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Unformatted text preview: ﬁﬂl’ﬁ‘R/f S’C’h’HfIFOHS‘ ftp #ﬁ/ #3 SECTION 64 I. (a) Let us write 1 1 1 1 1
2:——w———==«(1z~l~z"1'"zainn)=——}.—I~z~~2:2+~~
2+2 2 1+2. 2 z
The residue at z = 0, which is the coefﬁcient of —1—, is ciearly 1,
z (b) We may use the expansion 2:2 2‘s 25 C0$ZEIWZ+EE"EI+ to write (0<lzi<1). (IZK 00) 1 1 1 1 1
zc08[ijzz[lmmm+__.i_.u——nunE+..u):~:z_ii+i,i3.wu1_.n;§+.n 2m?” 42 z,“ 6! z z
(0 <lzl< no)
The residue at z = 0, or coefficient of 1—, is now seen to be ”312.
z
(0) Observe that
zwsinz 1 . 1 z3 z5 z; z4
=— z~smz “—“m z 2:w+—~—~~~ :m—nm+m 0<lzl<oci
z z( ) z[ [ 3! 5! ‘ 3! 5! ( )
Since the coefficient of E ' this Laurent series is 0, the residue at z n 0 is 0“
z
(d) Write
eotz_ 1 cosz
4 “mfg
z 2 sum
and recall that
2 4 2 4
z z z z (Izl< m) coszmlu§T+mw~ml—_+mmn 4! 2 24 95
Dividing the series for sin: into the one for cosz, we find that 0032 1 z z =———m+...
sinz z 3 45 (0<lzf<7r:). Thus 4 __ ....._____.... _—.__S.——~.. ————— +m (0<lz!<1t)n cotzwllzzsw 11111
2 z4z345 Note that the condition of validity for this series is due to the fact that Sims 2 0 when z = my: (:2 m 0,il,:t2,...). It is new evident that 60:2 has residue “2%" at z 2 0.
z
(e) Recall that
3 5
’ — E... E... no
smhzz+3i+sg+ (fzk )
and
1 2
Wml+z+z +m GER”)
1*2 There is a Laurent series for the function sinhz 1 . 1
“2*(w—1mzz)?(smhz)(1~zz) that is valid for 0 <lzi<1w To find it, we first multiply the Maclaurin series for sinhz I
d w
an 1~z (sinhz)[ 1 )=[z+£z3+u}_z5+.“)(1+zz+z4+n.) {”22 6 120
1
:z+~ 3i———~~5 H
6" 120g
3 1 s
z +~— +
62
£54.... wz+~z +~~ (0<izl<1)n 96
We then see that weight. 0 H 1
24(122) 23 6 z ( <Z< )“ sinh z 7 This shows that the residue of m......._,, at z m 9 is —.
z4(1m 2:2) 6 2. In each part, C denotes the positively oriented circle lat: 3. (a) To evaluate JG 6x138?) dz, we need the residue of the integrand at z = O. From z
the Laurent series
“PP2L}— ...32. 512:: Mimi: 1 z
32 “32(1 11+2! 3t+ “2,2 1¥z+21 “gt“ (0<iZE<°°), we see that the required residue is 1. Thus j exp?” dz : 2m(1) m —27ri. C 2 (6) Likewise, to evaluate the integral L 2:2 exp(~1~)dz, we must find the residue of the
z integrand at z = 0. The Laurent series which is valid for 0 (izk on, tetls us that the needed residue is 3— Hence 2 1 m . 1 um‘
L" expiridwmizi“? 4. Let C denote the circle lzim 1, taken counterclockwise. (a) The Maclaurin series e « 22"' — (lzl< on) enables us to write
71:0 ’2! M n M 1 1
ex z+w 3:sz emdzm e"z 5—d = ~— ”ex (—Jd ‘
L p( 2. Le: C We . z 2 . z p z. z (1;) Referring to the Maclauxin series for e1 once again, let us write ,, 1) ,, °" 1 1 '° 1 “k
z ex  =z —~—= " n=0,l,2,m.
pt: 25k! z" é? ( )
Now the l in this series occurs when n Mk 2: ml, or k=n+1. So, by the residue 2
theorem,
,, 1 ' . 1 jz exp — dz=2m (n:0,l,2,...). C :5 (n+1)! The final result in part (a) thus reduces to j; exp(z+—:— —)dz ZniZu—mw n=on’(n+1)' 99
5. We are given two polynomials ’ P(z)=ao+a1z+azzz+m+aﬂz" (aﬂqeO)
and
9(2) : b0 +blz+ (7252 +«+ Iowa“ (13," a o), where m 2 n+2.
