1 chapter 1 complex numbers from schaum's complex
TRANSCRIPT
1
Chapter 1 Complex Numbers
From Schaum's Complex Variables
2
3
4
Example 1-1: Roots of zn = 1 when n = 6.
Using nj
n e/2 and roots : 1,,0,
/2 nkenkjk
n K , then
since 3/6/2
6
jjee , roots are given by 5,,0,
3/6/2
6 K keejkkjk .
They are located on the unit circle separated by 60º (or /3).
Example 1-2: Find cube roots of -j8.
Since 2/3
28j
ej , roots 2,1,0,2
3/26/ keez
kjj
k
.
They can be given in terms of 3/2
3
je as follows:
2,1,0,30 kzzk
k .
5
Chapter 2 Functions, Limits and Continuity
6
7
8
9
Example 2-1: Let f(z) = z + z-1, z0 ( }0{\Cz ) then
222222;;
1)(
yx
yyv
yx
xxu
yx
jyxjyx
jyxjyxzf
.
Using polar coordinates z = rej yields
Example 2-2: Single-valued functions vs Multiple-valued functions
sin)(,cos)();sin(cos)sin(cos)()( 1111 rrvrrujrjrerrerefzfjjj
10
(a) w = f(z) = z2 is a single-valued function.
(b) zzfw )( is a multiple-valued function (two-value).
For example, 11 ; j1 ;
8/94/9)24/(
8/4/4/
222
2221
jjj
jjj
eee
eeej
If choosing 0 arg z < 2 to be "principal branch", then 8/2 j
e is a "principal value".
Note that the Riemann surface of zzfw )( has two sheets (0 arg z < 2and 2 arg z < 4), the origin is
the branch point and arg z = 0 is the branch cut.
(c) w = ln z = ln re j = ln r + j( +2k), k = 0,1,2,…
If define principal branch to be
),0(,ln rjrzLn ,
then the origin and = is the branch cut and the origin is the branch point.
Example 2-3: Show that )1ln(sin 21 zjzjz
jzzujuezuzuzu ju 221 1sincos;1cos;sinsin
zzjzjujzzju 122 sin)1ln(1ln
Example 2-4: Show that if f(z) = jz/2 in the open disk |z|<1, then 2/)(lim1
jzfz
Note that for z in the region |z|<1, 2
1
222)(
zjjzjzf .
Hence, for any z and any positive number ,
2
)(j
zf whenever 0 < |z-1| < 2≤
Example 2-5: Consider f(z) = z1/2 where z given by ej and ej(2
1lim)(lim 02/)(
00
jjeezf
; 1lim)(lim 2/)2(
00
jjeezf
Hence, the limit does not exist across the branch cut, which means “discontinuous”.
Example 2-6: Show that z
z
z 0lim
does not exist.
1limlimlim0
000
x
x
jyx
jyx
z
z
xyxz
; 1limlimlim0
000
jy
jy
jyx
jyx
z
z
yxyz
Example 2-7 (Transformation): Let w = f(z) = z2; u = x2-y2, v = 2xy.
(1,2) in z-plane is mapped to (-3,4) in w-plane. (1,0) is mapped to (1,0) and vice versa.
u = c1 in w-plane is mapped to c1 = x2-y2 in z-plane, and v = c2 in w-plane is mapped to c2 = 2xy in z-plane, as
shown below.
11
Likewise, x = c1, c1≠0 in z-plane is mapped to u = c12-y2,
v = 2c1y, i.e., u = c12-(v/2c1)
2 parabolic curve in w-plane,
and y = c2, c2≠0 in z-plane is mapped to u = x2- c22, v =
2c2x, i.e., u = (v/2c2)2 - c2
2 parabolic curve in w-plane.
If c1 = 0 or c2 = 0, then x = 0, y = 0 is mapped to (u = -
y2,v = 0) and (u = x2,v=0), respectively.
For example, x = 1 is mapped to u = 1-(v/2)2 and y = 1 is
mapped to u = (v/2)2-1.
Exercise 2-1 Repeat example 2-7 for f(z) =
(a) z-1 (b) (z+1)-1 (c) (z-1)(z+1)-1
(d) (z+1)(z-1)-1 (e) (2z-1)(z+1)-1
Exercise 2-2 Let f(z) = ln z, then draw the image of |z|=c1,
and = c2 in w-plane. (HINT: use polar form of z.)
