𝑥) = ∑∞ 𝑟
𝑟=0(−1) (2) r!Γ(−𝑛+𝑟+1)

since, if 𝑝 is an integer, then Γ(−𝑝) is infinite for 𝑝 > 0. So, we get terms in 𝐽−𝑛 equal to zero
till −𝑛 + 𝑟 + 1 < 1, i.e., 𝑟 < 𝑛.
(−1)𝑟 𝑥 −𝑛+2𝑟
Hence, we can write𝐽−𝑛 (𝑥) = ∑∞
𝑟=0 r!Γ(−𝑛+𝑟+1) (2)

∞
(−1)𝑛+𝑠 𝑥 𝑛+2𝑠
=∑ ( ) ,
(n + s)! Γ(𝑠 + 1) 2
𝑠=0

by putting 𝑟 = 𝑛 + 𝑠
∞
(−1)𝑠 𝑥 𝑛+2𝑠
= (−1)𝑛 ∑ ( ) = (−1)𝑛 𝐽𝑛 (𝑥)
s! Γ(𝑛 + 𝑠 + 1) 2
𝑠=0

2
2. Show that J−1 (x) = √πx cosx
2
 𝑥𝑛 𝑥2 𝑥4
Ans. We have 𝐽𝑛 (𝑥) = [1 − + +⋯]
2𝑛 Γ(𝑛+1) 2(2𝑛+2) 2⋅4(2𝑛+2)(2𝑛+4)
1
−
1 𝑥 2 𝑥2 𝑥4
Putting 𝑛 = − 2 , we get 𝐽𝑛 (𝑥) = 1
1
[1 − 2⋅1 + 2⋅4⋅3⋅1 + ⋯ ]
−
2 2 Γ( )
2

2 𝑥2 𝑥4 2
=√ [1 − + + ⋯ ] = √ 𝑐𝑜𝑠𝑥
𝜋𝑥 2! 4! 𝜋𝑥

2
3. Show that J1 (x) = √πx sinx
2

𝑥𝑛 𝑥2 𝑥4
Ans. We have 𝐽𝑛 (𝑥) = 2𝑛Γ(𝑛+1) [1 − 2(2𝑛+2) + 2⋅4(2𝑛+2)(2𝑛+4) + ⋯ ]
1
1 𝑥2 𝑥2 𝑥4
Putting 𝑛 = 2 , we get 𝐽𝑛 (𝑥) = 1
3
[1 − 2⋅3 + 2⋅4⋅3⋅5 + ⋯ ]
22 Γ( )
2

𝑥 1 1 𝑥3 𝑥5 2
=√ ⋅ [𝑥 − + + ⋯ ] = √ 𝑠𝑖𝑛𝑥
2 1⋅ 𝜋 𝑥 3! 5! 𝜋𝑥
2 √

4. Prove that 𝐽0′ = −𝐽1 .

Ans. Putting 𝑛 = 0 in the recurrence formula 𝑥𝐽𝑛′ = 𝑛𝐽𝑛 − 𝑥𝐽𝑛+1
5. Express 𝐽4 (𝑥) in terms of 𝐽0 and 𝐽1
48 8 24
Ans. 𝐽4 (𝑥) = (𝑥 3 − 𝑥) 𝐽1 (𝑥) − (𝑥 2 − 1) 𝐽0
1 dn y 1 dn
6. Prove that Pn (x) = = 2n n! dxn (x 2 − 1)n
c dxn

Ans.
 𝑑𝑦
𝑦 = (𝑥 2 − 1)𝑛 and then (𝑥 2 − 1) = 2𝑛𝑥𝑦
𝑑𝑥
Differentiating (𝑛 + 1) times by Leibnitz’s theorem on successive differentiation, we have
𝑑𝑛+2 𝑦 𝑑𝑛+1 𝑦 𝑑𝑛 𝑦
(1 − 𝑥 2 ) − 2𝑥 𝑑𝑥 𝑛+1 + 𝑛(𝑛 + 1) 𝑑𝑥 𝑛 = 0
𝑑𝑥 𝑛+2

