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Page 1
Edurev123
4. Laurant's Series
4.1 Find the Laurent series of the function ?? (?? )=?????? ?[
?? ?? (?? -
?? ?? )] as ?
?? =-8
8
??? ?? ?? ?? for
?? <|?? |<8
where
?? ?? =
?? ?? ? ?
?? ?? ?????? ?(???? -?? ?????? ??? )???? ;?? =?? ,±?? ,±?? ,…..
with ?? a given complex number and taking the unit circle ?? given by ?? =?? ????
(-?? =
?? =?? ) as contour in this region.
(2010 : 15 Marks)
Solution:
?? is the only point of singularity here. So, we can expand exp?(
?? 2
(?? -
1
?? ))) as Laurent
series valid for every value of ?? ?0, i.e., for ???? >0.
Thus, we have
???????????????????????????????????????????????????????????????|?? |?0
??????????????????????????????????????exp?(
?? 2
(?? -
1
?? ))=?
?? =-8
8
??? ?? ?? ??
when
?? ?? =
1
2????
? ?
?? (?? )
?? 2
?? +1
2
?? +1
????
where ?? is any circle with its center at the origin. Taking ?? as the circle with radius unity,
we have
?? ?? =
1
2????
?
?? ?
?? ?? /2(?? -
1
?? )
?? ?? +1
????
Putting ?? =?? ????
, we obtain
Page 2
Edurev123
4. Laurant's Series
4.1 Find the Laurent series of the function ?? (?? )=?????? ?[
?? ?? (?? -
?? ?? )] as ?
?? =-8
8
??? ?? ?? ?? for
?? <|?? |<8
where
?? ?? =
?? ?? ? ?
?? ?? ?????? ?(???? -?? ?????? ??? )???? ;?? =?? ,±?? ,±?? ,…..
with ?? a given complex number and taking the unit circle ?? given by ?? =?? ????
(-?? =
?? =?? ) as contour in this region.
(2010 : 15 Marks)
Solution:
?? is the only point of singularity here. So, we can expand exp?(
?? 2
(?? -
1
?? ))) as Laurent
series valid for every value of ?? ?0, i.e., for ???? >0.
Thus, we have
???????????????????????????????????????????????????????????????|?? |?0
??????????????????????????????????????exp?(
?? 2
(?? -
1
?? ))=?
?? =-8
8
??? ?? ?? ??
when
?? ?? =
1
2????
? ?
?? (?? )
?? 2
?? +1
2
?? +1
????
where ?? is any circle with its center at the origin. Taking ?? as the circle with radius unity,
we have
?? ?? =
1
2????
?
?? ?
?? ?? /2(?? -
1
?? )
?? ?? +1
????
Putting ?? =?? ????
, we obtain
?? ?? ?=
1
2????
? ?
2?? 0
?
?? ?? 2
(?? ????
-?? -????
)
?? ?? (?? +1)?? ·???? ????
????
??????????????????????????????????????????????????????????????? ?? =
1
2?? ? ?
2?? 0
??? ???? sin??? ·?? -??????
????
????????????????????????????????????????????????? ?? =
1
2?? ?
0
2?? ?cos?(?? sin??? -???? )???? +
?? 2?? ?
0
2?? ?sin?(?? sin??? -???? )????
Taking ?? =2?? -?? , we have
?? ?
2?? 0
?sin?(?? sin??? -???? )???? -(-? ?
0
2?? ?sin?[-?? sin??? -2???? +???? ]???? )
?=? ?
2?? 0
?sin?(?? sin??? -???? )????
So that ?? ?
2?? 0
?sin?(?? sin??? -???? )???? =0
?????????????????????????????? ?? =
1
2?? ?
0
2?? ?cos?(?? sin??? -???? )???? +0 (Putting value from (2) in (1))
?=
1
2?? ? ?
2?? 0
?cos?(???? -?? sin??? )????
?=
2
2?? ? ?
?? 0
?cos?(???? -?? sin??? )????
?? ?? =
1
?? ? ?
?? 0
?cos?(???? -?? sin??? )????
So, given expression can be expressed as ?
-8
8
??? ?? ?? ?? where ?? ?? is given by above
equation (3).
4.2 Find the Laurent series for the function
?? (?? )=
?? ?? -?? ?? with center ?? =??
(2011: 15 marks)
Solution:
We have
?? (?? )?=
1
1-?? 2
=
1
(1-?? )(1+?? )
??????????????????????????????????????????????????????????????????=
1
?? (
1
1-?? +
1
1+?? )?????????????????????????????????????????????????????????????(??)