-6 -4 -2 0 2 4 6-5
-4
-3
-2
-1
0
1
2
3
4
5
u
v
x=1 x=1.5 x=2
y=2 y=1.5 y=1
12
Chapter 3 Complex Differentiation
13
14
15
Theorem function f is an analytic function Cauchy-Riemann equations hold
Proof (from Churchill's book)
Necessity ():
16
Sufficiency ():
17
Re z ≤ |Re z| ≤ |z|; Im z ≤ |Im z| ≤ |z|
18
Example 3-1: Let f(z) = z2, z, then
zzzz
zzzzz
z
zfzzf
zzz22lim
)(2lim
)()(lim
0
222
00
Example 3-2: Let w = f(z) = |z|2, z, then
z
zzzz
z
zzzzzz
z
zzz
z
w
22
.
Now, consider z = x+jy,
Let
zzz
wyx
zzz
wxy
then,0,0
;then,0,0
;
which are not equal in general except when z = 0. Thus, f(z) is not analytic.
Example 3-3: Let f(z) = z2 = x2 – y2 + j2xy, then
xyyxyyxx vuvuxvyuyvxu ;2,2,2,2 AND zyjxjvuzf xx 222)(' .
Example 3-4: Let w = f(z) = |z|2 = x2 + y2, then
0at except ;0,2,0,2 zvuvuvyuvxu xyyxyyxx . (differentiable but not analytic)
Example 3-5: Let w = f(z) = (1+z)/(1-z), (a) find dw/dz (b) determine when f(z) is non-analytic.
(a) Method 1: using the definition of derivative yields
200
000
)1(
2
)1)(1(
2lim
)1)(1(
)1()1)(1()1()1)(1(lim
)1)(1(
)1)(1()1)(1(lim1
1
1
1
lim)()(
lim
zzzzzzzz
zzzzzzzz
zzzz
zzzzzz
z
z
z
zz
zz
z
zfzzf
dz
dw
zz
zzz
Method 2: using the differentiation rule yields
22 )1(
2
)1(
1
1
1
1
1
zz
z
zz
z
dz
d
dz
dw
.
(b) Since f'(z) exists everywhere except at z=1, f(z) is analytic except at z = 1, which is a "singular point" called "pole".
Example 3-6: Prove that (sin z)' = cos z.
(Method 1) zz
zz
z
zzzzz
z
zzzz
dz
d
zzzcos
sincoslim
sinsincoscossinlim
sin)sin(limsin
000
.
(Method 2) Note that sin z = sin (x+jy) = sin x cosh y + jcos x sinh y, then
xyyx vyxuvyxu sinhsin;coshcos , and
zjyxyxjyxjvuzf xx cos)cos(sinhsincoshcos)(' .
Example 3-7: Find f'(z) when w=f(z) = z tan-1(ln z).
Let s = ln z, then z = es and w=f(z) = estan-1s, thus,
2
1
2
1
)(ln1
1)(lntan
1
1tan)('
zz
zs
ese
dz
ds
ds
dw
dz
dwzf
ss
.
Example 3-8: In polar coordinates, C-R equations are given by rr vur
vr
u 1
;1
.
Now, consider )sin(cos1111 jrerrez jj ;
22
sin1;
cos1
rvu
rrv
ru rr
Using rr
jjvuezf )(' yields
22
2
22
1sincos)('
zr
e
rj
rezf
jj
.
Note also that z = 0 is a simple pole.
Example 3-9: Consider w = sin-1z. Since z = sin w, dz = cos w dw, and dw/dz = 1/cos w.
Note that 2122 1)(sinsin1sin1cos zzww , thus 21 1/1)'(sin zz
.
19
Chapter 4 Complex Integration
20
21
22
23
Proof of Cauchy-Goursat theorem
Example 4-1: Evaluate C dzz from z=0 to z=4+j2 along the curve C given by (a) z = t2+jt, (b) the line from z=0 to z=j2 and
then the line from z=j2 to z=4+j2.
