So tion tion
an al
S. Garr Garrett ett R. Campbell Campbell-Wr -Wrigh igh D. Levinson
Fund Fund ment ment ls
Comp Comple le Prenti Prenticece- Hal
2003
Anal Analys ysis is
rd
ed.
XE
IS
.1
he
lg
ra
le
1. -i=a+bi::=>a=Oandb=-l=>
(_i)2
(a
-lY
(2ab)i
')
from from th real real coun counte terp rpar arts ts Comm Commut utat ativ iv la fo mult multip ipli lica cati tion on (a
bi)(e
e -
ad)i (d
(e
Asso Associ ciat ativ iv la [(
di)(a
eb)i i)
fo multi multipl plic icati ation on
bi)(e
di)](e
Ii
(b
e-
e/))
(a
(d
(a
bi
di
Ii
a(de
el)]i
[(bc+ ad)e
ad)I]
b(de
ad)i](e
eI)i]
fi))
Dist Distri ribu buti tive ve law: law: (a
bi)[(e
(e
i)
e)
(d
Ji)] I)
(a [b(e
e)
bi)[(e
e)
(d
(bc+ ad)i] (a
3.
bi)(e
(a
i)
f)]i (b
bi)(e
af)i]
fi
a.
fi
(e
Ii
(e
and (a
a)
(e
i)
e+
<===>
f)i]
(d
(a
i)
<===>
XE
IS
.1
he
lg
ra
le
1. -i=a+bi::=>a=Oandb=-l=>
(_i)2
(a
-lY
(2ab)i
')
from from th real real coun counte terp rpar arts ts Comm Commut utat ativ iv la fo mult multip ipli lica cati tion on (a
bi)(e
e -
ad)i (d
(e
Asso Associ ciat ativ iv la [(
di)(a
eb)i i)
fo multi multipl plic icati ation on
bi)(e
di)](e
Ii
(b
e-
e/))
(a
(d
(a
bi
di
Ii
a(de
el)]i
[(bc+ ad)e
ad)I]
b(de
ad)i](e
eI)i]
fi))
Dist Distri ribu buti tive ve law: law: (a
bi)[(e
(e
i)
e)
(d
Ji)] I)
(a [b(e
e)
bi)[(e
e)
(d
(bc+ ad)i] (a
3.
bi)(e
(a
i)
f)]i (b
bi)(e
af)i]
fi
a.
fi
(e
Ii
(e
and (a
a)
(e
i)
e+
<===>
f)i]
(d
(a
i)
<===>
h. (e
fi)(c e - f d
i) and id
i,
fd)e
(I
ed)d
(I
:I
-2
a.
6.
h.
Oi
c.
(-2)i
-2
a.
(-2)i
-2i
h.
(-3)i
c.
7.
Then Then
a.
h.
Ii
c.O+
3'"
Oi
-7)i
Z2
fd)d.
ZlZ2
O.
+5i 14.
i. Re(iz)
15 i 4 A :
-b
(i4)A: i41r 41r
i4i+l
i4i+2
(i)
i41r•
i+
16.
b)
a.-I c.
d. 17 3i (4)+3
8i- (4)
6i
3( (-
i- (4)+3
8(1)
i?
(-i)
2(-1
(-2
real equa ions ar Re( z3
5z
Im(Z3
5z
3i)
Im(z
3i).
be 3alr
5a
3a 20.
a.
b-
z=-=-, 2i 5i
c. d.
±4i
8-
27
5i
-Imz
21.
3z
(1
(1
Z4
2i)Z2J
i)(l)
-z
3 0=
(1
i)[iz1
i, -2
i)
16
bi.
Suppose
b2>0 whenever
24. Suppose
bi. Im
-: whenever
bi and
25. Le
di. and ad Zl
and
-b and ad
Z2
he hy
th
cify that
bd
e)
be ar
real
a( -b
be
c,
ZlZ2
26.
is Re(L~l
j)
fo al
Zj
vi
itiv
integers
Zn
Re(zn) j=
j=
fo
0,
Disprove:
Im(z;).
1m Re[(a Re(a
bi)(c di)] i) Re(c i)
Im[(a
bi)(c dill i) Im(c i)
!m(a (F
tl
bd::f.
th
Z2)m
ir
::f.
btl.
and i.
