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.
.
.
.
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조E 디 converget absoluteconverge
태다 그리 ,onverge
RatioTest :
Root Test 에다 L
ExampleI
=
{
The sequence i y converges
,
i
∞ 끔
since m
=
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6
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-
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lor
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ay sertes f
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Maclaurin series :f ( z = 코 :
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Exam 1
fl =**, expand withcenterzo= - 2i
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1
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Example
2Madaurin series
=
ofE
ㅗ
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+ E + + E 3+ "
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F 턴무+ 러러뷰 t
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Exampeexpand Taylor
=
서
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,
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=따*
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ii
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Series
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.
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vetf be analytic within
=
theannulardomam ( r 1 t -z 1 R )
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ak =π
* 9
f( s 7
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=
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,
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θ powerranalytic , θ → principal
of
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.(
zya
EXa
ㆍ
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t 섭
mi +
! "
+z
f ( :a sin/ z4= YE- 1BIz + 15?-/ n ! 79 !
Exam
2 expand
f = 트 ( 1 + E E 4 z3 +
+
=
-
통1
-
E
z
-
2
z
-
'
티
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-
Z
4
b 기디티니티리
f 1 E = ( 14 호 ) )
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=
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디테니 f ( Z ( H+ 리터 )
테파
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1
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1
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을 expand 2 ( z - 3 )
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( 9 ) 0 시터 K 2
BOL 1 E 3 K 2
-
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(
b)
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2
는3
π
3 ( 2 + z 3)글 ) ( I + )
E
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()대뷰탈 ) 밖 (벌 ]
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F
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O
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8
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Exam
5
초 9 9z +9 z 주
+
+
지
expand * 터) , 아니레 2
-+ ,
HE ) E
f
=
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=
'
,
,
f= 터
fila =
더 = 이 +π
ㅢ
)
δ 다 - 를어를 ] 르를 ' +를 □
fl ) 터
t ←
1+ 현
=
= 한다2 + - →+
이□
)
4
2
2
filz) convergesfor LE 2 k γ
-
,
falE )convergesfor 1 니 미
Z
Exam
bexpande /z, O 4 EKD
e디 ++
3고
=+
e
: + 풀!
↑
Poles and zeros
.
Kemovable sngulariy → Principal part 0
pole ofordern lase nonzero coefficienta u
5 imple pole principlecontains I ermwiin9 -1
essential singularicy principle ontains D terms
이
키 = GtnTorO 니레이고 , ①is analytic at zo
=
.
:
-
.
:
.
:
.
rero zo
zevo
-
f ( zo )
=
(7
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of order n : f
( z) = ,
"
끝 + 파를 )주취( 홑
]
""
-
1)
' ( to =O " fiz) o
)
=
51 z 계 E Zo ] ( z ) φ ( ← ) isanalytic utz zo
n
-
=
,
틀다뚝 , !.
. .
= 한 +++트 ++
4
씀
+
.
l
= ofremovable sin
rinoip p artp
)
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-
Exam
은 5 i
그
EFZ
a ) O 리티디 f 1
6
glariey
「
」
56 Residues
6 - . someconse
nence ofthe Residue Theorem
.
-
6
Thecoefficient ay of Laurent series
jscalled residue ⇒ a - Res ( f( z) Z )
=
,
=
=
.
Res (
f ( ) , Zo
←
)d
.
머( E-t) 티
=:
Ifflt) gl ㆍ/ lz) botn analytic
hasa zero of
orderI at Eo then fhas simple pole zo
and Res ( f ( ←7 , z. ) g ( z. ] /h '( t. )
=
.
atz = zoand if h
,
=
caucly ' sResidue Theorem
iff isanalytic exeptatfinie
number of isolated smgular
Pomts 21 Z 2
,
Zk withinC
".
.