𝑑𝑛 𝑦 𝑑2 𝑧 𝑑𝑧
Putting 𝑑𝑥 𝑛 = 𝑧, we have (1 − 𝑥 2 ) 𝑑𝑥 2 − 2𝑥 𝑑𝑥 + 𝑛(𝑛 + 1)𝑧 = 0

which is Legendre’s equation and one of the solutions is 𝑧 = 𝑐𝑃𝑛 (𝑥)

𝑑𝑛 𝑦
Putting 𝑥 = 1, we have 𝑐 = (𝑑𝑥 𝑛 ) 𝑠𝑖𝑛𝑐𝑒 𝑃𝑛 (1) = 1------(1)
𝑥=1

Again, 𝑦 = (𝑥 2 − 1)𝑛 = (𝑥 + 1)𝑛 (𝑥 − 1)𝑛

Differentiating 𝑛 times using Leibnitz’s theorem, we have
𝑑𝑛 𝑦
(𝑑𝑥 𝑛 ) = (1 + 1)𝑛 . 𝑛! = 2𝑛 𝑛! ---------(2)
𝑥=1

From eqns. (1) and (2), we have 𝑐 = 2𝑛 𝑛!

1 dn y 1 dn
Therefore, Pn (x) = = 2n n! dxn (x 2 − 1)n
c dxn
1
7. Prove that ∫−1 𝑃𝑚 (𝑥)𝑃𝑛 (𝑥)𝑑𝑥 = 0; 𝑚 ≠ 𝑛

Ans.
𝑑 𝑑𝑦
The Legendre equation can also be written as {(1 − 𝑥 2 ) } + 𝑛(𝑛 + 1)𝑦 = 0
𝑑𝑥 𝑑𝑥
𝑑 2 𝑑𝑃𝑛 (𝑥)
Since 𝑃𝑛 (𝑥) is a solution, we have {(1 − 𝑥 ) } + 𝑛(𝑛 + 1)𝑃𝑛 (𝑥) = 0 --------(1)
𝑑𝑥 𝑑𝑥
𝑑 2 𝑑𝑦
again since, 𝑃𝑚 (𝑥) is a solution of {(1 − 𝑥 ) 𝑑𝑥 } + 𝑚(𝑚 + 1)𝑦 = 0
𝑑𝑥
𝑑 𝑑𝑃𝑚 (𝑥)
we have {(1 − 𝑥 2 ) 𝑑𝑥 } + 𝑚(𝑚 + 1)𝑃𝑚 (𝑥) = 0 --------(2)
𝑑𝑥
Multiplying equation (1) by 𝑃𝑚 (𝑥)and equation (2) by 𝑃𝑛 (𝑥) and subtracting, we have

𝑑
or, 𝑑𝑥 {(1 − 𝑥 2 )(𝑃𝑚 (𝑥)𝑃𝑛′ (𝑥) − 𝑃𝑛 (𝑥)𝑃𝑚′ (𝑥))} + (𝑚 − 𝑛)(𝑚 + 𝑛 − 1)𝑃𝑛 (𝑥)𝑃𝑚 (𝑥) = 0
1 𝑑
Integrating between the limits -1 to 1, we have ∫−1 𝑑𝑥 {(1 − 𝑥 2 )(𝑃𝑚 (𝑥)𝑃𝑛′ (𝑥) −
1
𝑃𝑛 (𝑥)𝑃𝑚′ (𝑥))} 𝑑𝑥 + (𝑚 − 𝑛)(𝑚 + 𝑛 − 1) ∫−1 𝑃𝑚 (𝑥)𝑃𝑛 (𝑥)𝑑𝑥 = 0
1 1
or,[(1 − 𝑥 2 )(𝑃𝑚 (𝑥)𝑃𝑛′ (𝑥) − 𝑃𝑛 (𝑥)𝑃𝑚′ (𝑥))]−1 + (𝑚 − 𝑛) (𝑚 + 𝑛 − 1) ∫−1 𝑃𝑚 (𝑥)𝑃𝑛 (𝑥)𝑑𝑥 = 0
1
or, (𝑚 − 𝑛)(𝑚 + 𝑛 − 1) ∫−1 𝑃𝑚 (𝑥)𝑃𝑛 (𝑥)𝑑𝑥 = 0
1
or ∫−1 𝑃𝑚 (𝑥)𝑃𝑛 (𝑥)𝑑𝑥 = 0