Page 3
Edurev123
4. Laurant's Series
4.1 Find the Laurent series of the function ?? (?? )=?????? ?[
?? ?? (?? -
?? ?? )] as ?
?? =-8
8
??? ?? ?? ?? for
?? <|?? |<8
where
?? ?? =
?? ?? ? ?
?? ?? ?????? ?(???? -?? ?????? ??? )???? ;?? =?? ,±?? ,±?? ,…..
with ?? a given complex number and taking the unit circle ?? given by ?? =?? ????
(-?? =
?? =?? ) as contour in this region.
(2010 : 15 Marks)
Solution:
?? is the only point of singularity here. So, we can expand exp?(
?? 2
(?? -
1
?? ))) as Laurent
series valid for every value of ?? ?0, i.e., for ???? >0.
Thus, we have
???????????????????????????????????????????????????????????????|?? |?0
??????????????????????????????????????exp?(
?? 2
(?? -
1
?? ))=?
?? =-8
8
??? ?? ?? ??
when
?? ?? =
1
2????
? ?
?? (?? )
?? 2
?? +1
2
?? +1
????
where ?? is any circle with its center at the origin. Taking ?? as the circle with radius unity,
we have
?? ?? =
1
2????
?
?? ?
?? ?? /2(?? -
1
?? )
?? ?? +1
????
Putting ?? =?? ????
, we obtain
?? ?? ?=
1
2????
? ?
2?? 0
?
?? ?? 2
(?? ????
-?? -????
)
?? ?? (?? +1)?? ·???? ????
????
??????????????????????????????????????????????????????????????? ?? =
1
2?? ? ?
2?? 0
??? ???? sin??? ·?? -??????
????
????????????????????????????????????????????????? ?? =
1
2?? ?
0
2?? ?cos?(?? sin??? -???? )???? +
?? 2?? ?
0
2?? ?sin?(?? sin??? -???? )????
Taking ?? =2?? -?? , we have
?? ?
2?? 0
?sin?(?? sin??? -???? )???? -(-? ?
0
2?? ?sin?[-?? sin??? -2???? +???? ]???? )
?=? ?
2?? 0
?sin?(?? sin??? -???? )????
So that ?? ?
2?? 0
?sin?(?? sin??? -???? )???? =0
?????????????????????????????? ?? =
1
2?? ?
0
2?? ?cos?(?? sin??? -???? )???? +0 (Putting value from (2) in (1))
?=
1
2?? ? ?
2?? 0
?cos?(???? -?? sin??? )????
?=
2
2?? ? ?
?? 0
?cos?(???? -?? sin??? )????
?? ?? =
1
?? ? ?
?? 0
?cos?(???? -?? sin??? )????
So, given expression can be expressed as ?
-8
8
??? ?? ?? ?? where ?? ?? is given by above
equation (3).
4.2 Find the Laurent series for the function
?? (?? )=
?? ?? -?? ?? with center ?? =??
(2011: 15 marks)
Solution:
We have
?? (?? )?=
1
1-?? 2
=
1
(1-?? )(1+?? )
??????????????????????????????????????????????????????????????????=
1
?? (
1
1-?? +
1
1+?? )?????????????????????????????????????????????????????????????(??)
First we will find Laurent expansion for ?? (?? )=
1
1+?? about ?? =1.
We write,
?? (?? )=??
8
?? =0
??? ?? (?? -1)
?? (???? )
where
?? ?? =
?? ?? (1)
?? !
But
?? ?? (?? )=(-1)(-2)……(-?? )(1+2)
-(?? +1)
=(-1)
?? ?? !(1+?? )
-(?? +1)
????????????????????????????????????????????????
?? ?? (1)
?? !
=
(-1)
?? 2
?? +1
????????????????????????????????????????????????????????? ?? =
?? ?? (1)
?? !
=
(-1)
?? 2
?? +1
From (ii),
?? (?? )=??
8
?? =0
(-1)
?? 2
?? +1
(?? -1)
??
=??
8
?? =0
(1-?? )
?? 2
?? +1
? From (i),
???????????????????????????????????????????????????????? (?? )=
1
2
(
1
1-?? +?
?? =0
8
?
(1-?? )
?? 2
?? +1
)
This is the required expansion.
4.3 Expand the function ?? (?? )=
?? (?? +?? )(?? +?? )
in Laurent series valid for :
(i) ?? <|?? |<?? ;
(ii) |?? |>?? ;
(iii) ?? <|?? +?? |<?? ;
(iv) |?? |<??
(2012 : 15 Marks)
Page 4
Edurev123
4. Laurant's Series
4.1 Find the Laurent series of the function ?? (?? )=?????? ?[
?? ?? (?? -
?? ?? )] as ?