It is straightforward to show that 1IP(1/z): 2"”2 +azz’" ”+022”4 +~n~+ a:z,,z""""'"2 z QUIZ)“ buz'" +btzm‘l+bgz’""z+m+bm (#0)" Observe that the numerator here is, in fact, a polynomial since mmn  2 2 0. Also, since
b," ¢0, the quotient of these polynomials is represented by a series of the form do +diz+ c1127" +w. That is, 1 P(l/z) 2 __4__.__.m:d +d + H,  £2 Q(l/z) o lz dzz + (O<Izl<R2),
and we see that t—~z12 Eurmix: has residueO 2:0. Suppose now that all of the zeros of Q(z) lie inside a simple closed contour C, and
assume that C is positively oriented. Since P(z)/ Q(z) is analytic everywhere in the finite plane except at the zeros of Q(z), it follows from the theorem in Sec. 64 and the residue just
obtained that Hz) 1 Pale) ~ ...
jCQ(z—)~dz= 2m" Res[? 90 To] am one. If C is negatively oriented, this result is still true since then Ifgld dz:""IC P(—z—)dz—0
CQQE) CQCE) SECTION 65 1. (a) From the expansion 2 3
gz=1+£+i+mﬁ+ut (izi<00), 1! 2! 3!
we see that zexp(l):z+l+~l—il+im1w+m (0<IZI<°°)‘
z 2! z 3122 100 . . . l .
The prmcrpal part of zexp(~) at the Isolated singular point z z 0 is, then,
2.
1 l l 1
2! z 3! z and z n 0 is an essential singular point of that function. 2 (b) The isolated singular point of is at z x —L Since the principal part at z 2 m1 1 + z
involves powers of 2 +1, we begin by observing that 2:2 =(z+1)2~2z~imtz+i)22(z+1)+i This enables us to write 2 +12% +
z zﬁwxw+nwg+$n
1+2: z+1 z+1 Since the principal part is —L~, the point z = l is a (simple) pole. 1 2+1
1 (c) The point 2: =0 is the isolated singular point of 5125», and we can write
2
sinz 1 :3 zs _. z2 Z4
z "E(Z“§+ET"'H]IW'€T+"§TM'H (0<]ZE<°°). The principal part here is evidently O, and so 2: L" 0 is a removable singular point of the function EEK“
z
(d) The isolated singular point of 93% is z = 0‘ Since
2.
cosz 1 2:2 2:4 1 z 23
~——m~1———+——~« =~+~mn 0< l<oo, ..
zz(2!4!)z2!4l (Z)
. . . 1 . . . (2052
the plinc1pal part IS w. This means that sz lsa(smtiple) pole of ————.
z z
I ——1 l .
= , we find that the rinci al art of at its
isolated singular point z = 2 is simply the function itself. That point is evidently a pole
(of order 3). (3) Upon writing 101 2. (a) The singular point is z =2“ 0. Since 1.... 2 4 6 3
Highly 1w(1+£m+£m+§m+m] bigwigs.“ z z 2! 4! 6! 212: 4! 63
when O<izl< 00, we have maland BzM§T=m§ (17) Here the singular point is also 2. = 0. Since z z
hiiﬁiijiiiimzs m .
1' 2.3 2! z2 3! z 4! sgz
3
when O<lzl<oo, we have mm3 and BzméFz—uj—
. . exp(2z') . __ .
(c) The smgular pomtof (z~l)2 IS z——I. The Taylor senes
_ 2 ... 2 3 .... 3
exp(?.2.')''=32”"“92 =ez[1+ 2(2' I) +—~———m2 (z I) + 2 (z D 4] (izk 9°)
1! 2! 3!
enables us to write the Laurent series
exp(22) __ 2 1 3.. 1 22 22
(zI)2 We [(z—1)2+1! z—1+2a+?f(z 1)+ (047' km)"
2 2 2
Thus m=2 and B=e 17:26? .
l
3. Since f is analytic at zo, it has a Taylor series representation
I Z Ir
f(z>:f(ze)+fl(,°)(z«zo)+féf9)<z—ze)2+~ (IznzakRu).