8. Prove that 𝑃𝑛 (1) = 1

Ans.
The generating function for Legendre function, of first kind is
−1
(1 − 2𝑥𝑧 + 𝑧 2 ) 2 = ∑∞ 𝑛
𝑛=0 𝑧 𝑃𝑛 (𝑥); |𝑥| ≤ 1, |𝑧| < 1------(i)
−1
Putting 𝑥 = 1 in equation (i) we have (1 − 2𝑧 + 𝑧 2 ) 2 = ∑∞ 𝑛
𝑛=0 𝑧 𝑃𝑛 (1)
or, (1 − 𝑧)−1 = ∑∞ 𝑛
𝑛=0 𝑧 𝑃𝑛 (1)

Equating the coefficient of 𝑧 𝑛 both sides, we have 𝑃𝑛 (1) = 1

′ (x)
9. Prove that nPn (x) = xPn′ (x) − Pn−1
Ans.

From the generating function for the Legendre’s function of first kind, we have
−1
(1 − 2𝑥𝑧 + 𝑧 2 ) 2 = ∑∞ 𝑛
𝑛=0 𝑧 𝑃𝑛 (𝑥); |𝑥| ≤ 1, |𝑧| < 1------(1)

Differentiating both sides of Equation (1) with respect to ‘𝑧’, we have

−3
(𝑥 − 𝑧)(1 − 2𝑥𝑧 + 𝑧 2 ) 2 = ∑∞
𝑛=0 𝑛 𝑧
𝑛−1 (𝑥)------(2)
𝑃𝑛
Again, differentiating both sides of Equation (1) with respect to ‘𝑥’, we have
−3
𝑧(1 − 2𝑥𝑧 + 𝑧 2 ) 2 = ∑∞ 𝑛 ′
𝑛=0 𝑧 𝑃𝑛 (𝑥)-----(3)

Multiplying both sides of equation (3) by (𝑥 − 𝑧) and from equation (2), we have
−3
𝑧(𝑥 − 𝑧)(1 − 2𝑥𝑧 + 𝑧 2 ) 2 = (𝑥 − 𝑧) ∑∞ 𝑛 ′
𝑛=0 𝑧 𝑃𝑛 (𝑥)

or, ∑∞ 𝑛 ∞ 𝑛 ′ ∞
𝑛=0 𝑛 𝑧 𝑃𝑛 (𝑥) = 𝑥 ∑𝑛=0 𝑧 𝑃𝑛 (𝑥) − ∑𝑛=0 𝑧
𝑛+1 ′ (𝑥)
𝑃𝑛
Equating the co-efficient of 𝑧 𝑛 from both sides, we have
′ (x)
nPn (x) = xPn′ (x) − Pn−1

10. Express 𝑃(𝑥) = 𝑥 4 + 2𝑥 3 + 2𝑥 2 − 𝑥 − 3 in terms of Legendre polynomials.

8 4 40 1 224
Ans. P (x) + P (x) + P2 (x) + P (x) − P0 (x)
35 4 5 3 21 5 1 105
𝑑2 𝑦
11. Apply the method of variation of parameters to solve + 𝑎2 𝑦 = 𝑆𝑒𝑐 𝑎𝑥, (𝑎 ≠ 0)
𝑑𝑥 2

Ans.
1 1
𝑦 = 𝑦𝑐 + 𝑦𝑝 = 𝐶1 𝐶𝑜𝑠 𝑎𝑥 + 𝐶2 𝑆𝑖𝑛 𝑎𝑥 − 𝑎2 𝑙𝑜𝑔|𝑆𝑒𝑐 𝑎𝑥|𝐶𝑜𝑠 𝑎𝑥 + 𝑎 𝑥 𝑆𝑖𝑛 𝑎𝑥, where 𝑐1
and 𝑐2 are arbitrary constants.
 𝑑2 𝑦 𝑑𝑦
12. Solve: 𝑥 2 −𝑥 + 4𝑦 = 𝑥 𝑆𝑖𝑛(𝑙𝑜𝑔𝑥)
𝑑𝑥 2 𝑑𝑥