24
Example 4-2: Let f(z)=y-x-j3x2, evaluate C dzzf )( from z=0 to z=1+j along the curve C given by (a) the line from z=0 to
z=j and then the line from z=j to z=1+j, (b) z = t+jt
(a) Ca = Ca1 (0->j) +Ca2(j->1+j); On Ca1 : x=0, dx = 0, dz = jdy, y : 0-> 1, 2
)(
1
0
jjdyydzzf
C ;
On Ca2 : y=1, dy = 0, dz = dx, x : 0-> 1, jdxxjxdzzfC
2
11)31()(
1
0
2;
22
1
2
1
2)(
jj
jdzzf
C
(b) On Cb, z = t+jt -> dz = (1+j)dt, t: 0->1, x=y=t, jdtjtjdzzfC
1)1(3)(
1
0
2
Example 4-3: Let f(z)=z2, evaluate C dzzf )( from z=-j to z=j along the curve C given by (a) half circle of radius 1, (b) the
line from z=-j to z=1-j, then from z=1-j to z=1+j and from z=1+j to z=j.
Example 4-4: Evaluate C az
dzwhere C is any closed curve C and z = a is (a) outside C, (b) inside C.
(a) Since f(z) = (z-a)-1 is analytic everywhere except z = a, f(z) is analytic inside and on Ca. Hence, the integral becomes 0.
(b) Deform the contour Cb to be a small circle centered at z = a and radius , then let z-a = ej with 0≤≤2,
22
0j
e
dej
az
dzj
j
C
Example 4-5: Evaluate
,...4,3,2, n
az
dz
C nwhere C is any closed curve C enclosing z=a.
Using the same approach used in example 4-4 yields
2
0
)1(12
00dej
e
dej
az
dz njn
jnn
j
C n
Example 4-6: Show that for C: z=3ej (0 ),
8
33
12
C dzz
z.
25
Chapter 5 Cauchy's Integral Formulas
Proof
26
For proof of (5.2), see Churchill’s Complex Variables and Applications, Chapter 4, Section 51.
Example 5-1: Let C: |z|=3, evaluate
(a) ,23
cossin2
22
Cdz
zz
zz (b)
C
z
dzz
e4
2
1.
(a) Since z2-3z+2 = (z-2)(z-1), (z2-3z+2)-1 = (z-2)-1 – (z-1)-1. Then use (5.1) with f(z) = sin z2+ cos z2:
4))1(1(21
cossin
2
cossin
23
cossin 2222
2
22
jjdzz
zzdz
z
zzdz
zz
zz
CCC
(b) Use (5.2) with n = 3, a = -1, f(z) = e2z, f’(z) = 2e2z, f’’(z) = 4e2z, f(3)(z) = 8e2z
22
4
2
3
88
!3
2
1
e
je
jdz
z
e
C
z
Example 5-2: Use Cauchy's integral formula to show that for C: |z-a| = r (r is a constant),
C nn
njdz
az 10
121
Use (5.1), (5.2) with f(z) = 1, then
C
jdzaz
21
AND
C nndz
az1,0
1(since f(n-1)(z) = 0, n>1)
Example 5-3: Suppose f(z) is given in terms of the following power series:
0
2
210 )()()()(n
n
n
n
n azaazaazaazaazf LL
Show that !
)()(
n
afa
n
n .
27
Chapter 6 Infinite Series
28
29
30
Proof of Taylor's Theorem
31
Proof of Laurent's Theorem
32
33
Convergence of power series -> Cauchy-Hadamard Formula
Theorem (Radius of Convergence) Consider a power series given by
0n
n
n za . Suppose that the sequence |an+1/an|,
n=1,2,… converges with limit L*. If L*=0, then, R=∞, which means power series converges for all z. If L*≠0, then
1
lim*
1
n
n
n a
a
LR .
EX Consider
0
23
)!(
)!2(
n
njz
n
n,
4
1
)22)(12(
)1(lim
)!22(
)!1(
)!(
)!2(lim
)!1(
)!22()!(
)!2(
lim22
2
2
2
nn
n
n
n
n
n
n
nn
n
Rnnn
.
Thus, region of convergence (ROC) : |z-j3|<1/4 (centered at z = j3).
EX Consider
0
!
n
n
nz
n
n,
enn
n
n
n
nn
n
n
nR
n
nn
n
nn
n
n
n
nn
11lim
)1(lim
)1(
1
1lim
)!1(
)1(!lim
11
.