(7)Zi"kZ1
fo al
ositiv integers s,
n. Recall
iv in
rs
-i)
32 29
_i)2
80
up os
40i
where
10
_i)3
_i)5
_i)4
41i
and
p'
fo some intege
contradiction (I
and q'
p.).
ty
of th comple P.
(-I)i
la ly
is io te elon
to P. and
Z2
Input a,
di.
c, (a
Print "zl z2 Se diff=( c, Print Se prod=(
c,
d)
I-/'
Print
c"2
d"2 is no quotient
Else
d)/ (denom), (b
Se quot=( (a Print Endif Stop 32. prod=( XE
IS te
al
.2 an
d, (a
d)
nt
tati nj gate
) / denom))
d)
give th familiar lternatively on coul establis that (Z Z2)/2 is Iz2)/21 2through and ..
2(1+i)+(-3i)+3(1-2i)+5(-6)
U- 11' ;:i
(!=.!.+~i)
-z
.I;
;.
"";
;;
z2)/21·
11-(-~
1-
(-~+V;i)-(-~-~i)
10
10.=
i)
~i =~
(4
4i)\2
(4
4iW
1.
this parabola
ci
J.
ongm
±l,
with
intercepts
±~
and
intercepts
±~v'5.
20
8.I(a+bi)-II={(a-l)2+b2 .-
(_h)2
)2
z-
Ir
I r a + rbil
bi)1
/(ra)2
(rh)2
112)
/r2(a
IR.ezl la [Im
12.
a..
Iz
.J
(ZI)
(a
(ala2
btb-z) a~
(a
btb-z)
(-a2bt
a1b,)i
a~+bI
-~6,i =b.
(a
c.
(a
2i
bi
2i
(a
i)
(a
i)
/-
Re
13. (z)2_Z2=O=?(Z-z)(z+z)=O=? either:
0=
=?
0= 14
IZIZ212
=?
(ZlZ2)(ZlZ2)
(ZlZl)(Z2Z2)
is
re imaginary.
IZll21z212 is
(z)m
vi
k.
Also,
(z
-k
(z)k i. Since
zk
z-
Izl2
1,
1-
17 Zo
Zo z2
O=?
-a
ai
hi
(zm)
ax+byec cIa
-I (-alb)
triang es
ca
and
ai
-bl--J(a
equation from poin cIa perpendicula as Equating cIa by)I--J(a
nt by
+b
Zr.
(-a
Zr
ib)I--J(a ={ [( [2ic
{(
ib)/)I--J(a
by)I--J(a
)}(
_b (b-ai)(x-iy)]/(b+ai) ut
Let
1co
lh
i--J3/2
nth
(ajD(y2 i--J3/2)2
-i--J3/2)2
Y2 -i--J3/2
-(1/
i- J3 )(ai )- (y2+
i--J3/2)(
1r Hermitian At
A.
u"
ut Hermitian. Hermitian.
\-
-a-
IS
1.
.3
to
ar
%1+Z2=3
h.
2=
1- 2i
1-2i
1- 5i
\--\~
2. I Z I Z 2 Z 3 1
I(ZlZ2)Z31
IZ211
lllz211 21z212 Izml
Izlm
for
k.
Also,
5.
a.
b.
5V26
c.
d. b.
a.
4cis
1-13
-5
a.. '2C1S
h.3~~~)
c.
d.
e. 2v 2c g.
1;
C1S
f.
Vii.
CIS
( - U121 r )
Suppose 1 % 2 1 1%
21
9. It is a.vectoroflengt by rotating
arg
and angleofinclina.tion a.rg % + < / > ; it is obtained
arg
/o~.
arg
Zl Z2
(l
11.
arg
Z2
arg
Zl
Z2
~2 cis(rrl4)~(26)cis(-4tan-I(l/5))
(l arg(l rrl4
-.
arg
Zl
(l
i)(24 rl 4tan-
-tan-I(11239)
(l/5)
(1/239).
37r
a. 7r 7r
7r
d.
Arg
Zl
-z,
Z2
Ill. If
to
(1-1)
(;
arg
7r,
: :
tan-
Arg
(;),
Argz2
Arg
Zl
whic
Z2
571
="3
-71
correspond
2' (1
If
1)
tan-
(;
rr
71,
arg If
7r
2(1)
argz.
7r
If
2(-1)
If
is undefined
If If If
JX
argz.
