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 Page 1


Edurev123 
5. Cauchy's Residue Theorem 
5.1 Using Cauchy's residue theorem, evaluate the integral 
?? =? ?
?? ?? ?????? ?? ??????? 
(2013 : 15 Marks) 
Solution: 
Using trigonometry, 
sin
4
??? ?=[
1
2
(1-cos?2?? ]
2
=
1
4
(1+cos
2
?2?? -2cos?2?? )
?=
1
8
[2+(1+cos?4?? )-2cos?2?? ]=
1
8
[cos?4?? -2cos?2?? +3]
?? ?=? ?
?? 0
?sin
4
??????? =
1
2
? ?
2?? 0
?sin
4
??????? ?=
1
2
? ?
2?? 0
?
1
8
[cos?4?? -2cos?2?? +3]????
?=
1
16
? ?
2?? 0
?[cos?4?? -2cos?2?? +3]????
?= Real part of 
1
16
? ?
2?? 0
?(?? ??4?? -2?? ??2?? +3)????
 
Let ?? =?? ????
,???????????????????????????? =???? ????
???? ????? =
????
????
. 
= R.P. of 
1
16
???
?? (?? 4
-2?? 2
+3)????
????
 
where ?? is the unit circle. 
???
?? ?? 4
-2?? 2
+3
?? ???? =2???? Sum of Residue of ?? (?? ) 
where ??????????????????????????????????????????????????????????? (?? )=
?? 4
-2?? 2
+3
?? 
Now ?? (?? ) has one pole at ?? =0 inside unit circle. 
                               Residue at ?? =0=lim
?? ?0
????? (?? ) 
????????????????????????????????????????????????????????????????????=3 
Page 2


Edurev123 
5. Cauchy's Residue Theorem 
5.1 Using Cauchy's residue theorem, evaluate the integral 
?? =? ?
?? ?? ?????? ?? ??????? 
(2013 : 15 Marks) 
Solution: 
Using trigonometry, 
sin
4
??? ?=[
1
2
(1-cos?2?? ]
2
=
1
4
(1+cos
2
?2?? -2cos?2?? )
?=
1
8
[2+(1+cos?4?? )-2cos?2?? ]=
1
8
[cos?4?? -2cos?2?? +3]
?? ?=? ?
?? 0
?sin
4
??????? =
1
2
? ?
2?? 0
?sin
4
??????? ?=
1
2
? ?
2?? 0
?
1
8
[cos?4?? -2cos?2?? +3]????
?=
1
16
? ?
2?? 0
?[cos?4?? -2cos?2?? +3]????
?= Real part of 
1
16
? ?
2?? 0
?(?? ??4?? -2?? ??2?? +3)????
 
Let ?? =?? ????
,???????????????????????????? =???? ????
???? ????? =
????
????
. 
= R.P. of 
1
16
???
?? (?? 4
-2?? 2
+3)????
????
 
where ?? is the unit circle. 
???
?? ?? 4
-2?? 2
+3
?? ???? =2???? Sum of Residue of ?? (?? ) 
where ??????????????????????????????????????????????????????????? (?? )=
?? 4
-2?? 2
+3
?? 
Now ?? (?? ) has one pole at ?? =0 inside unit circle. 
                               Residue at ?? =0=lim
?? ?0
????? (?? ) 
????????????????????????????????????????????????????????????????????=3 
??                                                ???? (?? )=6???? 
???????????????????????????????????????
1
16?? ??
?? ?? (?? )=
3
8
?? 
????????????????????????????????????????
0
?? ?sin
4
??????? = R.p 
3
8
?? =
3?? 8
 
5.2 Evaluate the integral ?
?? ?? ?
????
(?? +
?? ?? ?????? ??? )
?? using residues. 
(2014 : 20 Marks) 
Solution: 
Let 
?? ?=? ?
?? 0
?
????
(1+
1
2
cos??? )
2
?=
1
2
? ?
2?? 0
?
4????
(2+cos??? )
2
?=? ?
2?? 0
?
2????
(2+cos??? )
2
.
 