?? =-8
8
??? ?? ?? ?? for
?? <|?? |<8
where
?? ?? =
?? ?? ? ?
?? ?? ?????? ?(???? -?? ?????? ??? )???? ;?? =?? ,±?? ,±?? ,…..
with ?? a given complex number and taking the unit circle ?? given by ?? =?? ????
(-?? =
?? =?? ) as contour in this region.
(2010 : 15 Marks)
Solution:
?? is the only point of singularity here. So, we can expand exp?(
?? 2
(?? -
1
?? ))) as Laurent
series valid for every value of ?? ?0, i.e., for ???? >0.
Thus, we have
???????????????????????????????????????????????????????????????|?? |?0
??????????????????????????????????????exp?(
?? 2
(?? -
1
?? ))=?
?? =-8
8
??? ?? ?? ??
when
?? ?? =
1
2????
? ?
?? (?? )
?? 2
?? +1
2
?? +1
????
where ?? is any circle with its center at the origin. Taking ?? as the circle with radius unity,
we have
?? ?? =
1
2????
?
?? ?
?? ?? /2(?? -
1
?? )
?? ?? +1
????
Putting ?? =?? ????
, we obtain
?? ?? ?=
1
2????
? ?
2?? 0
?
?? ?? 2
(?? ????
-?? -????
)
?? ?? (?? +1)?? ·???? ????
????
??????????????????????????????????????????????????????????????? ?? =
1
2?? ? ?
2?? 0
??? ???? sin??? ·?? -??????
????
????????????????????????????????????????????????? ?? =
1
2?? ?
0
2?? ?cos?(?? sin??? -???? )???? +
?? 2?? ?
0
2?? ?sin?(?? sin??? -???? )????
Taking ?? =2?? -?? , we have
?? ?
2?? 0
?sin?(?? sin??? -???? )???? -(-? ?
0
2?? ?sin?[-?? sin??? -2???? +???? ]???? )
?=? ?
2?? 0
?sin?(?? sin??? -???? )????
So that ?? ?
2?? 0
?sin?(?? sin??? -???? )???? =0
?????????????????????????????? ?? =
1
2?? ?
0
2?? ?cos?(?? sin??? -???? )???? +0 (Putting value from (2) in (1))
?=
1
2?? ? ?
2?? 0
?cos?(???? -?? sin??? )????
?=
2
2?? ? ?
?? 0
?cos?(???? -?? sin??? )????
?? ?? =
1
?? ? ?
?? 0
?cos?(???? -?? sin??? )????
So, given expression can be expressed as ?
-8
8
??? ?? ?? ?? where ?? ?? is given by above
equation (3).
4.2 Find the Laurent series for the function
?? (?? )=
?? ?? -?? ?? with center ?? =??
(2011: 15 marks)
Solution:
We have
?? (?? )?=
1
1-?? 2
=
1
(1-?? )(1+?? )
??????????????????????????????????????????????????????????????????=
1
?? (
1
1-?? +
1
1+?? )?????????????????????????????????????????????????????????????(??)
First we will find Laurent expansion for ?? (?? )=
1
1+?? about ?? =1.
We write,
?? (?? )=??
8
?? =0
??? ?? (?? -1)
?? (???? )
where
?? ?? =
?? ?? (1)
?? !
But
?? ?? (?? )=(-1)(-2)……(-?? )(1+2)
-(?? +1)
=(-1)
?? ?? !(1+?? )
-(?? +1)
????????????????????????????????????????????????
?? ?? (1)
?? !
=
(-1)
?? 2
?? +1
????????????????????????????????????????????????????????? ?? =
?? ?? (1)
?? !
=
(-1)
?? 2
?? +1
From (ii),
?? (?? )=??
8
?? =0
(-1)
?? 2
?? +1
(?? -1)
??
=??
8
?? =0
(1-?? )
?? 2
?? +1
? From (i),
???????????????????????????????????????????????????????? (?? )=
1
2
(
1
1-?? +?
?? =0
8
?
(1-?? )
?? 2
?? +1
)
This is the required expansion.
4.3 Expand the function ?? (?? )=
?? (?? +?? )(?? +?? )
in Laurent series valid for :
(i) ?? <|?? |<?? ;
(ii) |?? |>?? ;
(iii) ?? <|?? +?? |<?? ;
(iv) |?? |<??
(2012 : 15 Marks)
Solution:
Given:
?? (?? )?=
1
(?? +1)(?? +3)
?=
1
2(?? +1)
-
1
2(?? +3)
(using partial
fractions)
(i) ????????????????????????????????????????????????1<|?? |<3
????????????????????????????????????????