Arg
y2
/..J
y2
Arg
-7r
Arg
71.
O.
(-z,)1
15..
1-
IZll
IZll
IZ21
n.. z,)
IZll
IZ21
I Z I I -l z 2 1
1%21), 21-1%11
Thus, I Z I I -l z 2 1
I-;~ IT vector f;
% 2 1 or
ZI
real n u m b e r
then ZI'!2
is real
lu
in
ZIZ2
cl%112.
%1'!i
.J ~'
arg
arg
arg( % 1 % 2 )
arg
arg
kr
By xa pl if Z -
kr Vector
ZI
and % 2 ) , which is true
1, %1
if
%1),
and only if
if and
1.
<===>
I~. arg
arg
arg
arg %It
vectors 2 and 2 -
and
,%3
Then
be
\ -
arg and z,. Let
3 -
21. s (
1'2sin ()2]
ri
(c
ri
r~
r~ ri
:=:::}
ll~
cis ()2)
ri
lcis ()2)
Im
r~
l2)
tan T2
)f 22. Assume
LZk k=l
~Llzkl k=l
n. Then
Zk
Zn
k=l
Ez·I+,zn' IZkl
IZnl
IZkl·
23.
m1 +m2+ mtZt
m2
Physica interpretation: If thre particles un
2,
and
Z3
circle
24. (See Exercise 14
Input
,y
y"2)
sgn(y)
arcoos(x/r) r,
ep
r,
to
Input r,t
cos(t), ep
sin(t)
(x,y)
to
"j(x,y)
25.
) )
iy
(x
Re 26.
1/(-
Z2 Z2·
If
Zl
icz,
leads Z2 1tI2
ZI ·Z =Re(ZIZ2)=Re(-ic(x and z, Z2.
-iY2)(X
(a) Im(ZIZ2)=Im((x -iYI)(X
+iY2))=X
(b) If ZI and Z2 ar If Im(ZIZ2)
parallel
Zl
CZ2
xl/YI
0, XIY2- 2Yl
I-\~
Zl Z)
icz-,
+iY2)=-CX
Im(ZIZ2)
x2/Y2
CX2Y2-X2CY2 ZI
CZ2
=0
is
EXERCISES 1.4:
1.
he Comple
Exponentia
a.
h. co
c.
ieco
cos sin 1)
sin( sin 1)
a. sin3
h. e3~+ cos
c.
si 2y
ie
-i1r/4
-e
b. 167re- i21r i31r
c.
i21r
4.
b.
2V2e i1
c. 5.
i1r/2
+ iY I arg(
::::V2e-i71r/12
2e i51r
7r/2-1)
:le
YI
iy
:e
arg eX iy :::: arg e" (e i9
sin
6.
: l eX l Y I
arg iy ::::
e- i9
i9
(ei9
e- i8
:i
e-i9)
2e i1
2i :e
X ( cos
sin
1r : :: : e X [c o s( y
b.
::::eX cis
:e
r)
:e
iy
:e
sin(y
cos
::::e
7 r ) : :: :
y)
isin(-y))
::::eX(cos(-y)
::::e
9.
Z)
Xc
sy :e : : : e n( x+ iy )
(ezt
10. z::::
::::
iy
: enz
e- nz
nz
~ith
2k7r,
O.
eO::::
I-I~
sin
:_e
11. a,c, 12.
2ri
is false because
e", sin f J ) 3
Im(
a.
cos J (
cos 8(
Im[cos
sin 6)
in
co
Im(
Im( cos
in f J ) 4
4cos
co
sin 6)
13.
a , s in '
sin 6]
~'..;,r ( . i 1 ~
cos 6 s
00II'
e-i28
i28
e-"), i2')
!(ei2
isin 1 1 1 ) ( C O S
[cos( -(
cos
-6)
2)
in(-8
cos
sin
14 Ye becaus
(cos
sin
(cos n6
f-lO
sine -(
sin s ( -6)
sin
6)
si 6) sin n6
cos
)]
nO
sin nO sin( -nO)
-nO) Xl
Zl
(cos Yl Xl
sin
(cos
de
os Ylco
sin 2)
in
lsi
cos
sin
2) Xl Zl
Zl
h.
