Set-1,

Q6()
Dec.-16,
...
(1) ...(2)
Set-2,Q5(b)]
..13
|Oct.JNov-18,
-1).
STUDENTS
Q3(b) (z
of
Paper-l1, powers
ENGNEERING
(Model
nagatlve
+ and FOR
21(2-1)?
31(2-)
zero 4(z-1)'e,(2-1posltive 30RIIAL
is,
o (z-1)
denominator is,
1) of
(z.- of ALSN-ONE
3! seriea
Theorem of powers -1)-3)
powers
its (z-)
(z-1
2! a B
equating off2).
aboutz
=1 in 2e4e in
in offz) 2(2-))e SPECTROAM
Residue G-h7-3
offz) z-12
by pole f)=-1)e-3)1)
expansion i:fi)e, B(2-
and obtained orderexpansion - e-1)(e-3)
+
Expanslons second Given
function
is, z=A(z-3)
arc Given
function
is, Laurent'series e
of) series Let,
=z-1· =
f(2)
:(¢-1y
ie,
=0
a
z=1.is f{z) Express Consider,
Seriespoles 1z=1, Laurent z=‘+] Expand and
z=] .
SolutiOn
1 5olation ’
The
The
UNIT-2 The Q19.
Q18.
JNTU-KAKINADA
proceedngs.
convergence.,
LEGAL
METHODS
faca
of to
region LIABIE
STATISTICAL is
the gullty
find found
Also
Anyona
AND serles. 1.
= act.
VARIABLES z
about CR:MINAL
Laurent
(1), =0
cquation series
a (1-1)-I|<0
COMPLEX as isa
f Laurent book
in is
4e centre
(3) z is this
and about expansion, 2e. convergence of
from a XeroxPhotocopying
(2) 1 as
= is, circle
equations
scrics isz 1,1
1,z=
circle öff2)
= Given
funtion
is, of of
e (2-1=
0
singularity
Substituting f(z) Laurent radiusregion 7.Expand Suncáon
is,
i(z)= Given
for
Bxpand centre WARNNG:
Wherc, the the
Solution
: Fron
The The Then, Thus, Souston:
2.12
Q16.
 expansion]
binomial STUDEN
Using
EKINEE
(":
FOR
+) 3OURNSL
C:Xz
G(z-2)
+B(X2-2)+
1) QLL-IN-ON
z(z+1Xz-2)
C\z+ 5
Theorem z+1-2 cquation
(1), 3(z+)
C
B"
A, + as,
B()(z-2) +0 (-1-2)+0
A(z+1Xz-2) 2+3-Q+0+
(2+1)
c(2) expresscd
5 SPECTRO
Residue +0 (2), 2z
fraçtion, (2),
equation (2),
(0-2)
1) cquation cquation values
in' "3(
+1)
+ +lX7-2), be 2z 0
and
partial z(2+1Xz-2)
Expanslons
A(r+1)(-2) 3=A(1)(-2) 1) 2=B(-1)(-3) S=
B(- C(2)(3) can 2 2
+
z(z+1X:-2) -A(0
in in n C z+3 (3)
intoz+3 'z+3 0 =-1 =0+ 2 Substitutingz=26 B,
A,
z(2 3
cquation
Resolvingf) z= 0+3 z -1+3 B 1<]z|<2
+3- Substituting Substituting Substituting
Sories [z]<1Then,
UNIT-2
(h)
()
Dec.-16,
Sut:2,
Q6b)
procedings.
LEGAL
2.
(21> face
(1) to
LIABLE
<|z<2
ls
gultly
1
(1) found
<1
Anyone
(z
)
where a>.
CRIMINAL
series
isa
(1)\ power book
(2), equation
(2), =in tis
1)
cquation cquation z[zlz-2)+I(z-2)]
of
- 1) is, z(-2z+z-2) XorowPhotocopyi
B(3 - àn of) 3
B(1 values z+3 z(z+1)Xz-2)
z('-z-2)
in + in + z+3 z+3
3 3=A(3-3) z+3 z+3
z=l-3) series
z 3=B(2) 1-A(-2)4,B Given
function
is,
Substituting |-A(1
Substizuing Substituting I=w
z-
Let, Laurent f(zj
Expand fO)
= WARING:
The Solution:
Q20.
 Solution :
Q21.
ING: Substituting Resolving (ii)
Given1<|z<<2.
functionis, in
B=1 Obtain Then, |z|
l=A
1=A(-1+ i)(z+ Then,
ying =A(2+1) I fünction
above
into Laurent's equation >
(0) 2= 2
(2+2\(%+I)
+t
B()) 1) in
-1
2)\(2 1
-) 22
(3)
equation

equation + expansionforfz)=+2)2+1)
+1)
+B(- (z+2) B can (3);
z+2 2 be
can
2
of
this (2),
1
+2).(2+2)(2+1)A(2+ fractions,"
partial expressed be
book cxpressed
wmrLEA
1)+B(2+2) z+1 Dec.-18,
Set3,
5
is as,
a
NAL as,
VANIADLcO
(2).. ()
.. QG(b)
act
yone
Solution : Evaluate
Q22.f(z) ANU1AISIICAL
ound Substituting ’
.. Substituting ’ Applying function
Giyen1<jz<2
i0<z-1|<1
s; (i) ()
A=-1 Then 14<2Fot Substinuting
uilty 1=A(-1)1=A(1 B(2-1)1=A(z-2)
B=1l=B(1)
1-A(0) + Substituting
Paper-111,
(Modet equation
is. z= partial
LE -2) z=2 and1<|z| <2
1 7-3z+2 the l-A-2+
1=1(-) +B
(0) z-2
+ in in fractions (3)
the + values METHODS
to B(0) the B(z-1) becomcs, K+1
face above above 1 in
Q3(bj ofA equation
1)
AL to
cquation, the. and +B
equation, NovDec.-17, | regionthe in
s. above (: B (-2
[JNTU-KAKINADA
By equation
in +2) (2),
equation, binormial
8et-2,
(1),
..(2) QB(b)) series]
(3)