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z ),
=
EK )
,
파
3
" 쳐
+
F터 "
Kes ( f ( z )
1) =
,
e 3 |= +
( bl
Res ( f ( z ) 0 )
,
를
+
,
㉝
고
t "
←
: Z3
3
=
EXam
theresidue
find
-
=
-
3
Res /f(z ) 1)= l =e
메
리텍
eπ
3 πil
,
,
.
=
=
=
Res ( f ( z), E2 ] =
FE
-Fi
설
F
Res ( f( E ), E )
FN 도TFii
=FE =
Res ( fiz Z4) 4
E
F + FIi
=
4
3
3
=
,
Examp
은4kesidue Theorem
.
-
1,
y
=
rect
1
=
,
,
-
'
Res ( f( z) 3 ) 천) p= 4
2 π il 4+¢ )= o
=
|
lm
=
iRes ( f( , 고 , E또 vs
*
.
)
iRes ( f( E) 지i2
(← , 21 )
f
Res (
=
l 러지
2t
P 블러리 어 = ππ 3 + 2 i 7
2
)
.
Exam
bkesidue Theorn
.e 45
z3dz , 디티 2
2 π iRes ( f( z) 0)
3 LE
=z π i Ʃ
터 5) )
( :e
.
=
,
E
고주 8=
|
"
숍
5
Z + ]3
5
EXamnKesidue Theoven
CIE =α
PCtanEdE
2 π ; j[ Res f← ,π ) + Res (f ← 품 )]
,
,
1
(
)
)
π)=
- 1
sm (
Res (f( z) -품] sin
[ - πY 1
3; ( (
1
Res ( f/ z) 는 ] π
12) -
=
=
,
n
=
Pctauzdt ππ( - 1 1)
=
2
puellEdzf
EXa 8
=
3
=
☆π
( ) ,E2 ) =
π
=
-
=
-
4π 5
kesidue Theorem
C 여티디
z
2π 5Res ( / ), 0 ) = 6 u
_
AB z xeoi; ED Z xe
/
dz
Z
(
I 19 Ia
김유
=
2π
'
⇒
:
)
|≤ ( K 리 ) (R 9 )
πR
_
R →D
:
⇒
.v
z= 9 , f
*
By the previous example 6 5 3
.
.
EFe π
upperplane poles
-i
)
i (4
⇒ Res ( f( z ,E
i
3 π14
*
-
도
, ←1
z 5-g ( z
.
,
.
=
=
O + i ( 플3 ] ⇒58
x+
9 dxzes
*
=
Exam5
xta dx
Jx =
dE
)
considerp ElE
C
=
2 Z + 2)
Yx Jok +/K fer +/= πiRes( f /z] ei,l + i )
2
+
=
dx
-π
iRes( flzJe; o )
p . 5% π
(x 르x+
x
2π iRes ( f/ z] ei:/ + i )
,
.
=
e(
e fIEleixz
ez , "
(
,
,
)
π
얇
eix
dx
P. U. 10
% xlx =
rx +z )
:
π+
δπ itz π ; [ - ( 1
- xfi
e
p .8
=
e
3
( Los / + i 세
) 의
sinxx
rx zdx
=
+
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-
=
(+ i)
+ i 7
,
-
wsi )
)
]
=
.
-
→
2
,
+
i]
[Res (4
+Res ( π뜯은 2 i ) ]
,-
π ot 여
4 ,
2
π ot
=π
Res ( 끝여
2i t
4,
4T
πo
땐
2i
Res [ 끝며
] =π
4
4
loth π
시 tsnh
( 2 π))
코음 π
란의 ⑫ 업 π
란
+ + 2 π4
4
22 π 44
+ = 228k
44 π= wth π
28 π
44 = xoth π
=
x
dx o xsinx
18 19
x 249
=
P틀
9 eizdz JiR +JR
2 π iRes (f(z) eiz 3 ;)
2π i
( Zez] 11 증3 i
.
8
dx .