Thus, region of convergence : |z| < e (centered at the origin).
Exercise Find the radius of convergence of
0
2
)!2(
)2(
n
n
n
z(=cosh z).
Example 6-1: Let 32
2j
nzn , check the convergence.
3lim jznn
; Need 0220 ,2
32
3;,,0 nn
jn
jnnzz n
00 ,
2nnn
34
Example 6-2: Find Taylor's series of the following functions about the origin.
(a) z
zf
1
1)( (b)
21
1)(
zzf
(c)
zzf
3
1)( (d)
21
1)(
zzf
(a)
,...1
6)(;
1
2)('';
1
1)('
4
)3(
32z
zfz
zfz
zf
Thus, f(n)(0) = n!, and
0
21)(n
nzzzzf L .
Example 6-3: Find a series representation of e z, cos z, sin z (about z=0)
Let f(z) = ez, then f(n)(z) = ez. Thus, f(n)(0) = 1, and
0
2
!21
n
nz
n
zzze L .
Now, using cos z = (ejz + e-jz)/2, then
0
22
00 !
)1(
21
!!2
1
2cos
n
nn
n
n
n
njzjz
n
zz
n
jz
n
jzeez L .
Likewise, using sin z = (ejz – e-jz)/j2,
0
123
00 !
)1(
6!!2
1
2sin
n
nn
n
n
n
njzjz
n
zzz
n
jz
n
jz
jj
eez L .
Example 6-4: Find a series representation of f(z) = sinh-1z (about z=0)
Use 21
1)('
zzf
, then find Taylor series of
u1
1and find indefinite integral of the Taylor series.
Example 6-5: Find Laurent's series of the following functions.
(a) z
ezf
z
)( about z=0 (b) 3
1)(
z
ezf
z
about z=1 (c) zz
zf
2
1)( about z=0 and z=1
(d) 21
1)(
zzf
about z=j (e)
22 3
1)(
zzzf
about z=0, z=-3
(a) Since
0 !n
nz
n
ze ,
{ 44 344 21L
Analytic
2
Principal
0
1
621
1
!)(
zz
zn
z
z
ezf
n
nz
(b) Let u = z-1, then
00
1
!
1
! n
n
n
nuz
n
ze
n
ueee ,
4444 34444 21L
4444 34444 21Analytic
2
Principal
230
3
3 !5
)1(
!4
1
6
1
)1(2
1
)1(
1
)1(
1
!
)1(
)1()(
zz
zzze
n
ze
z
ezf
n
nz
.
Example 6-6: Find Laurent's series of the following functions. (Essential singularity)
(a) 2
1cos)1()(
zzzf about z=-2 (b)
zezzf
/12)( about z=0
35
Chapter 7 Residue Theorem
36
Alternative formula for residue calculations at simple poles
Suppose z = a is a simple pole of f(z) = p(z)/q(z), where q(z) = (z-a)r(z), then
)(
)(
)(
)(
)(
)()()()()(Res
ar
ap
zr
zp
zq
zpazzfazzf
azaz
azaz
But q’(z) = r(z) + (z-a)r’(z) and q’(a) = r(a), thus
)('
)(
)('
)()()()(Res
aq
ap
zq
zpzfazzf
az
azaz
Example 7-1: Evaluate the residue at the poles of the following functions:
(a) z
ezf
z
)( (b) 3
1)(
z
ezf
z
(c) zz
zf
2
1)(
(d) 21
1)(
zzf
(e)
22 3
1)(
zzzf
(Compare results with example 6-5!)
Example 7-2: Evaluate the contour integral C dzzf )( when
(a) z
ezf
z
)( , C: unit circle about the origin (b) 3
1)(
z
ezf
z
, C: unit circle about z=1
(c) zz
zf
2
1)( , C: circle of radius 2 about the origin (d)
21
1)(
zzf
, C: unit circle about z=j.
(e) 22 3
1)(
zzzf
, C: unit circle about z = -3
37
Evaluation of Improper Real Integrals
(i) Integral of the form
)(
)()(;)(
xq
xpxfdxxf , where both p(x) and q(x) are polynomial functions with the
degree of q(x) greater than that of p(x). Also, q(x) does not have zeroes on the real axis.