Let g be deﬁned by means of the equation
_ ﬂz) 3(z Mw.
:5th 44 102
(a) Suppose that f (20) ¢ 0~ Then ﬂﬂ=z ;[fmu+ffd(z zﬁ+£§%%z»%f+u] This shows that g has a simple pole at 20 , with residue f (20). (b) Suppose, on the other hand, that f (2:0) m 0. Then 1 I 0 rr 0
8(Z)$ zwzif if )(z~zo)+i—§~Z(z—zo)2+..,]
=fﬁo)+f2(fe)(z_,20)+m (O<EzZUE<RD).\ Since the principal part of g at 20 is just 0, the point z x 0 is a removable singular point of g. 4. Write the function _ Saaz2
f(z) ~ (z; + a2? (a > 0)
as
3 M2) .. 81513252
ﬁg) (2: ~ at“)3 Where (ME)  (z + at)3 "
Since the only singularity of ¢(z) is at z = "ai, 45(2) has a Taylor series representation
¢(z) ¢(ai)+ ¢(a i(z) —ai)+ ¢”2(a )(z~ ~ai)z + (Exeaii(Za) about z z ai. Thus f(z):.~ 1.3[¢(ai)+¢(a1)(z ai)+¢”(m)(z of) + ] (Miranda).
(rat) 1 2. :; 6 4 __ 3 2 ___  2
I a 12 8a :5 and ¢"(z)—— 16c13(z2 4azz a )
(z +ai)5 103
Eonsequently, Mann—air, ¢’(ai)m~%, and ¢”(ar)m~i. This enables us to write 1
(Zeal)3 HZ)“ [“ali“§(z~ai)—%(ZWCIE)2sun] (0<lz—ai!<2a). The principal part of fat the point z = of is, then, i/Z a/Z azi SECTION 67 2
1. (a) The function f(z) = 2 +2 99(3) 1 has an isolated singular point at z :1» Writing f(z) 3 “mi.
7.  Z " where 93(2) m 22 +2, and observing that (25(2) is analytic and nonzero at 2 =21, we see
that 2 =1 is a pole of order m =1 and that the residue there is B a $0) m 3. (b) If we write 3
2 (15(2) z}
m n M: h = my
ﬂit) (234.1] I: ( 1)], w ere 915(5) 3
z t 2
we see that z = mi» is a singular point of 1? Since (M25) is analytic and nonzero at that point, f has a pole of order m m 3 there. The residue is an mum) m 3 21 16" (c) The function
exp z exp 2 22m! u(z—m‘)(z+m') has poles of order m :1 at the two points z = inf“ The residue at z a m' is and the one at zx—Iri is 104 U4
2. (a) Write the function m) = :1 (52b 0, 0 < argz < 22:) as
Z
f(z) m 533, where ¢(z) a z‘” = aim“ (Ezl> 0,0 < argz < 27:)" The function ¢(z) is analytic throughout its domain of definition, indicated in the ﬁgure below“
Also,
1 t .
—log(—1} ~—(Iui+in') . 7! ﬂ 1+ I
(—1)m(—“1)m m€34 =e" :e‘m mcos+isin~:~——~¢O,
‘0 a
This shows that the function f has a pole of order m z 1 at z m —1, the residue there
being
1+‘i
B x “"1 z: ——_
¢< ) ‘5
. . . Log:
b Wnte the functxon z a W as
( ) f ( ) (2:2 +1);
(NZ) Logz
. 2) == . where z ..—.. .
ﬂ (Z"£)2 ¢( ) (zH‘)2 From this, it is clear that f(z) has a pole of order m=2 at z =i. Straightforward differentiation then reveals that Res Logz — qb’(i) _ z=i (6+1)1 “ 3 105 (c) Write the function
112 :5
17(2):??? (lz>0,0<argz<27r)
as
(ME) 2‘”
2 =m where 2: ..~.—————_
f( > (H); M ) (m):
Since
. _ (z+i)z‘”2 __4zuz
lb (‘7‘) 2(z+i)3
and
my: __ “mm "J." J...“ H: mm .... 1 l
[e Z‘ﬁ’lwemzmkz’
U2 .