1
|?? |
?<1 and
|?? |
3
<1
??????????????????????????????????????? (?? )?=
1
2(?? +1)
-
1
2(?? +3)
?=
1
2?? (1+
1
?? )
-
1
6(1+
?? 3
)
?=
1
2?? (1+
1
?? )
-1
-
1
6
(1+
?? 3
)
-1
?=
1
2?? (1-
1
?? +
1
?? 2
-
1
?? 3
?·)-
1
6
(1-
?? 3
+
?? 2
9
-
?? 3
27
….)
?=?..+
1
2?? 3
-
1
2?? 2
+
1
2?? +
1
6
-
?? 18
+
?? 2
54
-
?? 3
162
…..
(ii) ?????????????????|?? |>3
?
3
|?? |
<1
??? (?? )=
1
2?? (1+
1
?? )
-1
-
1
2?? (1+
3
?? )
-1
=
1
2?? (1-
1
?? +
1
?? 2
-
1
?? 3
+?..)-
1
2?? (1-
3
?? +
9
?? 2
-
27
?? 3
…..)
=
1
?? 2
-
4
?? 3
+
13
?? 4
-
40
25
+?..
(iii)
0<|?? +1|<2
Put ?? +1=4
Then
Page 5
Edurev123
4. Laurant's Series
4.1 Find the Laurent series of the function ?? (?? )=?????? ?[
?? ?? (?? -
?? ?? )] as ?
?? =-8
8
??? ?? ?? ?? for
?? <|?? |<8
where
?? ?? =
?? ?? ? ?
?? ?? ?????? ?(???? -?? ?????? ??? )???? ;?? =?? ,±?? ,±?? ,…..
with ?? a given complex number and taking the unit circle ?? given by ?? =?? ????
(-?? =
?? =?? ) as contour in this region.
(2010 : 15 Marks)
Solution:
?? is the only point of singularity here. So, we can expand exp?(
?? 2
(?? -
1
?? ))) as Laurent
series valid for every value of ?? ?0, i.e., for ???? >0.
Thus, we have
???????????????????????????????????????????????????????????????|?? |?0
??????????????????????????????????????exp?(
?? 2
(?? -
1
?? ))=?
?? =-8
8
??? ?? ?? ??
when
?? ?? =
1
2????
? ?
?? (?? )
?? 2
?? +1
2
?? +1
????
where ?? is any circle with its center at the origin. Taking ?? as the circle with radius unity,
we have
?? ?? =
1
2????
?
?? ?
?? ?? /2(?? -
1
?? )
?? ?? +1
????
Putting ?? =?? ????
, we obtain
?? ?? ?=
1
2????
? ?
2?? 0
?
?? ?? 2
(?? ????
-?? -????
)
?? ?? (?? +1)?? ·???? ????
????
??????????????????????????????????????????????????????????????? ?? =
1
2?? ? ?
2?? 0
??? ???? sin??? ·?? -??????
????
????????????????????????????????????????????????? ?? =
1
2?? ?
0
2?? ?cos?(?? sin??? -???? )???? +
?? 2?? ?
0
2?? ?sin?(?? sin??? -???? )????
Taking ?? =2?? -?? , we have
?? ?
2?? 0
?sin?(?? sin??? -???? )???? -(-? ?
0
2?? ?sin?[-?? sin??? -2???? +???? ]???? )
?=? ?
2?? 0
?sin?(?? sin??? -???? )????
So that ?? ?
2?? 0
?sin?(?? sin??? -???? )???? =0
?????????????????????????????? ?? =
1
2?? ?
0
2?? ?cos?(?? sin??? -???? )???? +0 (Putting value from (2) in (1))
?=
1
2?? ? ?
2?? 0
?cos?(???? -?? sin??? )????
?=
2
2?? ? ?
?? 0
?cos?(???? -?? sin??? )????
?? ?? =
1
?? ? ?
?? 0
?cos?(???? -?? sin??? )????
So, given expression can be expressed as ?
-8
8
??? ?? ?? ?? where ?? ?? is given by above
equation (3).
4.2 Find the Laurent series for the function
?? (?? )=
?? ?? -?? ?? with center ?? =??
(2011: 15 marks)
Solution:
We have
?? (?? )?=
1
1-?? 2
=
1
(1-?? )(1+?? )
??????????????????????????????????????????????????????????????????=
1
?? (
1
1-?? +
1
1+?? )?????????????????????????????????????????????????????????????(??)
First we will find Laurent expansion for ?? (?? )=
1
1+?? about ?? =1.
We write,
?? (?? )=??
8
?? =0
??? ?? (?? -1)
?? (???? )
where
?? ?? =
?? ?? (1)
?? !