Can be found as
R
RRdxxf )(lim . If the limit exists,
dxxfdxxf
R
RR)(P.V.)(lim ,
where P.V. denotes “Cauchy’s Principal Value” of the integral. This
can be evaluated using Residue theorem as
0)(lim if )(lim
)( Res2)()()(lim)(
R
kR
CR
R
RR
kzzCC
R
RR
dzzfdxxf
zfjdzzfdzzfdzzfdxxf
Note that only Residues of poles in the top (upper) half plane contribute to this integral as shown in the figure.
Example 7-3: Evaluate the following integrals:
(a)
21 x
dx (b)
0 24
2
45
12dx
xx
x
(a) Since
jz
C jz zj
zj
z
dz
2
12
1
1Res2
1 22, and 0
1||1lim
22
RCR R
R
z
dz
R
,
21 x
dx
(b) Since f(x) is even,
dxxfdxxf )(
2
1)(
0.
Also, z4+5z2+4 = (z2+1)(z2+4) = (z-j)(z+j)(z-j2)(z+j2). Thus
24
3
22
)1)(2(
12
)4)((
122)( Res )( Res2
45
12
45
12
2
2
2
2
2
224
2
24
2
jjj
zjz
z
zjz
zjzfzfjdz
zz
z
xx
x
jzjzjzjzC
Note also that 04||5||
)1||2(
45
12lim
24
2
24
2
RCR RR
RRdz
zz
z
R
.
Therefore, 445
12
0 24
2
dxxx
x.
(ii) Integral of the form
)(
)()(;
sin
cos)(
xq
xpxfdx
ax
axxf ,where both p(x) and q(x) are rational functions with
the degree of q(x) greater than that of p(x). Also, q(x) does not have zeroes on the real axis.
Since
R
R
jaxR
R
R
Rdxexfaxdxxfjaxdxxf )(sin)(cos)( and ejaz=eja(x+jy)≤e-ay is bounded in the upper half
plane y≥0, the approach mentioned above can be used. (See more details in Churchill’s book)
Example 7-4: Evaluate the following integrals:
(a)
221
3cos
x
xdx (b)
dx
xx
xx
22
sin2
(a)
33
3
3
2
3
2
3
22
3
22
3
2223
221
Res21
ejej
jz
e
jz
ejj
jz
e
dz
dj
z
ej
z
dze
jz
zjzj
jz
zjzj
C jz
zj
.
38
Thus,
C
zj
edzezf
x
xdx3
3
22
2)(Re
1
3cos .
(b) jj
jz
jzjz
jzC
jz
ejejj
ejj
jz
zej
zz
zej
zz
dzze
)1(
11
)1(2
12
22Res2
22
1
1
21
2
.
Thus, 1cos1sin22
Im22
sin22
ezz
dzzedx
xx
xx
C
jz .
(iii) Definite Integrals involving Sines and Cosines
39
Inverse Laplace Transform
which is called a “Bromwich Integral”, where P.V. denotes the Cauchy’s principal
value.
40
Jordan’s inequality : RdeR /
0
sin
41
Examples
(a) F(s) = 1/s ; 0)(1/Res)(0
ttuzetfzt
zQ .
(b) F(s) = 1/(s+1) ; tzt
zezetf
)1/(Res)(
1(s-shift)
(c) F(s) = 12/(s3+8), z3+8 = (z+2)(z2-2z+4), thus, poles at 31,2,31 jj
23
12
3
12
8
12Res)(Res
223
tz
k
zzk
k
zt
zz
ztzt
zz
zt
zz
k
kk
kk
ez
z
z
z
e
z
e
z
ezFe
.