P 6.18
x9 dx
pvJ
x
*a
=
=
ot 2 π; =
=
(
-
|
려
4
4
-
+
2
2
)
( f /) ←s ) )
x5
dx
18 x
9
=
,
Weconclude thasa 2 ero of ordera lies
}
5n the disk 1 z 1 K Ʃ
+
또 ii
Res ( f( z) ,← ) =π π
;
=
,
)
⇒
지
=
4
용ㆍ주=
.
evaluate * +
, dx
we kpow
f( ) +
=
Exam은 8Find the 3 um 조
K
4ππ
Z ) Z 주 4 zero. fzi f( ←)
1
,
Z
4
P
p . .J∞ ( x위 x
) ( 주)dx 품
R
(
π iRes (
( =
=
)
z
/
18 51 ≤ 811 주 5
fi
by expanding the searchto cirlel 러
=
1 f( z 7 - g ( E 1 ≤ 81 타주 5 8 ( ) ( 5 < ) 9 1 f( E) 1
|
I*=SuR
E레기
2
은
Exa
the zerosof Eg
←) 9 - 8 z 45
loate
π
from the UL inequality
=
J 8형
( ×+11 dX =π
=
KE ) ( E 49 ) ) 1 고 | / | 1 고 a 1
드
≥ 11 z 고이며 - 91 리국의 ) ( K a )
( I . 1 - 70 when
i×+1)
25 & π
dx
π=- i
Res ( f( z) , + ) = z eai e
,
i (T
;)
;-4
g
π
+
fRes ( f( z) 3 i ) )
=
)
)
[ tJpffor JaBF2 iBes 에
.JuB
)
E
Λ
J
dx
웨
idx )=
er π
+|
+
,
]
+
dx
IAB J( xevi
eoidx= 로 x ,
3 i ) ( zt 3 i )
-
P ( Z 류) ( 49) I 22
dz
+
]x )
( x주 ) f Z ) ( 49)
(
계 dx
=
π
2π ;
=
=
i 7 ( Efi ] / E
-
:
2
=
=
(E
=
ez i
y(
JED JK/xeskitl
f ( z ) Y 1 E 17 ( z * 9)
=
담‰
+
2 π iBe )( f 11 ,- 1 )
:
evaluates( x주 ( ) (x ÷ 9)
|
=
ExamP
6 Integraion alonga Branchcnt
=
.
CAUCMY R dx
Exam
은2
(
+
8; v
금역농
viRes
2
π oot
evaluateJx ) dx
=
Zpr
K)
러 , K) (
), ZP5
j ), Res ( (
- aRes [ π
④ 설미테 ) 어 Joktfip fart/Aβ
=
.
=
z
-
x
고- )
Exam
=4
=
=π il .
λ= .우티 E f( ]
Z
-
(f (
Exam
은Residues Theorm
24
P dE, C 이 E - i 2
=
,
f( ← )
테
→ v
.
)
C
=
π
aπ
.. .
,
=
동
utt
and
, Ir
m
finite number of poles Ep) EP2
adt 용.
o④ 테
(
흙 ]로레ㆍ )P 코음 π
)
4 )
7 .
,
,
=π ,
1
ff haimple
pole mfinie
a
←
,
/ +
)ipes
dxvitRes
)
Res( f (E ) Ek )
Res ( f ( ), | =π,
2π :
'
,'
+
t
.
zπ
op
+ ) 어를
Px ( 2 + 조 z + - )) t .렸
b) C circle / EF 2
=
5
do
Exa
은 의 taluate
Ez = e
,
=
forall z on C, then fghave
samenurm of zaros in a
=
nResf1E ) 1 EK 1
i(Resifπ ← ) ]+Res( f ( z) 3))
ㅢ
Res ( f(z ) , ) =
레 4
2 π j2
.
toflxleiaxdx =jfix )cosaxdx+ ifaflx) sinaxdx
Exam
= lauchy
3
2j코
1
-
Fourier integrals
.
v
.