Ans.
𝑥 𝑆𝑖𝑛 (log 𝑥)
𝑦 = 𝑥 [𝐶1 𝐶𝑜𝑠 (√3𝑙𝑜𝑔𝑥) + 𝐶2 𝑆𝑖𝑛 (√3𝑙𝑜𝑔𝑥)] + , where𝑐1 and 𝑐2 are arbitrary
2
constants.
𝑑2 𝑦 𝑑𝑦
13. Solve: 𝑑𝑥 2 + 5 𝑑𝑥 + 4𝑦 = 𝑥 2 .
1 5x 21
Ans. 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 4𝑥 + 4 (x 2 − + ), where𝑐1 and 𝑐2 are arbitrary constants.
2 8

𝑑2 𝑦
14. Solve: + 𝑦 = 𝑥𝑐𝑜𝑠𝑥.
𝑑𝑥 2
1 𝑥
Ans. 𝑦 = 𝑐1 𝑐𝑜𝑠𝑥 + 𝑐2 𝑠𝑖𝑛𝑥 + 4 𝑥 2 𝑠𝑖𝑛𝑥 + 4 cosx, where 𝑐1 and 𝑐2 are arbitrary constants.

𝑑2 𝑦 𝑑𝑦
15. Solve: 𝑑𝑥 2 − 5 𝑑𝑥 + 6𝑦 = 𝑒 𝑥 𝑐𝑜𝑠𝑥.
𝑒 𝑥 (𝑐𝑜𝑠𝑥−3𝑠𝑖𝑛𝑥)
Ans. 𝑦 = 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 3𝑥 + , where𝑐1 and 𝑐2 are arbitrary constants.
2

d2 y dy
16. Solve in series the equation dx2 + (x − 1) dx + y = 0 in power of 𝑥 − 2.

Ans.

(𝑥 − 2)2 (𝑥 − 2)3 (𝑥 − 2)2 (𝑥 − 2)3

𝑦 = 𝑎0 (1 − + − ⋯ ) + 𝑎1 ((𝑥 − 2) − − −⋯)
2 6 2 6

which is the required general series solution; 𝑎0 , 𝑎1 being the arbitrary constants.

Complex (Differentiation)
xy
1. Show that lt does not exist.
z →0 x + y
2 2

Ans.
𝑧 → 0 𝑖. 𝑒. (𝑥, 𝑦) → (0,0) along the straight line 𝑦 = 𝑚𝑥
xy 𝑚
lt = 1+𝑚2 , which has different values for different values of 𝑚. So, limit
z →0 x + y
2 2

does not exist.

 𝑍̅
2. Prove that lim
𝑍→0 𝑍
does not exist.
Ans.
𝑍̅ 𝑥
Let 𝑧 → 0 along the 𝑥 axis. Then lim 𝑍 = lim 𝑥 = 1
𝑍→0 𝑥→0
𝑍̅ −𝑖𝑦
Next let 𝑧 → 0 along the 𝑦 axis. Then lim 𝑍 = lim = −1
𝑍→0 𝑦→0 𝑖𝑦
𝑍̅
lim 𝑍 has different values for different paths. So, limit does not exist.
𝑍→0
𝑥 2 𝑦 5 (𝑥+𝑖𝑦)
,𝑧 ≠ 0
3. Show that the function 𝑓(𝑧) = { , is not analytic at 𝑧 = 0, although
𝑥 4 +𝑦 10
0, 𝑧=0
satisfies Cauchy Riemann equations at the origin.
Ans.
𝑥 2 𝑦 5 (𝑥 + 𝑖𝑦)
𝑢 + 𝑖𝑣 =
𝑥 4 + 𝑦10
𝜕𝑢 𝜕𝑢 𝜕𝑣 𝜕𝑣
= 0, 𝜕𝑦 = 0, 𝜕𝑥 = 0, 𝜕𝑦 = 0. Hence C-R equation is satisfied.
𝜕𝑥
𝑓(𝑧)−𝑓(0) 𝑥2𝑦5
Now, = 𝑥 4 +𝑦 10
𝑧
𝑓(𝑧)−𝑓(0) 𝑥3
Let 𝑧 → 0 along the 𝑦 = 𝑥. Then lim = lim 1+𝑥 6 = 0
𝑍→0 𝑧 𝑥→0
𝑓(𝑧)−𝑓(0) 𝑥 2 .𝑥 2 1
Next Let 𝑧 → 0 along the 𝑦 5 = 𝑥 2 . Then lim = lim 𝑥 4 +𝑥 4 = 2
𝑍→0 𝑧 𝑥→0
Therefore, 𝑓(𝑧) is not analytic at 𝑧 = 0, although satisfies Cauchy Riemann equations
at the origin.