Let the contour ' ?? ' be the unit circle |?? |=?? with centre at the origin. 
Let ?? =?? ????
 then           cos??? ?=
1
2
(?? +
1
?? )
???? ?=???? ????
????
???? ?=
????
????
?? ?=?
0
2?? ?
2????
(2+cos??? )
2
=?
0
2?? ?
2????
???? (2+
?? 2
+1
2?? )
2
 
?=
1
?? ?
0
2?? ?
8?? (?? 2
+4?? +1)
2
????
????????????????????????????????????????????????? ?=
8
?? ?
?? ?
?? (?? 2
+4?? +1)
2
????
?=
8
?? ?
?? ??? (?? )????
 
where 
?? (?? )=
?? (?? 2
+4?? +1)
2
(??)
 
Now the poles of ?? (?? ) are given by 
Page 3


Edurev123 
5. Cauchy's Residue Theorem 
5.1 Using Cauchy's residue theorem, evaluate the integral 
?? =? ?
?? ?? ?????? ?? ??????? 
(2013 : 15 Marks) 
Solution: 
Using trigonometry, 
sin
4
??? ?=[
1
2
(1-cos?2?? ]
2
=
1
4
(1+cos
2
?2?? -2cos?2?? )
?=
1
8
[2+(1+cos?4?? )-2cos?2?? ]=
1
8
[cos?4?? -2cos?2?? +3]
?? ?=? ?
?? 0
?sin
4
??????? =
1
2
? ?
2?? 0
?sin
4
??????? ?=
1
2
? ?
2?? 0
?
1
8
[cos?4?? -2cos?2?? +3]????
?=
1
16
? ?
2?? 0
?[cos?4?? -2cos?2?? +3]????
?= Real part of 
1
16
? ?
2?? 0
?(?? ??4?? -2?? ??2?? +3)????
 
Let ?? =?? ????
,???????????????????????????? =???? ????
???? ????? =
????
????
. 
= R.P. of 
1
16
???
?? (?? 4
-2?? 2
+3)????
????
 
where ?? is the unit circle. 
???
?? ?? 4
-2?? 2
+3
?? ???? =2???? Sum of Residue of ?? (?? ) 
where ??????????????????????????????????????????????????????????? (?? )=
?? 4
-2?? 2
+3
?? 
Now ?? (?? ) has one pole at ?? =0 inside unit circle. 
                               Residue at ?? =0=lim
?? ?0
????? (?? ) 
????????????????????????????????????????????????????????????????????=3 
??                                                ???? (?? )=6???? 
???????????????????????????????????????
1
16?? ??
?? ?? (?? )=
3
8
?? 
????????????????????????????????????????
0
?? ?sin
4
??????? = R.p 
3
8
?? =
3?? 8
 
5.2 Evaluate the integral ?
?? ?? ?
????
(?? +
?? ?? ?????? ??? )
?? using residues. 
(2014 : 20 Marks) 
Solution: 
Let 
?? ?=? ?
?? 0
?
????
(1+
1
2
cos??? )
2
?=
1
2
? ?
2?? 0
?
4????
(2+cos??? )
2
?=? ?
2?? 0
?
2????
(2+cos??? )
2
.
 
Let the contour ' ?? ' be the unit circle |?? |=?? with centre at the origin. 
Let ?? =?? ????
 then           cos??? ?=
1
2
(?? +
1
?? )
???? ?=???? ????
????
???? ?=
????
????
?? ?=?
0
2?? ?
2????
(2+cos??? )
2
=?
0
2?? ?
2????
???? (2+
?? 2
+1
2?? )
2
 
?=
1
?? ?
0
2?? ?
8?? (?? 2
+4?? +1)
2
????
????????????????????????????????????????????????? ?=
8
?? ?
?? ?
?? (?? 2
+4?? +1)
2
????
?=
8
?? ?
?? ??? (?? )????
 
where 
?? (?? )=
?? (?? 2
+4?? +1)
2
(??)
 