Xl
eZ
eX
(COS(YI
+X
2)
sin(Yl
Y2))
Z2
(cos Yl (cos
sm sin Y2)' cos (cos YIco
) e
lsi
si YIcos
-ico Ylsin 2)
[COS(Y
6.
1nr iB
iO
cos t, sin
sin
271",w hic
[z
il
counterclockwise.
traverse
counterclockwise traverse
Iz
18.
give
e". traverse
h.
Y2)]
re iB
is cos
2)
(2
i)
tr
a. 200
lo
counterclockwise
0.4
ki
19
le2ki
20. (z
,1 .. ,n
1,
11' "1
e2(k+l)1I'i/nl
1)(1 z2
Z2
+...
circle. Th
ti
are
z")
zn
z"+1
z-
when
th
f.
equal.
Suppose
ill
O. Then
()
ill
z2
inB
ei28
1+c n2 and ill
-~ ': -
(cos
sm
: : "--, ; : --
2+
s8+ 2cosO
.sin
sin(
-z---------'--~----
sin(n
2cosO /2
1/2)0
.sin(n
2sin /2
1)0/2sin(n8/2) sin 8/
im s8
1- (cosO 1-
sin )n
nO)2
(1- cosO)
i(sinO
2- 2c
n(
cosO
)2
n8)/2
n())
sin (n()/2)
sin(n()/2)
0)j2
n.
Z2
.1
JeinOdO
22.
21r
JeodO i21rn
21[,
0, j a r ·
-n
"#
1.
li>t
cosm9
iv
rs
ZR
isin(n -1)9)][r(cos
sine
)9co
sin
)9]
cas(
sin nO
rm(cosmO
sin m9 sin
cos( -mOl 3.
sine -mO» is
mArg
2 k7 l
0,
n. arg(zn-l)
l)Argz
argz
argz 2k7l"
2k71"
7 ( co -- 7 7 1 " . .
/i;3
V\)-
18m_-
71r)
27
-64V3+64i
i)95
.J2)9S
(cos 9:11'
sin
9!1r)
~7(1_
2 ~
i)
isinO)]
-,
(.7r
b.
0,1,2,3,4 2k7r)
7r/2
C.
,k=0,1,2,3 7r
(.37r/4
f.
l+i
'
-
,
/4+~k)
(l+z) w,
27r
Iz
Iw
Iwl'5'
-rr/10.
10 -rr/20.
V'23
a.
b.
i,
c.
v'f'=Z
4ac
z=
4ac Jb 4ac jugates.
iJ-(b
4ac)
4ac
0,
-1
0.
1/4
exp
4) (z
(.11"
2z
2kr)
e (1 '1 1 ' 4 )
(k
e(5'II' 4)i)
11
(z
====>
Z5
11/n.
12.
Fo each
exp
1,1
0,3) 1,2) w, where
q1
-,
4),
e(2k1r/5)i,
1,2,3,4.
(.IJo
1,
is th constant dist nc
here
Arg
IZoll/n.
ifferenc in th ar uments of z:,lft for consecutive
Iz
~/n.
13
(2"'/3)i
=-"2+'2 -"2
w3
.v'3)
'T
IT
e('II'/2)i
-1)
14. (Z"')l/n.=
(lzl"'eim
argz
(mIJ
exp 1m
2h')
(-i)
O.
IzoI1/n.
(Izll/nexp
i(0+n
br
I z l m / n (cos :(
zm/n
(zl/n)m
isin :(
2b")
i)3/2= (V2)3/2 (3i/2)(-1r/4+2k7r>,
0,1,"""
0,1
0= 0~ + e2 1 tk i O O (z-I e 21 tk iI l0 0 ~ z (1 _e 27 tk i1 1 00 ) = ( e2 i t k il lO O + 1 ) /( e2 1 t k i / o o _I )= ( e 1 tk i / IO O + e "1 t k i / O O ) /( e1 t k i Il O O e 1 t k i/ IO O OO)
1 tk l O O )
k= 0 ,1 ,
,99.
will
w~
m£
w~
w~m-l)£
Wm Wm
-1
n. Then (o:j3)k
19.