z 1...,
Res 2 ’1' mm,
(atii)2 ‘3 0 3 2
3. (a) We wish to evaluate the integral
3
I 32 «:2 dz,
C (z“1)(z +9) where C is the circle I:  21 = 2, taken in the counterclockwise direction. That circle and
the singularities 2: =1, i3t’ of the integrand are shown in the ﬁgure just below. Observe that the point z = 1, which is the only singularity inside C, is a simple pole of the integrand and that
R 32:3 + 2 3Z3 + 2 1
68W 2 ”1,— x w.
ml (Eh1X2. +9) z +9 :21 2 According to the residue theorem, then, 323+2.(1) .
W :2 ... : ~
Jc(z1)(zz+9)dz 7” 2 7“ 106
(15) Let us redo part (a) when C is changed to be the positively oriented circle Izl m 4, shown
in the ﬁgure below. In this case, all three singularities 1:1, $3i of the integrand are interior to C. We
already know from part (a) that 323 + 2 1
Res—*—2—"— " "”~
z=i (z —1)(z + 9) 2
It is, moreover, Straightfbrwa‘d to show that
esmﬁﬂimu 3334.2 _15+49i
ma. (2 m 1x6 + 9) (z 1)(z + 30 W3. 12
and
Res 3z3+2 =__3E'E_2_.__. WISH—491.
1;"31. (z — 1)(Zz + 9) (Z W 1)(Z m 3i) gun41' 12 ' The residue theorem now tells us that z — +
(poo2+9) z ,2 12 12 3+2 . ' m '
chﬂgmd [1 15449: 15 491):6m', 4. (a) Let C denote the positively oriented circle 1zE= 2, and note that the integrand of the dz integrai I has singularities at z m 0 and z 3 ~61. (See the figure below.) C2.3(2:~i»4) To find the residue of the integrand at z n O, we recall the expansion and write Mimi; Him 23...:
33am) 42.3 1+(z/4) 423 . 1
Now the coefﬁcrent of « here occurs when n = 2, and we see that Consequently, 107 (12k: 1) (0 <le< 4). (17) Let us replace the path C in part (a) by the positively oriented circle I: + 2! z 3, centered
at —2 and with radius 3. It is shown below. To find the residue at —4, we write 1 _ 45(2) 1 W"z—~c—4)’ Where “2):?” This tells us that z = "4 is a simple pole of the integrand and that the residue there is M" 4) = ”1 l 64“ Consequently, dz . 1 l
———.—.2 —~ = .
Cz3(z+4) ”{64 64) 0 108
. coshn‘ d . . . . .
5. Let us evatuate the integral f WTiE, where C 15 the posmvely ouented Circle lem 2.
C z(z +1)
AH three isolated singularities z : Oil" of the integrand are interior to C" The desired
residues are “:1, cosh ﬁz cosh 7:3]
Res 2 m 2
==° 22(z +1) 2 +1 1:0 cosh 11:2: 6081} 712 1
Res 2 m , = "a
at! z(z + 1) 2(2: +16) ,7, 2
and
cosh m cosh m5] 1
Res 2 = . 3 '"
z=~r z(z +1) 25(2' “ I) We;
Consequently, Iwmzni(l+i+ij=4m‘
C z(z +1) 2 2 6. In each part of this problem, C denotes the positively oriented circle Ezlm 3" (a) It is straightfomaxd to show that ., (sin—2)2 1 (1) (3+22)2
If = M, th —— — z w,
ﬁg) z(z 1)(2z + 5) en z2 f z z(1 z)(2 + 52)
. . 1 1 .
This functlon “I f  has a sunple pole at z m 0, and
z z 2
J' M dz = 2751' ReS:";i'f('£):l L" 2751(2) m 97d“
C z(z "1)(22+5) 2:0 :5 z (b) Likewise, .. : 23(1~32) "I“ (i): z3
If”) (1+z)(1+2z4)’ the“ 22f z z(z+l)(z4+2)' The function “lgf (—1) has a simple pole at z 2 0, and we ﬁnd here that
z z ‘ 3
f i jJ—gj334 dz=2mRes[1—;~f[i)]=2m(——3~)=—3m.
C(1+z)(1+2z) F 2 z i 2‘ ...
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 Fall '07
 RickRugangYe

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