Now the poles of ?? (?? ) are given by 
(?? 2
+4?? +1)
2
=0??? =
-4±v16-4
2
=
-4±v12
2
=-2±v3 (twice)  
??? (?? ) has poles of order at ?? =-2±v3 (twice) 
Let ?? =-2+v3,?? =-2-v3 
Clearly |?? |>1 
Since |???? |=1?|?? |<1 
Hence, the ?????? pole inside ?? is ?? =?? of order 2 . 
????
?? ??? (?? )???? ?=??
??????
(?? 2
+4?? +1)
2
?=2???? ( residue at ?? =?? )
 
Now the residue at ?? =?? is 
lim
?? ??? ?
?? ????
(?? -?? )
2
?? (?? 2
+4?? +1)
2
=lim
?? ??? ?
?? ????
?? (?? -?? )
2
 
=lim
?? ??? ?
-(?? +?? )
(?? -?? )
3
 
=
-(?? +?? )
(?? -?? )
3
 
=
-(-4)
(2v3)
3
 
=
(4)
8(3v3)
=
1
6v3
 
???????????????????????????????????????
?? ??? (?? )???? =2???? (
1
6v3
)=
2????
6v3
=
v3?? 3v3
?? from (i)
?????????????????????????????? ?
2?? 0
?
2????
(2+cos??? )
2
=
8
?? (
????
3v3
)=
8?? 3v3
??????????????????????????????????????????????????????????????? =? ?
?? 0
?
????
(1+
cos??? 2
)
2
=
8?? 3v3
 
5.3 State Cauchy's Residue theorem. Using it, evaluate the integral 
?
?? ?
?? ?? +?? ?? (?? +?? )(?? -?? )
?? ???? ·{?? :|?? |=?? } 
(2015 : 15 Marks) 
Page 4


Edurev123 
5. Cauchy's Residue Theorem 
5.1 Using Cauchy's residue theorem, evaluate the integral 
?? =? ?
?? ?? ?????? ?? ??????? 
(2013 : 15 Marks) 
Solution: 
Using trigonometry, 
sin
4
??? ?=[
1
2
(1-cos?2?? ]
2
=
1
4
(1+cos
2
?2?? -2cos?2?? )
?=
1
8
[2+(1+cos?4?? )-2cos?2?? ]=
1
8
[cos?4?? -2cos?2?? +3]
?? ?=? ?
?? 0
?sin
4
??????? =
1
2
? ?
2?? 0
?sin
4
??????? ?=
1
2
? ?
2?? 0
?
1
8
[cos?4?? -2cos?2?? +3]????
?=
1
16
? ?
2?? 0
?[cos?4?? -2cos?2?? +3]????
?= Real part of 
1
16
? ?
2?? 0
?(?? ??4?? -2?? ??2?? +3)????
 
Let ?? =?? ????
,???????????????????????????? =???? ????
???? ????? =
????
????
. 
= R.P. of 
1
16
???
?? (?? 4
-2?? 2
+3)????
????
 
where ?? is the unit circle. 
???
?? ?? 4
-2?? 2
+3
?? ???? =2???? Sum of Residue of ?? (?? ) 
where ??????????????????????????????????????????????????????????? (?? )=
?? 4
-2?? 2
+3
?? 
Now ?? (?? ) has one pole at ?? =0 inside unit circle. 
                               Residue at ?? =0=lim
?? ?0
????? (?? ) 
????????????????????????????????????????????????????????????????????=3 
??                                                ???? (?? )=6???? 
???????????????????????????????????????
1
16?? ??
?? ?? (?? )=
3
8
?? 
????????????????????????????????????????
0
?? ?sin
4
??????? = R.p 
3
8
?? =
3?? 8
 
5.2 Evaluate the integral ?
?? ?? ?
????
(?? +
?? ?? ?????? ??? )
?? using residues. 
(2014 : 20 Marks) 
Solution: 
Let 
?? ?=? ?
?? 0
?
????
(1+
1
2
cos??? )
2
?=
1
2
? ?
2?? 0
?
4????
(2+cos??? )
2
?=? ?
2?? 0
?
2????
(2+cos??? )
2
.
 