4. Show that the function 𝑓(𝑧) = √|𝑥𝑦| is not analytic at origin, although the Cauchy-
Riemann equations are satisfied at the that point.
Ans.
𝑢 + 𝑖𝑣 = √|𝑥𝑦|, where 𝑢(𝑥, 𝑦) = √|𝑥𝑦|, 𝑣(𝑥, 𝑦) = 0.
𝜕𝑢 𝜕𝑢 𝜕𝑣 𝜕𝑣
Now at origin, 𝜕𝑥 = 0, 𝜕𝑦 = 0, 𝜕𝑥 = 0, 𝜕𝑦 = 0. Hence C-R equation is satisfied
at(0,0).
𝑓(𝑧)−𝑓(0) √|𝑥𝑦|
Now, =
𝑧 𝑥+𝑖𝑦
𝑓(𝑧)−𝑓(0) √|𝑚|
Let 𝑧 → 0 along the 𝑦 = 𝑚𝑥. Then 𝑓 ′ (0) = lim = 1+𝑖𝑚, which has different
𝑍→0 𝑧
values for different values of 𝑚. So 𝑓 ′ (0) does not exist, although the Cauchy-
Riemann equations are satisfied at the that point.

𝑥 3 (1+𝑖)−𝑦 3 (1−𝑖)
,𝑧 ≠ 0
5. Prove that the function defined by 𝑓(𝑧) = { is continuous and
𝑥 2 +𝑦 2
0, 𝑧=0
Cauchy Riemann equations are satisfied at the origin, yet 𝑓 ′ (0) does not exist.
Ans.
 𝑥 3 −𝑦 3 𝑥 3 +𝑦 3
Let, 𝑓(𝑧) = 𝑢 + 𝑖𝑣, where 𝑢(𝑥, 𝑦) = , 𝑣(𝑥, 𝑦) = , 𝑧 ≠ 0.
𝑥 2 +𝑦 2 𝑥 2 +𝑦 2
𝜕𝑢 𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢 𝜕𝑣 𝜕𝑢 𝜕𝑣
Now at origin, 𝜕𝑥 = 1, 𝜕𝑦 = −1, 𝜕𝑥 = 1, 𝜕𝑦 = 1. 𝜕𝑥 = 𝜕𝑦 , 𝜕𝑦 = − 𝜕𝑥
Hence C-R equation is satisfied at(0,0).
𝑓(𝑧)−𝑓(0) 𝑥 3 −𝑦 3 𝑥 3 +𝑦 3 1
Now, lim = lim [𝑥 2 +𝑦 2 + 𝑖 𝑥 2 +𝑦 2] . 𝑥+𝑖𝑦
𝑧→0 𝑧 𝑧→0
𝑓(𝑧)−𝑓(0) 1
Let 𝑧 → 0 along the line 𝑦 = 𝑥. Then 𝑓 ′ (0) = lim = 2 (1 + 𝑖),
𝑍→0 𝑧
𝑓(𝑧)−𝑓(0)
Let 𝑧 → 0 along the line 𝑦 = 0. Then 𝑓 ′ (0) = lim = (1 + 𝑖),
𝑍→0 𝑧
Now 𝑓 ′ (0) is not unique. So 𝑓 ′ (0) does not exist, although the Cauchy-Riemann
equations are satisfied at the that point.