(i) (1) UNIT-2

Solution : Q23.
.0<z-1|<1.'Substituting
)1<<3
2>3(ü) Regionis, function
Givenis, for Find If1<|2|<2thenl<<z|a=lnd|<2, 1<<2 -
f2)
f)= Series
(i) (2-1Y+..-I
--[1+(¢-1)+
(¢-1y+ -1|<1
1< the (1+ Expansions
Laurant' 4,
z|<3 s (z-1) t-De-2) B
values
(1) +(¢-+(e-19+...
E [: in and
z|>3. series 1 equatiohResidue
Frpm
Nov./Dec.-17,
St-3,
Q6(b) of binomial (2),
f(2) TheórerH
=2
expansion]
-
R 4z,+3
..(1)()2|>3 (
can
Substituting Substitutingz=l
Substituting 1=A(-2) ’ Resolving
be <|z|<31<z|<3
As, I
Gobtained The 1=A(1-3) =A(z-3) 1
S2) = =2B 1=A(3
1
f)= 1)*z-3)
(2-2-3) equation
Laurent z
1 the -3) =3
as, corresponding +
series B(3
+ in
cquation
+ inB(z-
equation
(1)
B(1 into
S - -
expansion 1)
1) (z-1)2 partial
1) 3)
(3), (3),
values fractions,
of2),
in B
cquation
in .17
the -.(3.)
region (2),
2>
 Solution Represent
: Q24.z)=
Substitutingz= 4(-2)
Substituting 4z+3=A-3)(+)+B The function
Givenis, lz|1(i)() fE) =
VWARNING: Substituting
40)+ ze-3)(2+2) Resolving
=Ap-)0+2)+BS)3
B=1 +)+CO3-3)
B=1 43)+3
15 -5=0+0+10C
C=
+3 t-3)X+2) 7=0,3,-2.
4z+3 4z+3 possible
2-l-)
2a-)
15=0+ C- =A(-2-3)(-2+2)+
3=4-6)
+0+0 3=A f)singular
hotocopying +C-2)\-2-3) =-2 z (0-3)X0
z=0
into
B(15) 3 in
4z+3
zí2-3)(2+2)
in in Az- partial 2< 4z+3
cquation
+0equation equation +2)
points
|z<<3 ULN
3)X=+ fractions,
+B(2%+2) of/)
(1). (O),z(-3)(z+) 2)+ OctJNov-13,
Set-i,
Q8(bj
of (), (00+)+C(0X03} Laurent's
this B(- B(:)\+2)+Cele-3) are, () seriesin
book
+ C(
2)(-2 AINIADLCO
z k4>3.for
is -3)
a +) >3
CRIMINAL
-2) -
(1) HIVU

act. (ii)
Anyone 1z|>3(üi) SIAIIO
Substituting
Thus, 2<|z]<3
For
found IAL
<|z 2 --*-*+-6)*-)
quilly 2<|z,|z|<3 4z+3the
|<3. WeVUS
is. corresponding
LIABLE
-t-**B.) 2(z+2) 1
to
face UNIU-nAnIINAPA)
valucs
roceedings.
LEGAL
in
equation

..(3) (1).

UNII2
Solution :Q25.

Givenis, Obtain
funçtion
Equation
Substituting
=-1 .:B Resolving senes
Substituting Comparing Substituting’l=0+
1=A(-2+1=Az+2)
For A=10=4+1
l=4-1+2)(-1+1)+B
(-1+1)+
C(-1+2)
I=A C=1 expansion
Laurerit's (3)
ExpansSIons
H|> (0) z-1 B-1)+0 z (z+2)'
(z+1)
above 22
can
be
2 the coefficient
+
=-2
2) (z+ function 1 cxpressed
CTRUM e1 valucs
in'B(0e) quation (-2+1) in
) z+2
cquation(z+22+1
ana

of C + +B(z+1)+ ijto Kesiaue

of as,
4, in (1 y + partial
B cquation (2), B(-2+
(2), for
and f(z) ineoram
-1 C Cz fractions,
ALLHA-ONE in (2),
1)+C-2+2) +2) =
equation (2+2)(z+1)

(1),
JOuRNAL
[z|>2. in

FOR
INEERING

STUDENTS
[:
binomial
series]
Using
Dec.16,
Set-4,
Q6[b)

...(1)
. (2)
NG:
Solatibn :
Substituting equation
From
(4), equation
From
(3). 26. 2.20
Substituting Comparing ’
2C+A =12B+
C---) A+B=0’
A=-B Resolving function
Givenis, () Find
g
the
corresponding
2(-2E)-B=1
-4B-- =!B
=1
SB
equations
(e+2)+1)tC =
l=(A+l=A2+
CI=A2+4+

0 thie
1 z|<1
Laurent's
cqustion
conesponding
C-2B B)=+(2B 1)+
B+2Bz ()
serles
z+2,+! (3) (B| (1)
of into
this nd +Cz+2) 1<|z[<2
book values C+2C+A +
(4) coefficients 2+1 partial COMPLEX
f(z)
+
ls n0 in Cz
a in equation +2C fractions, tt2c+1)"
AL equation
on VARIABLES
(5), both |zl>2(it)
(2),
act. sides,
one
AND
ound
STATISTICALMETHODS.(JNTU-KAKINADA)
guily
is
BLE

to
face
GAL

ngs. Oct.ov.10,
Sot-3,
QS()

...(5) (3). . ;
(2) (1)

(iü>2
i) 1<Iz|<2(UNIT-2
i)
Solution: C27.