-
-
π
:
=
Evaluate ⑭터 3
, dz
a ' Cx =θ x = 4 ,y =
I
=
14 ,
,
Cr
(
이
대 *2π [ Res "
"
=
Res ( f( 21 ) F
-π ii
3 *"
"
Ak
piv
=
enuil4
e5 i
π
4,
CR
=
E
= 4 e
.
2t
.
Exam
은 키 ( 4 +1 )
3
4
Suppose f hasa
simple pole on the
θ
-
copverges , p. bas same ralde
the contour
iffflx 1dx divergessmaystillhasa R . K
E Ctrei ① O ≤ Q ≤π
if f ( x ) is ven ⇒ f (- x) f (x )
.
lcrf z7 dE =π iRes( f( z ) C )
) dx
then / ofixldx = jpJ 8f(x )dx =* tix
= No -µp
π 5
|
dz
f. f ( z ) dt forf ( z) dz +/f(x) dx= π iaRes ( f( ← ) ,K ) jff* * Oon Li
zeros , nsideci Np poles
No
p . Yf (x) dx Jf( x / dx=2 j 2 Res ( f( ) , π )
jf 1 f1 E) g ( E ] | 1 f ( z 1 holds
'
f1 ) = 티터 ] ( E 3 )
Res ( f( z), 3 ) lim ( E 3)f1 z) =☆
1-
o,
…
*
-
eire ; sina - eid ei
(
Res (
find theresi-
34 =
Cauchyprincipalvaluepfix ) dxpflxdxrealaxis. Ifor is
.
=
Exa
o
:
d
J π F( C 034, 3 .θ ] da = PF( ( E + 러 ) ," E - E 터) ] 를
.
If thas simple pole at z Eo then
- Zo ) f ( z )
Res ( f( ← ) , Z ) = eist
(E
n at E zo
oforder
If f hasa pole
.
Trigonomeric
.
2
2
b 6 some sequence
-
6 7 applications
-
.
(
=F
Laplace transformL 15 )f estf eldt
π
for t ≥ O
II. LSF (s13koJtiReseF(s1 ds
ifsF( s ) , 3 boundedoncr a 3 →∞ ,LF(s ) = ? Res ( eseF(s ) 3k )
poles
( α) =⇒ 무 fixk( a x) dx
tvansformF
Integral
(s )
8
.
=
,
R
3
,
=
.
,
f ( x)F (α개( 1 α x ) d α
supposef is piecewBeontinuouson [ 0 ,∞ )and of
exponentialorderfort , TThnyhfieyexizefor Re (s ) [
,
.
Fourier Trapsform e 3 fix) =&fixleiaxdx
y
=
F( α )
+ F (xJy πS∞ F ( ajejaxda f x )
3
=
laplace Transtorm
Lh 13=f8estdt=∞]양e되 t
Exam
은I
=
=le 중
아
"
e
-
sb
bx
-
=
e
( osbytisipby )
e - , owhen b →∞ xyi 13
3
)
=
Exa
2IverseLt
evalnateL-1533 Rels ) 70
L
fi = 1 $533 =Res ( est5 0)
,
)
,
음 33 .
또 .32
e
,
되 te 3t= Ʃ t
고
Exam
은3
InverseLT
)"
evaluaeL페)
,Re ( s ) 3
( 5-3y
< d)
jvjfrtria)
당;
(5
t
)
ㆍ더
=
-3
fe= LS 떼
]( 3 3 ) 3
- 20
t -8
=
-
Res (es)
) 1 ]+Res ( es
37
)
(3 - 3
( s- 3 )
'
=
es
lin쁘
+
37 ) 3 - 3 )
= etr +Ʃ
3esc
이
e 3 ( t)
when t
,2
EXa
은
4
fix
=
FTofflxFel × ,
0
gexxx ≈
eiflx) =jexeiaxdx
3
fqexeiaxdx= 1
1 ×"
5 ea . ex
+ I과
I
dxmae
이뭉 . (
=
]
=
e
더
처 -∞ [ e brosbat jesinba- l ]
-
=
*
j [O+어]
I1
=
Fai
2
=π
aifixfi =F (a Fa
Exam
은5
)
;
IFTofaFFaz
&SF (×)3π]
ejaxdx
r IHzyerizxdt
Cupper)
jzx
e
i
) ex
2π; Res ( π
I1+ z )
-
=
"
rItzy
eizxdz
-
=
,
izx
( F⑭ )
iJ
=ex
2π
[π
iRes
(1+ ㆍ ,
=
e
)
&13F( a)35 exx ?