𝜕2 𝜕2 2
6. If 𝑓(𝑧) is analytic function of z, show that (𝜕𝑥 2 + 𝜕𝑦 2) |𝑓(𝑧)|2 = 4|𝑓 ′(𝑧) | .
Ans.
𝜕2 𝜕2
( 2 + 2 ) 𝑢2 = 2𝑢(𝑢𝑥𝑥 + 𝑢𝑦𝑦 ) + 2(𝑢𝑥2 + 𝑢𝑦2 ) = 0 + 2(𝑢𝑥2 + 𝑢𝑦2 ) = 2(𝑢𝑥2 + 𝑢𝑦2 )
𝜕𝑥 𝜕𝑦

𝜕2 𝜕2
( 2 + 2 ) 𝑣 2 = 2(𝑣𝑥2 + 𝑣𝑦2 )
𝜕𝑥 𝜕𝑦

𝜕2 𝜕2 𝜕2 𝜕2
( 2 + 2 ) |𝑓(𝑧)| = ( 2 + 2 ) (𝑢2 + 𝑣 2 ) = 2(𝑢𝑥2 + 𝑢𝑦2 + 𝑣𝑥2 + 𝑣𝑦2 )
2
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
= 4(𝑢𝑥 + 𝑣𝑥2 ), using C-R equations.
2
2
= 4|𝑓 ′(𝑧) |

𝜕2 𝜕2 𝜕2
7. Show that (𝜕𝑥 2 + 𝜕𝑦 2) = 4 𝜕𝑧𝜕𝑧̅ , with the help of this relation show that if 𝑓(𝑧) is an
𝜕2 𝜕2 2
analytic function of z in any domain, then (𝜕𝑥 2 + 𝜕𝑦 2) |𝑓(𝑧)|𝑝 = 𝑝2 |𝑓(𝑧)|𝑝−2 |𝑓 ′(𝑧) | .
Ans.
𝑥 + 𝑖𝑦 = 𝑧 and 𝑥 − 𝑖𝑦 = 𝑧̅
𝜕 1 𝜕 𝜕 𝜕 1 𝜕 𝜕
= ( − 𝑖 ), = ( +𝑖 )
𝜕𝑧 2 𝜕𝑥 𝜕𝑦 𝜕𝑧 2 𝜕𝑥 𝜕𝑦
2 2 2
𝜕 1 𝜕 𝜕
= ( 2 + 2)
𝜕𝑧𝜕𝑧̅ 4 𝜕𝑥 𝜕𝑦
8. Find the analytic function 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦); 𝑧 = 𝑥 + 𝑖𝑦, where 𝑣(𝑥, 𝑦) =
𝑒 𝑥 (𝑥𝑠𝑖𝑛𝑦 + 𝑦𝑐𝑜𝑠𝑦).
Ans.
𝜕 2𝑣 𝑥 (𝑥𝑠𝑖𝑛𝑦
𝜕 2𝑣
= 𝑒 + 2𝑠𝑖𝑛𝑦 + 𝑦𝑐𝑜𝑠𝑦), = −𝑒 𝑥 (𝑥𝑠𝑖𝑛𝑦 + 2𝑠𝑖𝑛𝑦 + 𝑦𝑐𝑜𝑠𝑦),
𝜕𝑥 2 𝜕𝑦 2
𝜕2𝑣 𝜕2𝑣
𝑎𝑛𝑑 + 𝜕𝑦 2 = 0. Hence 𝑢(𝑥, 𝑦) is harmonic.
𝜕𝑥 2
By Milne’s Method 𝑓 ′ (𝑧) = 𝑒 𝑧 (𝑧 + 1)
Therefore, 𝑓(𝑧) = 𝑧𝑒 𝑧 + 𝑐
 9. Determine the analytic function 𝑓(𝑧) = 𝑢 + 𝑖𝑣 whose imaginary part is 𝑣(𝑥, 𝑦) =
𝑒 𝑥 𝑠𝑖𝑛 𝑦.
Ans. 𝑓(𝑧) = 𝑒 𝑥 (𝑐𝑜𝑠𝑦 + 𝑖𝑠𝑖𝑛𝑦)
10. Prove that 𝑢(𝑥, 𝑦) = 𝑒 −𝑥 (𝑥 𝑠𝑖𝑛 𝑦 − 𝑦 𝑐𝑜𝑠 𝑦)is harmonic. Find analytic function in
terms of z.
Ans. 𝑤 = 𝑖𝑧𝑒 −𝑧 + 𝑖𝑐
1
11. Prove that 𝑢(𝑥, 𝑦) = 2 𝑙𝑜𝑔(𝑥 2 + 𝑦 2 ) is harmonic and find its conjugate harmonic
function 𝑣(𝑥, 𝑦) such that 𝑓(𝑧) = 𝑢 + 𝑖𝑣 is analytic.
Ans.
𝜕 2𝑢 𝑦2 − 𝑥2 𝜕 2𝑢 𝑥2 − 𝑦2 𝜕 2𝑢 𝜕 2𝑢
= , = , 𝑎𝑛𝑑 + =0
𝜕𝑥 2 (𝑥 2 + 𝑦 2 )2 𝜕𝑦 2 (𝑥 2 + 𝑦 2 )2 𝜕𝑥 2 𝜕𝑦 2
Hence 𝑢(𝑥, 𝑦) is harmonic.
𝑦
Conjugate harmonic function 𝑣(𝑥, 𝑦) = tan−1 𝑥 + 𝑐