Comparing Resolving function

Givenis, Find Series
44-C
4B-D=0 =1
B+D=0 A+C=0 () the Expanslons
=z(A+ 2-12+4)
2z=A =(Az+ +4
z -1
equation Laurent's
[2|<1
còefficients
the
4A2
+
B)(?+4) (1)
-)e+4) 2
ee 1
and
M C)+2(B series
+ into 1<|z|<2(i) Residue
B=-DA=-C Bz+
on AztB.Cz+D
2-z
+ partial of
both + 4B(Cz f(z) Theorent
NE D) + + fractions,
sides, z(44 + C2D\
+ (2?-1)i22+4)
-C)+ D-1)
-C:-D |z|>2.(1Ë)
AL 4B-D
for
FOR
IIC
(isdel

Papor-N,
NTS
Q3(b)

|Oct./Nov.18,

2.21

Set-4,

(2).. (1).. C6(a))

(.6.) .(5.) .(4.) ..(3)
 1<|z[<2(i)
2.22
WARNING:
|z]<1 Substituting From Substituting Substituting
From
Xerox/Photocopying
equation4(-D)-D=0 equatiaa 4(-)-C-1
the - -SC=1
D=0 SD=0
cquatioa equation
corresponding(),B (3).4
04-T =- (4) =-( ()

this
of ..)---.) (0)= cquation
in COMPLEX
in
cquation
values 0
book 2-1
in (6), (5),
is
a
VARIABLES
CRIMINAL equation

(2),
act
Anyone AND
(ii3 ) Solrio Q28. |z|>2(ii)
STATISTICAL
fouind Subsituting Subsútutingz=3 » Resolving
0<-3|<3 Begionis, *0<|2-3|<3.
function
Givenis, fzFind
2--18-A(2-9)
7(3)-93)- 2-9%-18=A(z+ an>3
d =
7(0-9(0)
guilty
C=1 C(18)18
=
LIABLE
is z=0 equation 72'-9z-18
z-92
7z-9z-18
-9zLaurent's
the
b-*j). -
z-3
z+3 72' 72-9z-18
z -9%-38 METHODS
(o 18= - in
cquation 18=
equation
in z(-9) 2-9z (1)
face 3)(-3) into n
+
LEGAL (0 A(0)+B(0):+ B(2-32)+ partíal theserles
-9) (3), (3),
rsgions
(JNTU-KAKINADAJ
+B()¢-3)+
procaedings. ’AF2 C¢ fraction;
Xpansion
C(18) +3) 1z|
C)+3) >
(1).. 3
-3) (2)
. and of

() UNII
Solution Q29.
Singular
Point 2.2 7=2+B+1 ’
’ Bquating
0<z-3|<3
The then The Substituting
A DafineTYPES IEz-3<3, L¡urcnt
B=4
7=A+B+C Senes
f)--tne-9
singular Laurent
7.945 81 Z.zt3 th»
CXpansions
singular z-3 =3 z-3+6'
z-3+3z-3 2 series the coeffiicnt
series
point OF corresponding
expansion
'expansion
or
SINGULARITIES:
point 4 of ana
SPECTRUMsingularity z?
and
3 off) cquation
in
KeSIaue'
of) values
of isolated 2 .c., in I(laoram
ALLIN-ONE a (i.e.,
equation cquaion (3),
function 2 cquation
singular
ISOLATED-POLE
/z) (4) (2),
JOURNAL (4)
is in
defincd point. for0 4 the
in +.....for
|z|>3 the
region
POR as Give <-3] region
a 2|>3
ENGINEERING point one <3. 0:<|z-3|<3
<\2flor 3 can
where example OF be
ORDER oblaincd
Jz) can
STUDENTS fails for be
each. oblaincd as,
to
be
m.-ESSENTIAL
analytic.
as,
for
\2-3|< 3
-3|<3for(z (4..)
 singularity
off).
Exampie is Essentiai
Singularity
referred but Example
singularity.
removable z=o,
Singularity
RemovabBe Solution :
WARNING;of Since,
Xerox/Photocopying Li Q30. cannot
Example then Singular
Iff2) Here but Iffz) isPoint.
Isolated Sxample .
Az) to f) Refer Define
example.singularity
no
as is z Lr Define Three be theother
is an an 0
cists.I sin z f) is defined point Ifa ’
infinite isolatedinfinite is sin z single Only isolated At singularity
function
callcd is exist, removable removable z f)= z,
1 a valued Reróvable =0,andz=+3i,fE) at is
fünction with referred
function, 1 essential series removable then singulaiies these
z(2²+9) z+3 - within f)
funcion one three
function the OR as bas analytic, is fails
which Stigularib?
singuiarityexample. singularity
this z singularity singularity points. singularity
anthe
= singularity. off) isolatedcircle to
book then is which be
0
is is not are fails
around at
at analytic
a an offz. the definei znot is Dec.-s6,
Set-&,
Q1(g) z=0,z=3i,2= singular z
CRIMINAL essential point, andgivean and to z=, 0#
ois be atz
defned analytic the
zz, at Enown eSsential point point and =)
z=0 also
adt.
as at -3i. as off() z
Anyone there
=,
zerO.to it
Sofution :
fourd 12-2Let, funciion Since Q32.
non-isolated
singularity.
esscntial zeroto
Giyenis,
Consider, .Zeros function
Givenis, Sotution t Q31.
guity z=0 there (i) functions.
followingthe Find :.2) ’sìnu(e-a)=0i.e, polesThe
.. z=0ie, are function
Givenis,
FInd
LIABLE
is is are sinz.COSZ (z+1)sin z the
sinZ- z=0 TZN:
n=0,#1,±2 .... sin sin (-a)
a z=0 obtained -n;
to
removable no
negative is
nature has
a the
is
Tz
z=0
= offe) S)=cotz naturo
the
a double n=0, sin rz>zsin)?
face singularity by and limit (or) are
equating nr COs
LEGAL singularity.powers (or)(z-) obtained and
Iocation pole of 1,+2,
the z=a locatlon
procedings. offz). the above
of at ..... 0 by
z. denominator of =a
equating
slngularlties poles
andz Papor,
Q4(a)
Model the of
its
denominator singularitios
off2) o
of is
a