Let the contour ' ?? ' be the unit circle |?? |=?? with centre at the origin. 
Let ?? =?? ????
 then           cos??? ?=
1
2
(?? +
1
?? )
???? ?=???? ????
????
???? ?=
????
????
?? ?=?
0
2?? ?
2????
(2+cos??? )
2
=?
0
2?? ?
2????
???? (2+
?? 2
+1
2?? )
2
 
?=
1
?? ?
0
2?? ?
8?? (?? 2
+4?? +1)
2
????
????????????????????????????????????????????????? ?=
8
?? ?
?? ?
?? (?? 2
+4?? +1)
2
????
?=
8
?? ?
?? ??? (?? )????
 
where 
?? (?? )=
?? (?? 2
+4?? +1)
2
(??)
 
Now the poles of ?? (?? ) are given by 
(?? 2
+4?? +1)
2
=0??? =
-4±v16-4
2
=
-4±v12
2
=-2±v3 (twice)  
??? (?? ) has poles of order at ?? =-2±v3 (twice) 
Let ?? =-2+v3,?? =-2-v3 
Clearly |?? |>1 
Since |???? |=1?|?? |<1 
Hence, the ?????? pole inside ?? is ?? =?? of order 2 . 
????
?? ??? (?? )???? ?=??
??????
(?? 2
+4?? +1)
2
?=2???? ( residue at ?? =?? )
 
Now the residue at ?? =?? is 
lim
?? ??? ?
?? ????
(?? -?? )
2
?? (?? 2
+4?? +1)
2
=lim
?? ??? ?
?? ????
?? (?? -?? )
2
 
=lim
?? ??? ?
-(?? +?? )
(?? -?? )
3
 
=
-(?? +?? )
(?? -?? )
3
 
=
-(-4)
(2v3)
3
 
=
(4)
8(3v3)
=
1
6v3
 
???????????????????????????????????????
?? ??? (?? )???? =2???? (
1
6v3
)=
2????
6v3
=
v3?? 3v3
?? from (i)
?????????????????????????????? ?
2?? 0
?
2????
(2+cos??? )
2
=
8
?? (
????
3v3
)=
8?? 3v3
??????????????????????????????????????????????????????????????? =? ?
?? 0
?
????
(1+
cos??? 2
)
2
=
8?? 3v3
 
5.3 State Cauchy's Residue theorem. Using it, evaluate the integral 
?
?? ?
?? ?? +?? ?? (?? +?? )(?? -?? )
?? ???? ·{?? :|?? |=?? } 
(2015 : 15 Marks) 
Solution: 
Cauchy's Residue theorem states that in a given region, integral of a function along the 
closed curve is equal to 2???? times sum of its residues. 
Given, 
?? ?=??
?? ?
?? 2
+1
?? (?? +1)(?? -??)
2
???? ,?? :|?? |=2
?=??
?? ??? (?? )????
 
In the given region, |?? |=2,?? (?? ) is analytic everywhere except at ?? =0,-1,?? . ?? =0 is a 
pole of order 1 . 
?? =-1 is a pole of order 1 . 
?? =?? is a pole of order 2 . 
 