12. If 𝑢 = 4𝑥𝑦 − 𝑥 3 + 3𝑥𝑦 2 , verify that u is harmonic function and obtain its conjugate
𝑣(𝑥, 𝑦) such that 𝑓(𝑧) = 𝑢 + 𝑖𝑣 is analytic.
Ans.
𝜕 2𝑢 𝜕 2𝑢 𝜕 2𝑢 𝜕 2𝑢
= −6𝑥, = 6𝑥, 𝑎𝑛𝑑 + =0
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑥 2 𝜕𝑦 2
Hence 𝑢(𝑥, 𝑦) is harmonic.
Conjugate harmonic function 𝑣(𝑥, 𝑦) = −2x 2 + 2𝑦 2 + 𝑦 3 − 3𝑥 2 𝑦 +𝑐

13. Find the analytic function 𝑓(𝑧) = 𝑢 + 𝑖𝑣, 𝑧 = 𝑥 + 𝑖𝑦, where 𝑢 − 𝑣 = (𝑥 − 𝑦)(𝑥 2 +
4𝑥𝑦 + 𝑦 2 ).
Ans. 𝑓(𝑧) = −𝑖𝑧 3 + 𝑐
14. Find the bilinear transformation which maps the points 𝑧 = 1, 𝑧 = 𝑖 and 𝑧 = −1 into
the points 𝑤 = 𝑖, 𝑤 = 0 and 𝑤 = −1.
(𝑖−1)𝑧+(𝑖+1)
Ans. 𝑤 = , a bilinear transformation.
2𝑧

15. Find the bilinear transformation which maps the points 𝑧 = 1, 0, −1 onto the points
𝑤 = 𝑖, 0, −𝑖 and find the fixed points of the transformation.
Ans. 𝑤 = 𝑖𝑧 and 0 is the only fixed points of the bilinear transformation 𝑤 = 𝑖𝑧.

𝑧+𝑖
16. Show that the transformation𝑓(𝑧) = 𝑧−𝑖 maps the interior of the circle |𝑤| = 1 𝑖. 𝑒. |𝑤| ≤ 1
into the lower half plane 𝐼(𝑧) ≤ 0.
Ans.
4𝑦
Hints. |𝑤|2 − 1 = 𝑥 2+(𝑦−1)2