(
、
Σ
n
1
-
.
.
conformal Mapping
f
forevery
(Z)
ω=
Ciand Cr
Appendix 포
is conformalat Eo
pair of smooth curves
inD ntersectngatzo
W= 2
B
B π in A
.
angle magnitication at a cricical point
if f ( z) = "( z == f = 0 andf( ) # 0
thenp between any 2 smoothcurves
W
f
C
z
jntersectat Eo is inreased bya factor
of n by complexmapping w f( z )
B
녀
A
B
= t + (2 - tyi
E (t)
=
t + (tl) i
=
) ti
=
W = SInt
B
'
A
F
P
=
'
=
wi < 1 ) 며
ωI = ω() 1 - i
θ , p equal
mag
F
"
러
ena lmp
p
mappings
al Theentire funcron fizFez ; s
conformal at everypointn the complex
plane smce fi= ㆍ Foforall zin c
, Theentire funcron
fIE = z is
conformal if zFo since g <E) 2 z
,
.
2
'
' B
디
,
A
Exam
은
sconformal
mappings
Find all pointeswhere the mappingis conformal
KE ) = EE 4 ω= SMEi 3 a
0
n
= O ,f 1 ,t 2 .
a
A
"
iid
s E
A
'
'
F
제A
J
Exam
4
usinga Table of Mapping
i
i
→
}
A
Exam
5
mappingofabove( 이의
serips upper half plane s vircle
π z12
=
g f ( Fee
e
Z
π
1
the negative real axi3 is mapped
onto the ciroularare from l to 5
on the orde
.
,
D
C
ω
11
!
)'
z
.
'
B
A
'
'
'
'
A
W=
B
'
(
'
D
'
eTyEte-Yz
exye e TYZ
-
-
E
D
ㅢ
ㅌ
A
'
BCDiE
"
'
F
'
명
'*
Y= π
다 e,
1
ㅢ
A B
"
다
"
=
'
"
CPEFGi
ω= Tπ
1
'
C
C
E
㉖
,
'
D ω= 0 os )
쯤
B
"
"
,
FG
'
|
F
B
ntorm 에
zF (2n + ) π 2,
π
B
F
-
"
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=
(
1
|
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-
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'
w=
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,
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밑
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e
π;
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B A
C
w= e
겉
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용+(
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BY =π A
'
'
"
김A
BY A
□
Exam
2contormal
W= O
A
1
ㅢ
p
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α
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'
,
notequal im sense
mappingforall
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w=
D
'
im
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/
다
π TE
w = coshz
A ω=
C
ω= a (2 =1)
A
디
A
A
B
jntersect wo - i
ai
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13
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P
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B
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c
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.
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wile = - ( t | ) i
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wi
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,
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'
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a3 l
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'
B
[ [z 리 >" 주3 " (1 YE ) ]
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-
B
'
C
1
ω=
BA Z
Ei / e ) ) + ( t + r ) i = 의
Elt)= l + ti 1 + i
if the mapping =
'
⑰
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t
mtersece to
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ㅢ
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e
ππ
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'
B
A
B.
E =itw( 1 | '
Examp
은I
2
‰
{
x
D
i
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'
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1
F
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-
weeidz
z
)
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P
If fisan analytic function in domain
D containingzo and if f( ) # o then
wfiE ) is a conformal mappiagatzo
crjtical pone to if f ( Eo = o
.