Solution
(a) : Q33. (iüi)
UNIT-2
function
Givenis, Find i.C.,Zeros Given Since
tanztan z+2
the
type z=is tanzl=
cos coS are
obtained
function z=2 the f)-(+2+1) Series
2 z number
Sin sin - is
of SinzcOS z
is,
an z-2 Expansions
zz=0 by
SPECTRUM singularity simple a equating
essential
40(2-2)*
120(z-2)*
ofnegative
-f(z)=(b)
62-2)1 sin
pole. the singularity.
powets
(*3)sn and
LLIN-GNE of denominatc Residue
the
.9.. function of 2ig-
(2-2) Theorom
JOURRAL of(2) 2) |)
is
to infinite. .......

zero.
FOR (b Soution
(a)
Q34. negative
ENGINEERNG negative
(b
function
Givenis, Consider, function
Givenis, (c) (c) Find (b) (a:.z=1
) powers
Thc Let, Given .. Sincc,
z=0 S)=,z=0 Givez=0,zfdentlfy
=1. the IdentYfy ldentify above z-!=
powers
isexists.S2) an singularity of
functiortis, therc
sinz is (2 has ) of
STUDENTS the example the the the a serics (z-I)"
removabie .0 2 pole 1). - an z.are
singularity singularity
singularity has cssential infinite
of OR of
for the order finile
singularity. Non-isolated funcion singularity number
(four) 4.
Nov./Dec.-17,
Set-2,
Q1(0) of of
of f(z) f(z)
f{z)= D¡c.-16,
Set-3,
Q1(g) of
f(z) = = 1crms at terms
singulariiy. z(z-1) e~ z
=
et*, at containing 0,
Containing
z= atz
2.23
a C. =0.
 nator
zero.to
Solution :
Q37.
WARNING: The function
Givenis, (C)
Find zero.to
ie., poles The Az) function
tho Givénis, IKesf)l,., (b
(z ofz) Residue'offz) residue has The
otacopying - pole z=0i.e.,sin
Resfël,.(Resje. a function
Givenis,
1)?= poles
can
and
B--0}:) off2) pole,
Resfl,-1
z=0, cosec
S)= z
Residue
off:) = Residue
(Res
0 be lim = offz)
determined resídue
at
z cos(0) lim fz)),..,=Coeficient
off)
at atz=a tn, sin
lcosz are v(2).
of
=0 z
|a)m-)--aK)
0) 0of
for order ) 2r, 1 obtained at
this of zero. is =0z
by atz=0is1.
book equating Dec.-16,
Set-1,
Q1(g) m"3' 3n... + by is
2)z-4)2 order given
isa equating of
CRIMINAL
denomi pole inby,"
is its Laurent'sexpansion
givea dencriinator
BCL.
Anyone as,
(tt
denominator
zero.to
found Solution:. Q38.
(ResAtz=-2
Residue ’
fe),... [Resfl,.-,(k-4
ResidueThe
z=2 At The function
Givenis,
guity
residue Z=-2, Find [JNTU-KAKINADA)
ls
A) 2
z+2=0, (e+2)
i.e.,-4=0 poles
z-2=0 Herc The vinOIVAL
VItITUUS
’
LIABLE tResidue
he
(-a)A) hoff) has
[Resf)l,,= ofA)
m= residue(z-1) z
(-1)
two (2-2) pole 2, 1,
to [Modal off2) a offz) is 1
faco (e+2) , =az
single
at
=0 carn and
at
=ele
*e 1.
atz*asecond
a (-
LEGAL be Paper1,residue 1)
(simple poles determined, the =0
eedings. z pole fornh
order
= Q4(a).|
2, of z=
pole) z f(z) pole
=-2. Dec.-16, 1 order
is by =-4) is of).
given e. pole
equating Set-2,
by, is
Q1(g)1 given
its
by,