tteetfj
ezFeezFetttjzt
jz
tzt
z3sin33cos)(;
2
31)(Res;)(Res
2)31(
31
2
2
m
(d) F(s) = e-s/s ; )1()(2
1
2
1
2
1)(
1
)1(
tuudzz
e
jdz
z
e
jdz
z
ee
jtf
C
z
tC
tz
C
zzt
(delay 1)
(e) F(s) = e-2s[(s+1)2+25]-1
)2(5
)2(5sin)(
5
5sin
225
5151515125)1(Res
25)1(Res
25)1(2
1
25)1(2
1
25)1(2
1
25)1(
)2(55
)51()51(
251
251
2
2
2
2
2
2
2
21
tute
ue
j
e
j
ee
jj
e
jj
e
z
e
z
e
dzz
e
jdz
z
ee
jds
s
ee
js
e
tjj
jjz
jz
z
jz
C C
ztzztj
j
ssts-
LLLL
Partial Fraction Expansion
(a) Distinct Roots (Simple poles)
Consider f(z) = P(z)/Q(z) where Q(z) is a polynomial of degree n given by anzn+ an-1z
n-1 +…+a0 having distinct
roots 1, 2, … n. Want to express f(z) as follows:
n
k k
k
n
n
n
k
kn
z
c
z
c
z
c
z
c
za
zP
zQ
zPzf
12
2
1
1
1
)(
)(
)()(
L ; ck : constant
Evaluating contour integral 1
)(2
1
Cdzzf
j where the contour C1 encloses only one pole, 1, of f(z) yields
1)(Res)(2
1
11
czfdzzfj zC
.
It follows that )(Res zfckz
k
.
Example Consider
21
3
121
3
13
1
29103
1)( 321
2
23
2
z
c
z
c
z
c
zzz
z
zzz
zzf
3
1
9/30
9/10
)3/5)(3/2(3
19/1
)2)(1(3
1)(Res
3/1
2
3/11
zz zz
zzfc
12
2
)1)(3/2(3
11
)2)(3/1(3
1)(Res
1
2
12
zz zz
zzfc
15
5
)1)(3/5(3
14
)1)(3/1(3
1)(Res
2
2
23
z
z zz
zzfc
42
(b) Repeated Roots (Higher order poles)
Consider f(z) = P(z)/Q(z) where Q(z) = (z-)mR(z), i.e., is a pole of order m. Want to express f(z) as
rest The
)(
)(
)(
)()(
2
21
m
m
mz
d
z
d
z
d
zRz
zP
zQ
zPzf
L ,
where “the rest” represents the group of other terms associated with other roots of Q(z) (poles of f(z)).
Evaluating contour integral C
dzzfj
)(2
1where the contour C encloses only one pole, , of f(z) yields
11
1
)()!1(
1)(Res)(
2
1dzfz
dz
d
mzfdzzf
jz
m
m
m
zC
.
Likewise, taking
C
dzzfzj
)(2
1 yields
22
2
)()!2(
1)(
2
1dzfz
dz
d
mdzzfz
jz
m
m
m
C
Following the same approach, one obtains
z
m
km
km
k zfzdz
d
kmd )(
)!(
1
Example Consider 1)1(
1)( 1
3
3
2
21
3
2
z
c
z
d
z
d
z
d
zz
zzf
11
10
1
1
0
2
3
z
z
zd ; 1
1
10
)1(
1
1
2
1
1
)!23(
1
0
2
2
0
2
2
zzz
z
z
z
z
z
dz
dd
21
2
1
2
2
1
)1(
)1(2
)1(
2
)1(
2
1
2
2
1
1
1
)!13(
1)(Res
0
3
2
22
0
2
2
2
01
zzz z
z
z
z
z
z
zz
z
dz
dzfd
21
111)(Res
1
3
2
11
z
z z
zzfc
Exercise
1.1 Represent the following complex numbers in polar form:
(a) 2,2 jj (b) 86 j (c) j
j
1
1 (d)
3/22
223
j
j
1.2 Represent the followings in x+jy form:
(a) )3
sin3
(cos4
j (b) cos(-1.8)+jsin(1.8)
1.3 Find the roots of (a) z4+j=0 (b) z8-1=0 (c) z6+8=0 (d) z8-9=0
2.1 Regarding the following functions:
(A) z-1 (B) (z+1)-1 (C) (z-1)(z+1)-1
(D) (z+1)(z-1)-1 (E) (2z-1)(z+1)-1
(i) Sketch the images of the following lines in w-plane
(a) x = 1 (b) x = 1/2 (c) x = 3 (d) y = 1 (e) y = 1/2 (f) y = 3
(ii) Sketch the images of the following lines in z-plane
(a) u = -1 (b) u = 1/2 (c) u = 3 (d) v = -1 (e) v = 1/2 (f) v = 3
2.2 Show that
z
zz
1
1ln
2
1tanh 1
.