Now, Residue at ?? =0 : 
lim
?? ?0
?
?? (?? 2
+1)
?? (?? +1)(?? -??)
2
?=
1+1
1(-??)
2
=
2
(-1)
?=-2
 
Residue at ?? =-1 : 
lim
?? ?-1
?
(?? +1)(?? 2
+1)
?? (?? +1)(?? -1)
2
?=
1+?? -1
(-1)(-1-??)
2
=
-(1+?? -1
)
1-1+2?? ?=
(1+?? -1
)
2
?? 
Residue at ?? =?? : 
Page 5


Edurev123 
5. Cauchy's Residue Theorem 
5.1 Using Cauchy's residue theorem, evaluate the integral 
?? =? ?
?? ?? ?????? ?? ??????? 
(2013 : 15 Marks) 
Solution: 
Using trigonometry, 
sin
4
??? ?=[
1
2
(1-cos?2?? ]
2
=
1
4
(1+cos
2
?2?? -2cos?2?? )
?=
1
8
[2+(1+cos?4?? )-2cos?2?? ]=
1
8
[cos?4?? -2cos?2?? +3]
?? ?=? ?
?? 0
?sin
4
??????? =
1
2
? ?
2?? 0
?sin
4
??????? ?=
1
2
? ?
2?? 0
?
1
8
[cos?4?? -2cos?2?? +3]????
?=
1
16
? ?
2?? 0
?[cos?4?? -2cos?2?? +3]????
?= Real part of 
1
16
? ?
2?? 0
?(?? ??4?? -2?? ??2?? +3)????
 
Let ?? =?? ????
,???????????????????????????? =???? ????
???? ????? =
????
????
. 
= R.P. of 
1
16
???
?? (?? 4
-2?? 2
+3)????
????
 
where ?? is the unit circle. 
???
?? ?? 4
-2?? 2
+3
?? ???? =2???? Sum of Residue of ?? (?? ) 
where ??????????????????????????????????????????????????????????? (?? )=
?? 4
-2?? 2
+3
?? 
Now ?? (?? ) has one pole at ?? =0 inside unit circle. 
                               Residue at ?? =0=lim
?? ?0
????? (?? ) 
????????????????????????????????????????????????????????????????????=3 
??                                                ???? (?? )=6???? 
???????????????????????????????????????
1
16?? ??
?? ?? (?? )=
3
8
?? 
????????????????????????????????????????
0
?? ?sin
4
??????? = R.p 
3
8
?? =
3?? 8
 
5.2 Evaluate the integral ?
?? ?? ?
????
(?? +
?? ?? ?????? ??? )
?? using residues. 
(2014 : 20 Marks) 
Solution: 
Let 
?? ?=? ?
?? 0
?
????
(1+
1
2
cos??? )
2
?=
1
2
? ?
2?? 0
?
4????
(2+cos??? )
2
?=? ?
2?? 0
?
2????
(2+cos??? )
2
.
 
Let the contour ' ?? ' be the unit circle |?? |=?? with centre at the origin. 
Let ?? =?? ????
 then           cos??? ?=
1
2
(?? +
1
?? )
???? ?=???? ????
????
???? ?=
????
????
?? ?=?
0
2?? ?
2????
(2+cos??? )
2
=?
0
2?? ?
2????
???? (2+
?? 2
+1
2?? )
2
 
?=
1
?? ?
0
2?? ?
8?? (?? 2
+4?? +1)
2
????
????????????????????????????????????????????????? ?=
8
?? ?
?? ?
?? (?? 2
+4?? +1)
2
????
?=
8
?? ?
?? ??? (?? )????
 
where 
?? (?? )=
?? (?? 2
+4?? +1)
2
(??)
 
Now the poles of ?? (?? ) are given by 
(?? 2
+4?? +1)
2
=0??? =
-4±v16-4
2
=
-4±v12
2
=-2±v3 (twice)  
??? (?? ) has poles of order at ?? =-2±v3 (twice) 
Let ?? =-2+v3,?? =-2-v3 
Clearly |?? |>1 
Since |???? |=1?|?? |<1 
Hence, the ?????? pole inside ?? is ?? =?? of order 2 . 
????
?? ??? (?? )???? ?=??
??????
(?? 2
+4?? +1)
2
?=2???? ( residue at ?? =?? )
 
Now the residue at ?? =?? is 
lim
?? ??? ?
?? ????
(?? -?? )
2
?? (?? 2
+4?? +1)
2
=lim
?? ??? ?
?? ????
?? (?? -?? )
2
 