Solution Q40. Residne

z=0 At Solution : Q39. UNIT-2
[Resf()),-0= L =(z-0)f(z)
[Resf:)),.o
The The function
Givenis, Find L-Hospital'
Applying
nle,s Límit (Resf()),,-a)f(e) The function
Givenis, Find
(Resf()),..= residue
(z-aMe) u pole th[Resf(2)),-
e =1. is residue, the Series
S) of
fz)
ofz=0
is,
=sinz
COSZ 2
resldue
cos
0-0sin
=
I-0 Lt 0-
cos (sinz)
i0
-0(sinz the
0
2’0
of sin
fz) +2cos z1+e*
residue
Expansions
Ofe)+1+ z'-2sin COS
+ dz form. at of
CTRIM 24 at of z
2=ais
givenas, f(Z) 0+
1+eze' +z(-sin
+ 1+ze* +e
(sinz =az z zCOSz and
tz + for Ite
eoRzsinz
= cos z
+ cósz) cosz) z simple Residue
+
Nov./Dec.17,
Set-3,
Qi(9) 0 cosz z) sinz
+
N-ONE
0+ cos zcosz) pole Q7(b) atz0.
Dec.-18,
Set-3,
Theorem
at z
z0. is
given
URNAL
as,
zeroi.e,to Solution : Q41.
FOR
The Where, Let, function
Givenis, atzFind
1.
ERING Herc, The
division,By
Buto(z), /2)),..
[Res residue z=1 (-1)' 0= pole z)= the
2)=z+5+ t- (Res n
z-]=0' offz)
[Res)),-,=0
residue (Resf(),.
is (2-1'(z-2X2-3)
Sz fe)),., and 4 off2) pole a z1 (2-2X(z-3) (2-1)
6+ (2-2X2-3)
is
DENTS 2-2z-3z+6 a obtained of.f(z) =
192-30Sz-25z+S2'-6z
30 =4-)i(e-1
-5z+6
=l atz=a of
order cos(0) (0°)sin I-0
(z-2X2-3)19z-30 u
for 4. by =
COsz \
(z-0) Z
mh denominator
equatingits (z-2)(23)(2-1)4 Sinz
order Dec.-16,
Set-4,
Q7(b) -0
pole coSz
z-0
sinz)
is 2.29
(e-l given
(2)..
...(1) by,
 fa),.(Res .
sides,
both
fa)l,.1-2(e-3)Substinuting
WARAING: (Res Differentiating Differentiating
)-1+8z-2)*-27(2-3² Diierentiating Substituting Substiuting Sotving
")= Conparing By Consider,
"-)=0+8[-2(2-2)]-27|-2(2-3) d2
-3*-2 A=-8,B-27 partial
-6-1,606101
Xerox/Photocopying J01 16 --16-3(2-2)*]+
54[-3-3)]
=48(2-2)-162(2-3
equation
=-16(z-2}+
54(2-3) )
=8[-2(2-2)]-
27[-2(2-3)] G-2Xe-3)
)=z+5quation 19-30 equations -34-2B
34
i9-30 -3019z
(2-2)(2-3) 48 above the +A+B= 19 the 192-30= fraction
above above 2B y:-3)(-2)
(*-3) (z-2Xz-3)19z-30
values =30 =-30 coefñcients
48(7) equzion cqustion (4) = nethod,
166 16 in eqiationz-2.2-3 (6) ofA and (4A(g-3)
+
equation (z-3? in A(z-3)+
B(z-2)
of with cquation and (5), B-3d
of
this 162 wib with B "and B(-2) +
book espect respect in 28
(1), respect (2), equation
is constant
a 16 to
CRIMINAL 'z, to
'z, to (3)
z, teras
. .(5) .3)
act. (6) on
Anyone off(z)
zeI0. 1o SolutioD 1
Q43. zero.
Sotution Q42.
Rof()l,...
found Theresidue ; The fupction
Givenis,residues Find The The function
Givenis,
Find
guiltly z=iand z=-1 and-i e i.e.,
Thz*-1, poles residue pais :
the the
is off(2) poles (z+ of O i.e=0
, a)
LIABLE isa poles
at
--0) off) has poles
z=-iáe poloff(z) 1(+1) (z+1)'(+) f(z) these
at e
pole are
are 2'-2z of at|=o and
lo z of obtained poles. i2) at obiained
face 1
a ar e z=0
r order - =0
fosimple = for of rosiduas
LEGAL
n 21,i by (z+1)(2 0-2e) n equating
by
order poles. and equating z'-2z order order Dct./Nov,-18,
Sot-1,
Q7T{b)
proceedings. of
pole -i. pole 4. its f(z)
denominator. +1) denominator
is given
is
given and'the
as, (1..) by,
to

Atz1UNII-2
z=i At
[Resf()),--(2-1)!d2
(Resf()),.
(-)e).
[Resfa,.,
Resfe)l,., Th¹ Series
Resf(2),.,4 residue
Expansions
of2) +2-1)-2
2(-1?
- ((+1?
I+2i -l-2 (/+1°((+)
(+1+2)2i [O-249) (-9 atz=0 (X2r-2)-(*°-22)2zdzd. d
and
-l-2i
PECTRUN
-_H21. (z+1'(z+)
(a-2z)
f) for
Residue
simple (e+9°(e²+i)! (z--1)f(O)
(?-2) 4
heorem i
LLAN-ONE pole
is
given
J0URNAL as,
At
z=
zero.f2)10 Soution Evaluate
Q42.
FOR [Res
The integral
Givenis, (i) 4 -1The
2'142i
NEERING |+1-ij=2
z*-1+
i.e.,
fz)=Let, : residues f()l,.-=
poles |z+1-i|=2
|Z+1+i|=2 (Resf(2)),*A
(Resa),..i1
[Res/(P)),.,=
L(2-a)f) z-1+2i
.z+2z+
are
and
Z-3 t
2i --|+2ior-
obtained 2z+5z+ 24245 z-3 1-24 of 1-2B42i(-21) 1-2i
2i(|--1+2i
2i(1-+-21)
(z+1'(z+i)(z-i)
(ti) I
(z-(-
the=-1+2iV2-4()(5)
is -2+-2 S J)
UDENTS pole
=0 -3 respectively. at 1-2 (2'-22)
I-1+21 and 2(1) by using poles 1-2i ))f()
i-2-12 H31
4
which
-l-2i 1-2i cquating dz
z (-22)
residue 1-2i
lies OctJNov,-18,
Sot-2,
Q7(a) where =-
2C1+20)).(z-3)
(2+
Z-3
1, (-H°-4-)1
2i. 3+1-2)(2-3)
(z+1+2)-2)z?+22+5
inside are
the denominator
the
theorem.
the
C
is and i
+1+2i
-1+2i
the poles *4- 20 -
circle, circle are i
2i-3 of
 C:<z-
1|=1 its
denominator Soirtiog : Evaluate
Q45.
ARNING:
The Cis integral
Givenis,
AE) poles the bounded
Xerox/Photocopying has (-}-0 ie, toregion
[Resf,..y z=-]-2i
Thus,
zero. of
triple
a equafion d+2+5dz=
by
theorem,
residueBy z+I+|=2
theorem,
Resid°e By
bounded
z-1|=1 eáz
[Biodel
þole (1) =4-2i is
at can by
Paper-41; =i+2} 21 -4i tbe
z+2z+s-2) J)dz J
of z z-1|=1 using n2.+) pole
this be
2ri =
= Q4a)1 where
obtained which
book residue (E-f1-2)%e-3) -i-)r
is
inside
lies
and Doc-16,
+l+2)-3) ik 2xresfa), - [sum
a Cis +2i\2+
1-2i) of
CRIMINAL by theorem. +2-+ 5 inside 2i the
equatíng Set-2, the resid1zes]
the
. . Q7(b)] region circle
act. (1)
Anyone Solution :
Q46.
found (Bter, Find
integral
Givenis, zl=3.
whereEyaluate
dz,
guilty
Ont polesthe Theorem,
Residue By
is m-3Here, a The
LIABLE Poles by
Z
and residue CoS z residue
to aresldues
nd dz=
f)dz J
face Resjhhe) OR -H-200-0)-2-0--1
2ri(-1)
=-2ri = -H-2ai~-~on~) -;2 **-z(oos2)
-
off)
LEGAL theorem
2ni 2ni -*sinzcosa) + -t.
Oct./Nov.-18,
Set-3,
Q7(n) of zma at
proceedings. f2)=-1Ý(2+2) Res (sum
for
2)),.n residue)of sin COsz-sinz z lz- norder
-zz sinz(-1) + sinz)
cos
2) sinz] - +(- osz] + pole
is
sin2)] given
as,