=π
=
curvesci '& isatfizo ) im both
magnjtude and sense
.
B Cg
A
W E + Eo
the angle between ci & C2 jmage
.
o
많
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A
'
Ʃ [ Ln (z +) +L미터) ]
17 2 linear Fractional Transformation
.
.
fjnearfracional tvansformation
( ak (
=
=
터
①
A
td10
d
dC
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π탸
E +
+
,
a
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{
,
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adi ( ( )
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)
EF α
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온
커 =( 오 )
1
,35 chwarr- christoffel Transformation
7
.
Schwarr Christoftel formula
-
=
ac bdto
=∞
Circle- PresesvingProperty: If Lis
a cirdeinthe E plane and if Tis a
linearfractional transformation
then the mage ofc undertiserther
(
"
2
-
a
cirale or a lme in the plane
ω
.
ifand only if
E
dk is on C
cFoand the pole
if w f ( 2) is alinear fractsonal
,
transtormation MapS E ,E 3 →ω , w
The mage is a lime
=
.
=
.
1E
Z
then 3⑳ 1
=
s
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10 ) =
=
( 3 + ri ) / 1
=
,
,
i , Tli)
=
α
3 + 2 i, T ( ∞
=
주
0 + 17 / ( 0 - i ) =
(
T ( 1 fi )
/ z ;) , points O l + i i , ∞
.
-
=
2
EXa
2
Image
ofa cirole
unitcirles ?
T ( z) E + 리터
construct a mappingfrom
theapperhaltkplare
tolefe U 3 -1 ≤ v = 1 )
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is on the uniecirclec 4 a line
)
= , Tli = 코 i ⇒ U
)
=
ㅢ
황
Ʃ
7 ? T ( E) =+ 미터
Thepole is por on C ⇒ circde
T ( 21
4 and T - 2)
Cω - = α
=
tolefe
α =3 α2
=
=
t)=
=
f 1E
=
ω=
ω
) ' 템=
(
(
터
( ati> E + lt ,
maps 1 i ,
= ALLE ) ⑫+ osHz ] + B
tt + ) = i , +)
=
여 A =π , B = 0
mappingfrom
a
1
x1
=
0
,
x2 =
1
=α 2 =α 3 =π
5
미테
f
(
ds
이 =3 ⇒ B
=
(5
-
usings chNar2
Exam
4
,
-
-
t
-
틀
÷ 칙=!
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T (= J 1
Examp 6 conseruct LFT
construct a CFT maps i: 1, ∞
on y = x 1 to li i , t or 1 ) 의
-
-
렸스
값 ☆ * =l
. 러
다마
= , ⇒ ω= f )=
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Alz+ )
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let loapproach
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E
E
A (1 +Ʃ )
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)
E
f ( ) ALZTLnZ + B
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=
1, 0, 1
\
를
=
지려
A 고려
리 ⑫]
)
( a1/π -
LFT
lon circle |= 1 ) to-
,
).
theupperhalkplane
) z + γ+ T
1
Exam
5conseruct
1
11 zy
- i ,5) i ⇒ B O, A
to lefe ( U=π 1 - LU = O )
-
Solvngfor w
=
construct a mappingfrom
_
=
1 & X1
1 A P
fi 11 A16 3
- 1793=
ds
J( = 코' / 금
*(s143
보 ! 기걸겨, 값)
2 TF
,
+
성 :)
construota
-
닭 i = 이렉 *
A
)
=
T
=
51)
E+B
B
bareoncl
using matrices
31 ( T ( z 1 )
"
-
A
2
SL( Ei]I; 터), ( ←) 로터)/( z + )
( ! π)
Aisn-
=
AJ
* ( = 통페
)"
EXa
께
4
1
=
construct a mappingfrom
theupperhalt - plane
α
E)
=
,
=
또
-
fHE)
toleft
터를 러 터테F π
(
=
theupperhalplare
→
=
;
→
소
IE =AI테 ] 리터 )=" =- Ai
construct
Image ofa cirole
πF
-i
ω =
,
apply the Theorem with X,
(는 ]
Exam
은
3
2
,
Exa
은
3
usingschnar
the interior part ⇒ test T ( 0 )
cirdel =
플
α 1 =α2 =
=
T (- 1
usingsihnarz
Exam
2usingschwarr
Examp
은 I linearfractional
π ) = ( 2 E + 11 /
Exam
은I
=
youa
=
x + Oi , glz ) xeeulxltiπ
xa 0, values D - 1 g Eπ DLH 너
E
=
=
$x
-
1D
f( z )
-
=
1
,
-
,
-
=
z,
=
αnl )
π
E - xn ?