(o:j3)mn
(a) F { z )
1/
o:mnj3mn
z o lk a rg (Z - z o ) 2=
1/(~-
(o:n)m(bm)n
1/
1.
g (Z -Z o
3= )+
Z = (±--J2
2i
called sum., diff, based on exerci 31 ec io 1.1. Also define subroutines called po ar an rectangu ar b a s e d on exercise 24 sect on 1.3. De in compsq t{x, as follows: Input x,y Set (r, t)-polar(x, y) Set newr=sqrt(r), newt=t/2 Set (newx, newy)=rectangular{newr,newt ut ut ew Stop qu
a. pr
en
Input b r b i cr, ci et (discr di crim i) prod(ln-, bi, br, bi )prod cr, ci et toprootr op oo i)=compsqr disc im disc im et (zlr zli)=quo (-ln toprootr, -b ar, ail Set (z2r, z2i)=quot( toprootr, ar, ail (x,y) (zlr, zli); "which is (r, t) pola (z r, zli) (x,y) =";(z2r,z2i)j "which is (r,t) ="; polar{ z2r, z2i) Stop (a) ±(3+i)
21.
(b ±(3+2i (e) ±(1+3i)
(d ±(2-i)
XE 1. Let
IS ZI
.6
(f ±(3-i)
anar Sets
be in
Choose a. point
zol· in Iz
%0 %0 %0 rp
set.
(c ±(5+i)
zll ZI
R. Then ZI
lZl-wl zll fI
2.
(1
c.O
Iz-21
7r/4
Irnz :::; 2)
1-30
c,
5. a, 6.
(1
a.
i)
b. and d.
for all real
and
=2
e.
iy and
f.
iy for all real
-1
7. a, b, c, d, 8. a,
z,,} is
{Z17 max I Z j l ,
where
p,
n.
1%01
IZoI P+Po
12 Since
every
S. contains lo
.t
in S. Thus .to
o,
S.
:: :}
is in C\ C\ is op n.
¢::::}
S.
{::::::>
S. Then S. Because
is losed,
C\ 15.
8n
BuT
T.
S,
If
S, is
T.
if
p, (in
T. Hence 17
T.
T) that
o. Counte ex mple Iz
-1
o,
l,
Im
and
and
T,
iv S. Likewise and T. Therefore
and and
T.
8y {z
{z
IS
2} becaus
in is
20.
v(z,y)":" u(z,y) -z1l at all lJ lJ 8v 8v 8y u(z,1I)
Then
v(z, y) c.
1-"3
c,
l)
log(x
Let Let
Zk
in in
%Ic
i,llr
to for Ie
Let
D/c+l.
i 1 J k to
IH
or on al wi hout leavin
eg
to
iY
iY
iY
illk to iYk+l Dk+l, and om x~ iY'*l to X'*1 iYk+! Dk+l' Xk+i,lk to X k + 1 + iY k + l al gmen ou ea D k + l (and
by horizontal
). D,
gmen
iY
23.
continuum.
(a)
iy
2.
and xo)t
iY
Xl
YI
Yo)t
b.
ch
xo Zb
Z2
in
u(xI,
U (X 2
ar me rizati
xerc se
.7 (xi, X2.X3)=(0 1,0 (xi, X2.X3)=(12/101 -16/101, 99/1(1
(XI.X2.X3) (-12/25,16/25, (XI.X2.X3)=
-3/5 _1)
l)
X/(X2 1I£... [2X/(X y2 +1) 2Y/(X
(KI.K2.K3)=
-l)/(X2
(Xnl xni, xn-) (x X2.-X3) (b) -11~ (XXI.XX2.XX3) (-Xl -X2.-X3) z+ 1I~1/~(1 ,X must as
hr
gh -11~
In
w, -llw.
-1I~.
I=
,1~
{[2x/( +lzI2)]2 [2y/( +lzI2]2}.5 21zl/( +lzl'). =1 plane. This give th rati of sides: X[z,oo]/~( +lzI2)= {2Izl/(1+lzI2) li I. Solvin
yields X[z,oo]= 2/~(1+lzl\ lis Iz-wl2 l+lz +1+lwI 2~(1+lzI2)~(1+lwI2)cosa. coso [2+lzI2+lwI2_lz-wI2]/[2~(1+lzI2)~(1+lwI2) In
,w
thes solu ions he comp ex conjugat of is indicated IZ-WI
(2/ (1+lzI2»2+(2/~ (1+lwI2» -2 {(4)/[2 (1+lzI2) (1+lwI2)]}coso;
IZ-WI
2Iz-wl/~(1+lzI2)~(1 +lwl").