At
z=-2 At
citcle
"The z off2)
zero. to UNII-2
=
residue 1
esf:)1,.-1f:) Here, |z
r=Resfl),..ion-h
Theorem.
)"0)1| »"(-2-1
>r(et2z-1(z+2)fa),+2)f2)(Res » The Thus, Where, i.c., Thé Let, Wherc,
m residueoffz) 3z-2is Serles
=2, offa) f-)l,..,(Res f2)),..-
-a)fe)Res both poles
(2 (-3)? (-2)? z A)
Z=1,2=-2
= (z-1)= (z-1} offz) Cis Expansions
a= al th»
(2+2)
(z-)'
a isI has
SPECTRGNM 1 =az 4 2 poles. simple
pole apoles (2+ are
the
at circle
for 4 z=a 0, obtaincd
z pole of at (z 2)
pole a for Land
= order
z
=
+2) =0 |* and
2 |=
ALL-N-GNE of _imple z 2 and l =0 by 3 Residue
m =- cqgating
ordcr pole 2 =-2 z Theorem
lie denominetor
is is inside
JOARMAL. given given
as, as, the

FOR Soution :
Problem Theorem
Residue Q47.
ENGANEARNG integral
Givenis, For State From
answer z-1)(2-2) the Caúchy's
refer residue )d
STGDENTS z
Residue
-

Unit-I,
tz, 2ri = (3) J+4 (0+2)²
1'+40)
Q35, nhero
theorem. (Sumtheorem, (2+2)' |(2+2)()-e2)
(z+2)
Topic:Cauchy's C of
the is and residoes) (z+2)?
circl
evaluate
Residue j:-24= 2.33
 Atz=1 zero.to 2.34
Atz= 2

-From Res : Here, The Resf)],.,=0

’ Where, »ie, The Let, Where,
WARNING: 22-1)(z-2)7
(z-1)(-2)'
f),.,=-1
residue
Aa)1,.,=0 z=l z=lis az-1,z=2
[Res z=lz=2isapole j) (z-1l)
poles
Cithse
Cauchy's m=2=2, a =0,(2-2)²=-10 off)
z lies lies has
d| off) (z
olocopying Lu
outside outside simple
pole.
(z-1)(-2)?
polcs
-2)? are circl-2e)|z
residuce z-1-z(z-1)(0)-
z(1)
-2Ti -2ai
Zri 2zi - e-b0-t-1)
2-1)
tz=a
the order
C
of at
z
=
oblained
0=
VARIASLES
COMPLEX
AND
[0 (Suntheorem, for and l
(2-1)° circle. and by
+(-1)) of pole
a z= 2.
z
equating
of residues) 2 =2.
this - lies
of
book m inside its
is
order denoninator
a C.
CRIMINAL given
as,
act .Atz=2 zero. to Solution': Evaluate
Q48.
Anycne z=] At STATISTCAL
. off)are A)Let
LeThepoles =, Where, Givenis, circle
ir.tcgral
lound f2)l,.m-
The Here, (Res Tbe The Thus, Where,
The
f(:)),
[Res,= -[Resf)),.residue
, residue z has )
residue residye Cis zl=3
2-1, Aa),..-u-)la)[Res both 2 two
(sin r
)2ucoSR-2nz -
guily
2 is z=1;z=2
(2-1'(2-2) =0 the
(z-1)-2)
1
offz) (sin offz) off[z) the az-l=0,z-2 =0 (siz²
n+ using +cosnz)
rz (JNTU-KAKINADA)
| simple poles (slnnz' r
LIABLE
is dz off2)
=2. m e**cosTz')
(z-1)" polcs (z-1'(4-2) circle + (2-1)2-2)
METHODS
(sin at z=afor1
at e-l'(e-2) at at lie pole z=l at obtainedby (z| cosrz) resldue
10 (z-2) re² ez
z=2 z=a
inside T2)cos =3
face
LEGAL costa') + lis is-givenas, for and and
theorem.
(a-2) given pole a sin simple
a the z=1 z=2. denominator
cquatingits
sin (2-1) 47+ circlelz|=3 is where
esdings. as, of cos4t
)-(in (z-1)'(-2) e-af)) order pole pole
a

ce m is of
C
is given is
+ços') given order the
as, as, 2.

Solution : Q49. .:.