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fiz) onto an unbounded poly gonal region with
ismapped by
gan
jnterjoranglesayi α
51 z Al( Z x) 에 (E -x ) ( α 1! … ( E
w
-
α
-
,
x, cxsci . < xn
…
베패 ( ε x ) 저
)
f( E) A (E -x
~
DND
E +L 미 + 1
-
→
i
U
-
axis
=
0
-
)
απ - 1 dz +
xn )
B
Poisson Integral Formulas
7 4
.
.
General tormala for theupper Half plane
-
3 / ve
였더짧 = (
po
3 ubject
plx
이력
:
D
O-
, Y0
xc ∞
oLxCx1
.
1
KI X1 KKX2
:
kn , XnLXC ∞
⑥ ( x y) kn +π 코 n ( ;-1 kj)Arg ( E x5 )
=
,
-
-
f1 z) k- 돌코 ( K 51 - k5) Ln ( z x5 )
=
-
Poj 35 on megralform 에
ispecialcase ofPirichlet , Kn 0 )
301 ve
1 DLxLD y 70
.
였더 = -
=
subject to $( x I
1
,
O ,
,
k
=
=
K
-
D <
x x1
<
xikxaxa
:
O , xH < XL ∞
φ ( x y= 2 건뷰 [ Arg ( i)- Arg( z - 15 ) ]
.
친공먼/*)
며 yade
×
=
☆5∞ x-ψ ( )
=
t 0 )
(
.
,
t 4
Unie dj3 k
dt
ya
the ralues fleio) on the unit circle
θ
E ei give a piecewise continuous &
=
bounded funceion for
-
π≤ O ≤π
)ap
습
Then ( x , )=π]π flei
y
다
i3 a solution of ehe Dichirlet
m the open unitdisk ( 1 시
θ
withboundary φ (osθ ,sinb ) flei )
=
solving piichlet
Examp
은그
옮 += ,
30 lve :
0
subjece to p ( x 0)
,
Identifying Ko
=
φ (x y) 5 * (
=
.
+
55
2,
=
- D 2x ( ∞
-
,
0 cx 3
,
3 Kx 2 D
2 , k β = - 1 ,k
1
2
-
,Y 0
DLXLO
=
2
5
(1 ))Arg ( E
-
0)
+π ( - -5 )Arg ( E 3 )
-
1
=
Arg( z )- * Arg ( E 3 )
5+ *
-
solving poi3son
Exam
은2
30 lve
:
옮 옮 =ω
+
, - DExo
,y 0
, ) ; '1
역
subjece to plx
α
∞
'
뷰/응이
)
주 yp 뷰 !합 ,
주 ypdt
=뷰] 없
yds=뷰(Jyrds J,ypds)
φ
(x. y )
=
=
J=
yads =급ta ( 용 ) +(, yd 3jen( 3y+ c
/
(
φ ( x y) =× [ ta 서 y )- ta에(y) ]
1) + y
+( ← x
)
.
( x + 174 y
EXamP
은2
30 lve :
solving poisson
+=
,
x 주y 씨
subjecttoplos θ sin θ) 예비
,
-
,
π LD ≤π
fleit) =ψ ( ost smt) 디디
φ ( x y ) *% 퓨디 시티음
ie - pdt
,
.
=