3.1 Are the following functions analytic? If so, state the region.
43
(a) f(z)=z3 (b) 2
)( jzzf (c) zzzf )( (d) f(z)=z+1/z
3.2 Find the derivative of (a) sinh z (b) cosh z (c) 1/z (d) ze (e)
2ze (f) (z-1)(z+1)-1
from the definition of and by using f'(z) = ux + jvx.
3.3 Let f(z) = u(r,) + jv(r,), z = rej show that the Cauchy-Riemann equations are given by
rr vur
vr
u 1
;1
. Then show that rr
jjvuezf
)(' and use it to find (ln z)'.
4.1 Find
jz
zdzz
1
3along
(a) z = ej, 0 . (b) C1: (1,0)->(0,0) and C2: (0,0)->(0,j) (c) C3: (1,0)->(1,j) and C4: (1,j)->(0,j)
4.2 Find
jz
zdzz
1
0Re along
(a) x = y (b) C1: (0,0)->(1,0) and C2: (1,0)->(1,j) (c) C3: (0,0)->(0,j) and C4: (0,j)->(1,j)
4.3 Evaluate C z
dz
42
, where C is the square contour of side 2 centered at the origin without using Cauchy-Goursat
theorem.
4.4 Use Cauchy-Goursat theorem and Cauchy’s integral formula to evaluate C zz
dz
22
, where C is given by
(a) |z|=4 (b) |z-3|=1 (c) |z+2| = 1 (d) |z+1|=5 (e) |z-3|=4 (f) |z+j|=2
4.5 Let C be the square contour of side 4 centered at the origin, evaluate
(a)
C
z
jz
dze
2/ (b) C z
zdz
12 (c) C z
zdz4
cosh (d) C zz
zdz22
)1)(1(
sinh
using Cauchy’s integral formula.
5.1 Find the center and radius of convergence of
(a)
0
4n
njz (b)
0
2!n
nn
jzn
n (c)
0
2
!2
2
n
n
n
z
5.2 Find the first 4 terms of Taylor's series of
(a) Ln(1+z), z=0 (b) cosh z, z = 1 (c) 1/z, z = 1 (d) sin(z2), z=0 (e) ez, z = 1
5.3 Find Laurent's series of
(a) 3
2
)1( z
e z
; z = 1 (b) 2
1sin)3(
zz ; z = -2 (c)
2)1(
cosh
z
z; z = 1 (d)
2
2sin
z
z; z = 0
6.1 Find residues at all poles of
(a) zz
z
9
9153
(b)
6116
42523
2
zzz
zz (c)
99
93523
2
zzz
zz(d)
1
3223
2
zzz
zz(e)
485
3223
2
zzz
zz
6.2 Find Laurent's series of zz
z
9
9153
, then find
C
dzzz
z
9
9153
, along (a) |z|=1 (b) |z-3|=2
6.3 Find residues of functions in 5.3 (a),(c),(d), then evaluate C dzzf )( , where C is given by |z| = 2.
6.4 Evaluate the contour integral of functions in 6.1 (b)-(e) along
(a) |z|=4 (b) |z-3|=1/2 (c) |z+1| = 1 (d) |z+1|=5 (e) |z-4|=4
6.5 Find
(a)
0 41 x
dx (b)
0 221 x
dx (c)
0 6
2
1 x
dxx(d)
1
sin2 xx
xdx(e)
0 24 34
cos
xx
xdx (f)
0 4 16
2sin
x
xdxx
(g)
2
0 cos2
d(h)
2
0 sin45
d(i)
2sin1
d(j)
0 2cos4
d
6.6 Find the inverse Laplace transforms of
(a) bs
eas
s
22(b)
22)(
as
as (c)
52
22 ss
s (d)
4
24
3
s
s (e)
22
sincos
as
babs
44
(f) 222
2
as
as
(g) 222
22
as
as
(h) 222
32
as
a
(i)
22
cossin
as
babs
(j) 222
22
as
as
using the Bromwich integral. Also verify the results using Laplace transform table.
6.7 Find the partial fraction of
(a) xx
x
9
9153
(b)
6116
42523
2
xxx
xx (c)
xxx
xx
44
93523
2
(d) 1
3223
2
xxx
xx(e)
485
3223
2
xxx
xx