From
". Evaluate theorem.
residue
Thez=*in
nator
zero.to The LetJz), integral
Givenis, EvaluatY Cauchy's
r,=2r+1
= -
Res residue poles 4 2r+1 (I-2)(21()
(Resfa)1,., r(sin f)dz
(sin )|2r(-)-
are of12) [Nodel
(z-1)'
(-2) uzt (2-1)'-2)
n2 residue
z=
of
double
can '+dz e.
Paper-l;
cos +cos 2rci = cosa(1)-2n(U)sinz(1)'-
PECTRUN ofapoles be OR
g TLz") (Sumtheorem, 2n(0)
Tm-IT:u m
obtaincd Q4(b)
whare dz -dz dz= ) (1-2)
order which Oct./Nov.-18, | where 'm =4zi+r) (1 2ui=2ri(1
-
of -0
2ri residues) +
by Dac.16,
Sel-1,
Q7(b). C 41i (-1) sinr(1)y'
-AN-ONE pole lie
inside
cquating C isz=4by +)
(1 (2 + (r,
27+r)
is
Set-4,
:zj +2)
casx(1) +
(-a given = +
C its 1)
as, denomi Q6(b)] 4. using
JOURNAL
fe) z=
At
Ti

FOR
z=-T0 At (Resfal,.
NEERING
(Resf)),.T
Uletin)}

= 4 4n(T+)164r(-1)-4ti(-1)
16r -
4T
-Are+ 4rieM)
(-0-)'.e-2e(--n), 2.
L!
UDENTS +4ni(-1)
Ar(-1)
21 d
2.1 (-2n)
16 (-in)

["e=(-1); 1=
(2n)
ARNING: inside
z=0lies
theorem,
ResidueBy Solution :
Evaluate
Q50. 2e
[Res/)l,-=
(-ay) hThus, twhasA)
o
z(e-3)3 2e'dz z(z-3) )1,.,=3[Res Where, integral
Givenis,theorem. .
ocopying rjd J the
circlez2.
residue Cis
By
(0-3) 3
22H0(2-3) z(2-3)s0u. residue
simpleiz-3) [MRodel
=2nin) 2nisum = -2
2e.
2e off)
2e'. the
circle zz-3)z theorem,
poles Paper-N,
of ofresidue) at at z=2
this =az =0and z QA{e)} 2ri
=
book is where
given NovJDec.-f7, l4r (Sum
is z=3 4
FIMINAL
a C:<=2by of
by, residues
of
wiich Sst-1,
acl. only Residue offe)
QTb))
Anyone
Atz=- in
At Atz=0
found Residue
theorem
From z off()
zero to Solution : Evaluate
= Q51. dz
6=-2ri.
guity tMe)de (Resf(2),-
y(-m) in
[Resf),(Resf));-i=W0)Resf)),.,= The TheWhere,
(z-ae)residue
L polesThe Where, integral
Giventhoorem.
is, residue
uslng
is sinhz poles _inh
LIABLE n Cis
2i =2ri
z= is z off2) sinhz '
off) zero =
to (Sun cosh(-ir) -in) -cosh(in)
COS.T 0,
in, or
0 1 the
sohz-
9
face x(1- cosh(0) 0) atz - an are circle
LEGAL oftheresidues 1
mainteger r obtained
1-1) lie |24
where
is
inside
eedings. givei
cosh
by
atte i¢ as, the equating Nov.Dec,17,
Set-2,
a7(a) C
is
circle
imerior cosxl
= the
[zl=4 denominator clrcle
pols) lz
=
4

Solution: Q53. the Solation : Evaluato

Q52. .UNIT-2
circle
Thus, Where, Wher
. Th ’ f)=Let, Wherc, GivenCauchy'
tihsorem.
residuesEvaluate ’ Since, function
GivenCauchy'
is, s
two Jl=l the In Serdes
bot|=0,2=
h s)z=1,z=t 2-1)
(z-1)sinhz=0
n nn integral Residue 1=above
theorem,
residue
From
f)2-1)
poles sinhz Cis 6 is,
has is cothz [Model and
the zero
=0, cothz the expansion
residue Expênsions
oTsindz
two are, circle cothz
G.
. the,residue
poles ate l ot sinh cosbz Papor-l, f)de
poles an theorem.
CTROM lie simple intege. z=0 z dz z=0is 1
inside at Q4(b) -2zi(1) = and
z= =2. where 2ri the is Residue
the polea. 1,2=0. Nov/Dec.-17
thepole
(Sum coefficieat Nov./Dec.17,
Set-3,
Q7(a)
whore
IMONE circle C
: ofresiiues which C
Theorern
z Set-4, = of usil=1ng :
|=2. 2 lies
OURNAR Q7(a] using poles at insie