Complex (Integration)
1.
(2,4)
Evaluate ∫(0,3) (2𝑦 + 𝑥 2 )𝑑𝑥 + (3𝑥 − 𝑦)𝑑𝑦along the paths
i) the parabola 𝑥 = 2𝑡, 𝑦 = 𝑡 2 + 3
the straight line joining (0,3) and (2,4).
33 97
Ans. i) 2 ii) 6
2. Evaluate ∫𝑐|𝑧|2 𝑑𝑧 around the square with vertices (0,0), (1, 0), (1,1) and (0,1).
Ans. −1 + 𝑖
3. Using Cauchy’s integral formula, evaluate ∫𝐶
𝑐𝑜𝑠 𝜋𝑧
𝑧 2 −1
𝑑𝑧 where C is the rectangle with
vertices 2 ± 𝑖, = 2 ± 𝑖.
Ans. 0 easy
𝑧
4. State Cauchy’s integral formula. Use it to evaluate ∫𝐶 𝑧 2−1 𝑑𝑧 where C is the circle
|z| = 2.
Ans. 2𝜋𝑖
𝑒 2𝑧
5. State Cauchy’s integral formula. Use it to evaluate ∮𝐶 (𝑧+1)4 𝑑𝑧 where C is the circle
|𝑍| = 3.
8𝜋𝑖
Ans. 3𝑒 2
𝑒 3𝑧
6. Evaluate ∮|𝑧|=1 (4𝑧−𝜋𝑖)3 𝑑𝑧 .

3𝜋𝑖
9𝜋𝑖
Ans. 𝑒 4
64
𝑠𝑖𝑛 𝜋𝑧 2 +𝑐𝑜𝑠 𝜋𝑧 2
7. Evaluate ∫𝑐 (𝑧−1)2 (𝑧−2)
𝑑𝑧where C is the circle |z| = 3.

Ans. 4𝜋𝑖 + 4𝜋 2 𝑖
𝑧+1
8. Expand 𝑓(𝑧) = (𝑧−3)(𝑧−4) in Taylor’s series about the point z=2. Find also the region
of convergence.

3 11 27 59
Ans. + (𝑧 − 2) + (𝑧 − 2)2 + (𝑧 − 2)3 + ⋯
2 4 8 16
1
9. Find a Laurent’s series for𝑓(𝑧) = (𝑧+1)(𝑧+3)
valid in 𝑖) 1 < |𝑧| < 3 and 𝑖𝑖) 0 <
|𝑧 + 1| < 2.
Ans.
∞
(−1)𝑛 (𝑧 + 1)𝑛−1
𝑖𝑖) ∑ 0 < |𝑧 + 1| < 2
2𝑛+1
𝑛=0

1 (−1)𝑛 1 𝑧 𝑛
𝑖) 2 ∑∞
𝑛=0 − 6 ∑∞ 𝑛
𝑛=0(−1) (3) , 1 < |𝑧| < 3
𝑧 𝑛+1
𝑧2
10. Determine the poles of the function 𝑓(𝑧) = (𝑧−1)2 (𝑧+2) and the residues of at each of
the poles.
 5 4
Ans. 𝑅𝑒𝑠(1) = 𝑎𝑛𝑑 𝑅𝑒𝑠(−2) =
9 9
𝑐𝑜𝑡 𝜋𝑧
11. Find the poles and residues at each pole of the function 𝑓(𝑧) = (𝑧−𝑎)2.
1
Ans. 𝑅𝑒𝑠(𝑎) = −𝜋𝑐𝑜𝑠𝑒𝑐 2 𝜋𝑎 𝑎𝑛𝑑 𝑅𝑒𝑠(𝑛) = 𝜋(𝑛−𝑎)2

3𝑧 2 +𝑧−1
12. Evaluate ∫𝑐 (𝑧 2 𝑑𝑧 ; c: |z| = 2.
−1)(𝑧−3)

5𝜋𝑖
Ans. − 4
2𝜋 𝑐𝑜𝑠 2𝜃𝑑𝜃
13. Use residues theorem to evaluate the integral ∫0 5+4 𝑐𝑜𝑠 𝜃
.

𝜋
Ans.
6
2𝜋 𝑑𝜃
14. Evaluate ∫0 1+𝑎 2 −2𝑎 𝑐𝑜𝑠 𝜃
,0 < 𝑎 < 1 using contour integration method.
2𝜋
Ans. 1−𝑎2
𝑧𝑐𝑜𝑠𝑧
15. Evaluate ∮𝐶 3 𝑑𝑧 where C is the circle |𝑧| = 1.
(𝑧−𝜋⁄2 )
Ans. −2𝜋𝑖
𝑒𝑧
16. State residue theorem and use it to evaluate ∮𝐶 (𝑧 2 2)2 𝑑𝑧where C is the circle |z| =
+𝜋
4.
Ans. 0