But
?? ?? (?? )=(-1)(-2)……(-?? )(1+2)
-(?? +1)
=(-1)
?? ?? !(1+?? )
-(?? +1)
????????????????????????????????????????????????
?? ?? (1)
?? !
=
(-1)
?? 2
?? +1
????????????????????????????????????????????????????????? ?? =
?? ?? (1)
?? !
=
(-1)
?? 2
?? +1
From (ii),
?? (?? )=??
8
?? =0
(-1)
?? 2
?? +1
(?? -1)
??
=??
8
?? =0
(1-?? )
?? 2
?? +1
? From (i),
???????????????????????????????????????????????????????? (?? )=
1
2
(
1
1-?? +?
?? =0
8
?
(1-?? )
?? 2
?? +1
)
This is the required expansion.
4.3 Expand the function ?? (?? )=
?? (?? +?? )(?? +?? )
in Laurent series valid for :
(i) ?? <|?? |<?? ;
(ii) |?? |>?? ;
(iii) ?? <|?? +?? |<?? ;
(iv) |?? |<??
(2012 : 15 Marks)
Solution:
Given:
?? (?? )?=
1
(?? +1)(?? +3)
?=
1
2(?? +1)
-
1
2(?? +3)
(using partial
fractions)
(i) ????????????????????????????????????????????????1<|?? |<3
????????????????????????????????????????
1
|?? |
?<1 and
|?? |
3
<1
??????????????????????????????????????? (?? )?=
1
2(?? +1)
-
1
2(?? +3)
?=
1
2?? (1+
1
?? )
-
1
6(1+
?? 3
)
?=
1
2?? (1+
1
?? )
-1
-
1
6
(1+
?? 3
)
-1
?=
1
2?? (1-
1
?? +
1
?? 2
-
1
?? 3
?·)-
1
6
(1-
?? 3
+
?? 2
9
-
?? 3
27
….)
?=?..+
1
2?? 3
-
1
2?? 2
+
1
2?? +
1
6
-
?? 18
+
?? 2
54
-
?? 3
162
…..
(ii) ?????????????????|?? |>3
?
3
|?? |
<1
??? (?? )=
1
2?? (1+
1
?? )
-1
-
1
2?? (1+
3
?? )
-1
=
1
2?? (1-
1
?? +
1
?? 2
-
1
?? 3
+?..)-
1
2?? (1-
3
?? +
9
?? 2
-
27
?? 3
…..)
=
1
?? 2
-
4
?? 3
+
13
?? 4
-
40
25
+?..
(iii)
0<|?? +1|<2
Put ?? +1=4
Then
0?<|?? +1|<2 P 0<|4|<2
?? (?? )?=
1
(?? +1)(?? +3)
=
1
4(4+2)
?=
1
21
(1+
4
2
)
-1
?=
1
24
(1-
4
2
+
4
2
4
-
4
3
8
….)
?=
1
24
-
1
4
+
4
8
-
4
2
16
+?.
?=
1
2(?? +1)
-
1
4
+
?? +1
8
-
(?? +1)
2
16
+?..
(iii) ???????????????????????|?? |<1
Then
?? (?? )?=
1
2(?? +1)
-
1
6(1+
?? 3
)
?=
1
2
(1-?? +?? 2
-?? 3
….)-
1
6
(1-
?? 3
+
?? 2
9
-
?? 3
27
…)
?=
1
3
-
4
9
?? +
13
17
?? 2
….
4.4 Expand in Laurent series the function ?? (?? )=
?? ?? ?? (?? -?? )
about ?? =?? and ?? =?? .
Solution:
Given that
?? (?? )=
1
?? 2
(?? -1)
About ?? =0, the Laurent Series is given by
1
?? 2
(?? -1)
?=
1
-?? 2
(1-?? )
=-
1
?? 2
(1-?? )
-1
?=-
1
?? 2
(1+?? +?? 2
+?? +?.)
?=-(
1
?? 2
+
1
?? +1+?? +?? 2
+?.)
Let ?? -1=?? ??? =?? +1 and
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Oct 31, 2024 Last updated |
Document Description: Laurant's Series for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation.
The notes and questions for Laurant's Series have been prepared according to the UPSC exam syllabus. Information about Laurant's Series covers topics
like and Laurant's Series Example, for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Laurant's Series.
Introduction of Laurant's Series in English is available as part of our Mathematics Optional Notes for UPSC
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Information about Laurant's Series
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Additional Information about Laurant's Series for UPSC Preparation
Laurant's Series Free PDF Download
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The content of Laurant's Series is prepared as per the latest UPSC syllabus.
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