zero.to Solution:
FOR Evaluate
Q54.
From Thus, Where, Since The
NEERNG polTheesLet,f) Where, Givenis, theorm.
integral rasidue
using
=
Cauchy's 10)+0
fe)),.,=o R-[Res residue
y2)
the (Resresidue
R,
I7sinz le-I) R,=coth(1)
cosh(0)
(0-1) funcionfz) lim =
=Jimcoshz] si-nhz - f)),..- offz)
coth(1) liml(e-1)E-l)
of2)
C
is dz sinz' zy z-!fcoth z co Shdz = offz) =
=(Res
2Sinz the residue coshz (z-a)f),Aa)
TUDENTS are circle dz cosh(0) coshz cothz sinh z
fe)),.,= atz=
dz dz2ni(sumof at lin
obtained = = -2ri(R, = is
+ + pole }
z|= |
2ni(coth(1)-1)
,where 2ni(coth(1)
2ri(coth() théorcm, [: sinh(0) sinhz,. of
coshz (z-aA)) is1
cosh(0) z
= =0 form,
the lin[(z-1) given
by 1 + 0
equating Qct./Nov.-13,
Sot-4,
Q7{a)
isC
R,) residues) given
is as,
the - (=) + =1, f2))
1)
its circle sinh(0) as,
denominator 2..37
(z|=1 =
0)
 (1) circlein 2)
to Model
Paper-N,
Q4(b) shown ..
2.41
..(4) (4),
integration semi residues)
cquation as
of R
comprising
+
to of(Sum
in residues] -R
valucs contour from to,
contour Figure’0 cbanges|ni STADENT
corresponding of axis ()dz =
|re)da
(Sum
of From
rcsidue
theorem, rcal (1)
From
equation
(1), method
x'dx closed
2zi with
equation=
integral
Given
is, Sf)dr ENGINEER
the a -’
evaluate! is along Then,
`ubstituting the C
Where,
R
below.
R figure As
Use Solution
: Let, C,:|z|=
FOR
Q58.
1009RRAA
..(2) ...(3)outside
the
circle
C, ...(1) its
semi-circke equating f(),- ALL-NCNE
ineorem
[Res (2),
a by lies equation
f)),.,=
(z-a)
L
residues)f2)l,-+as,
KeSIUe
of obtained is.given SPEIRUM
consisting
R cirle) it
to circle) theorei,
of 1: as
integral, fromR the exist in
UTIU be (Sum [Res =.a (3)
contour can the outside not equation
f:)],.,
KpaISIOrIS
residue2ri
contout diameter ore, inside 2ri atz = does
f(z) = f2)
the 2*-i(lies
Cauchy'dzs
poles2=(lies of Restz)).~,
+1
=0 Substituting R’o,
As
the is of denominator
zero
to iLe.,?
Se) Jra)de residùe
Res [Res
"C"and poles simple
SUIs
Consider R
Where From
radius The The The
UVIE the
of
)'2+
h)X-i) -By2ib) (3),;
equation
-3
1+x face
LEGAL
procoedings.
(o
ta'Xe+10) in (1),
ib) välues atb cquation integration
+ as,
xb corresponding writen
(ib) +a° {Resf(,,2Ho-b) la+bEa-b) in fuaion
(ib)' (4) Contour be
equstion can LlABLEto
the GiveD
integal
is, integrei 1+x* tven
Sutshing Substtuing iby en is
Evaiuate above ic guilty
fa)
So|ation: Where, found
The
Q57. Ayoss
..
(2) radiusfigure. denominator acL,
the as, ÇRiLSINAL
C,of the in given (z+iaXz-
iaX=?+6')
ie
C: circlein ib is -a
2ia(6²
(ia+iaX(ia)+b)
within shown equating andz= pole a
semi is
[Resf()),..,=
t(-a)f()
simple book
theorenm, residues Ras
of becomes, by ib ia (z+ia)(z²
+
b²) this
consisting =- z=
-Rto obtained for WARNING:
Xerox/Photocopy
of
of Figure Le,
(+ z poles pales,
a)+b=0
residue (Sum ib, a
from (2) 1= simpBe (Resf(2)),.
-io) 2i(a-b')
contour equation are two z (ia
Cauchy's -2zi =-ia, at
axis off(2) only off(z)
4
z has z-plane.
the real z=ia, these, residue
Prom Where, Therefore, poles f)
Cis with off()
to
zero,
along of of
The Out half The
ia
z=
At
"R upper z=
ibAt
 WARNING: Substituting
The The
Photocopying (20n',2 resicue f) poles
the
valuc -16 has ii
(2o'124(ei?=16a1
-24aj'-2ait48{ai)'
16a)-12a)»%2a1
+32%0)
-24ai) off=) taO of)
of -{(ai (ai+ third are
r,
in
l(e+an'3+aij'(22)+(z+ai'
ai'3(ait+ (2)-6(+ane z=a
al
obtained
of equation (2ai ai)'(2ai) order
this (2a)2 for polcs
book
(2),
-3ai' -el'G+au-a) by
is -12%a) (a)'(ai 2ai order z=ai at cquating
a +t
CRIMINAL (ai+a)°(2) pole
(ai+ai) a'+ (2+ai)² andz<-ai
is
denominator
6(ai given
act.
Anyona +-
ai))6(ai+ as, Qut
ofj2)
of
found ai) these, to
(ai)'-6(ai+ zero.
gulty only
is one
LIABLE pole
ai)
'ai] z=
1o a
face lics
LEGAL in
the
ceedings, upper

half
of
the
z-plane.

'Solukion :
ofihe.z-plane. offz)
zero to along
R as9.
theorem,
residue
From theorem.
Ae)l,-(-a)f) The Out .) ic, The Then, withWhere, integral
Givenis, Evaluate
residuc of z=i, 2i poles (+1)(2?+4)f)=Let,
these, + (?+ equation -R real
axis Cis
has = off()
(+4) 1)
of2), contour =
only simple 4 - (1) from 2ri
arç 2ri
atz=a the obtained (Sum changes Figuro comprising
-R (Sum
CTRGM poles poles to
0.
ofresidues). 10,
+Rofresidues) ,
for z= at
simple z by shown
as
of
dx
i,2=z=2i,2i1, = equating +R
semi using
NONE
lie ’X incircle
pole in Figure.
is, the denominator
C, residue
givenupperhalf z|= .(1)
URNAL
-2i z (..2)
as, |Atz=i
z=2i At
FOR Evaluatod.
Solution : Q. ", (2-)f()
,Res
f2)),.F
RING Substítuting (Res (z+)z-X:
integral
Givenis, Lt. +4)zs
theorem,
residue
From Foreven
A-)1,.y5 6i (z+)
functions the 3i +1)(z= (z²
ENTS values <(2)*+1)(2i+
(4 21) +4)
u
the of
+)4/2, (z4+1)(z
(25 (2-21)f() (2n
1 definite r, 4i (?
andr,
(-3)4(-1) (2-2)2²
(z+2iXz-21) +4)
integral
in (41) 2)+
cquation (22)(-1+4)
can 31
be
written (2),
i)..
as,