=lim
?? ??? ?
-(?? +?? )
(?? -?? )
3
 
=
-(?? +?? )
(?? -?? )
3
 
=
-(-4)
(2v3)
3
 
=
(4)
8(3v3)
=
1
6v3
 
???????????????????????????????????????
?? ??? (?? )???? =2???? (
1
6v3
)=
2????
6v3
=
v3?? 3v3
?? from (i)
?????????????????????????????? ?
2?? 0
?
2????
(2+cos??? )
2
=
8
?? (
????
3v3
)=
8?? 3v3
??????????????????????????????????????????????????????????????? =? ?
?? 0
?
????
(1+
cos??? 2
)
2
=
8?? 3v3
 
5.3 State Cauchy's Residue theorem. Using it, evaluate the integral 
?
?? ?
?? ?? +?? ?? (?? +?? )(?? -?? )
?? ???? ·{?? :|?? |=?? } 
(2015 : 15 Marks) 
Solution: 
Cauchy's Residue theorem states that in a given region, integral of a function along the 
closed curve is equal to 2???? times sum of its residues. 
Given, 
?? ?=??
?? ?
?? 2
+1
?? (?? +1)(?? -??)
2
???? ,?? :|?? |=2
?=??
?? ??? (?? )????
 
In the given region, |?? |=2,?? (?? ) is analytic everywhere except at ?? =0,-1,?? . ?? =0 is a 
pole of order 1 . 
?? =-1 is a pole of order 1 . 
?? =?? is a pole of order 2 . 
 
Now, Residue at ?? =0 : 
lim
?? ?0
?
?? (?? 2
+1)
?? (?? +1)(?? -??)
2
?=
1+1
1(-??)
2
=
2
(-1)
?=-2
 
Residue at ?? =-1 : 
lim
?? ?-1
?
(?? +1)(?? 2
+1)
?? (?? +1)(?? -1)
2
?=
1+?? -1
(-1)(-1-??)
2
=
-(1+?? -1
)
1-1+2?? ?=
(1+?? -1
)
2
?? 
Residue at ?? =?? : 
lim
?? ??? ?
?? ????
(?? -1)
2
?? 2
+1
?? (?? +1)(?? -1)
2
?=lim
?? ??? ?
?? ????
·
?? 2
+1
?? (?? +1)
?=lim
?? ??? ?{
-(?? ?? +1)
?? 2
(?? +1)
-
(?? ?? +1)
?? (?? +1)
2
+
?? ?? ?? (?? +1)
}
?=
+(1+?? ?? )
(+1)(1+??)
-
(1+?? ?? )
??(1+??)
2
+
?? ?? ??(?? +1)
?=
1+?? ?? 1+?? -
1+?? ?? 2
+
?? ?? ??(?? +1)
?=
1-?? 2
+
1+?? ?? 2
=
2-?? +?? ?? 2
??
?? ??? (?? )???? ?=2???? ( sum of residues)
?=2???? (-2+
(1+?? -1
)?? 2
+
2-?? +?? ?? 2
)
?=2???? (-2+
?? +???? -1
+2-?? +?? ?? 2
)
??
?? ??? (?? )???? ?=???? (?? ?? +???? -1
+2-4)
?=???? (?? ?? +
?? ?? -2)
??=???? (?? ?? +
?? ?? -2)
 
5.4 Show by applying residue theorem that 
? ?
8
?? ????
(?? ?? +?? ?? )
?? =
?? ?? ?? ?? ,??? >?? 
(2018.: 15 Marks) 
Solution: 
Consider ?
-?? ?? ?
????
(?? 2
+?? 2
)
2
+?
?? ?
????
(?? 2
+?? 2
)
2
 
Now, 
? ?
?? -?? ????
(?? 2
+?? 2
)
2
=? ?
?? ?? ????
(?? +???? )
2
(?? -???? )
2
 
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