Student Solutions Manual to accompany Complex Variables and Applications

170 Pages·2003·3.49 MB·English

by  James Ward Brown

Student Solutions Manual
for use with
Complex Variables
and Applications
Seventh Edition
Selected Solutions to Exercises in Chapters 1-7
by
James Ward Brown
Professor of Mathematics
The University of Michigan- Dearbom
Ruel V. Churchill
Late Professor of Mathematics
The University of Michigan
1B
Higher Education
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Table o/ Contents
Chapter 1 ......................................................................................................... 1
Chapter 2 ....................................................................................................... 22
Chapter 3 ....................................................................................................... 3 5
Chapter 4 ....................................................................................................... 53
Chapter 5 ....................................................................................................... 7 5
Chapter 6 ....................................................................................................... 94
Chapter 7 .. . . .. .. .. .. .. .. .. .. . . . .. . . .. . .. .. . . . .. .. .. . . . . . .. .. . .. . . .. . . .. .. . .. . .. .. . .. . .. .. . . . . . . . . . . . . . . . .. .. . 118
1
"COMPLEX VARIABLES ANO APPLICATIONS" (7/e) by Brown and Churchill
Chapter 1
SECTION2
l. (a) (,fi -i)-i(l-..fii) = ..fi -i-i -..fi =-2i;
(b) (2,-3)(-2,1) = (-4 + 3,6 + 2) = (-1,8);
~) (10,0>(1",
(e) (3,1)(3,-1)(} 1 = ~) = (2,1).
1
2. (a) Re(iz)=Re[i(x+iy)]=Re(-y+ix)=-y=-Imz;
(b) Im(iz) = Im[i(x + iy)] = Im(-y + ix) = x = Rez.
2 =
3. (1 + z) (1 + z)(l + z) = (1 + z) · 1 + (1 + z)z = 1 · (1 + z) + z(l + z)
2
= l+z+z+z2 = 1+2z+z .
4. If z =Ú ± i, tllen z2 2z + 2 = (1 ± i)2 - 2(1 ± i) + 2 = ± 2i - 2 + 2i + 2 = O.
-
S. To prove tllat multiplication is commutati.ve, write
Z1"2 = (x1,Yi)(XvY2) = (x,xz -Y¡Yz, Y1X2 +X1Y2)
= (XzX¡ -yzy¡, YzX¡ + XzYt) = (Xz,Yz)(X¡,Yt) = Z2Z¡,
6. (a) To verify the associative law for addition, write
(z1 +z2)+Z3 =[(Xi,Y1)+(x2,Y2)]+(x3,Y3)= (x1 +x2,Y1 + Y2)+(x3,y3)
= ((X1 + Xz) + X3, (Y1 + Yz)+ Y3) =(X¡+ (x2 + X3), Y1 + (Y2 + J3))
= (X¡,y¡) + (Xz + X3, Y2 + y3) = (xpy¡) + [(X2,Y2) + (X3,y3)]
=z1 +(z2 +z3).
2
(b) To verify the distributive law, write
z(z1 + Z2) = (x,y)[(x1,Y1) + (x2,Y2)l = (x,y)(x1 + X2,Yi + Y2)
= (XX1 + XX2 - YY1 - YY2, YX1 + YX2 + XY1 + .xy2)
= (xx1 -yy¡ +xx2 -yy2, YX1 +xy1 + YX2 +xy2)
= (xx¡ - YYi, YX1 + .xy1) + (xx2 -yy2, YX2 +xy2)
= (x,y)(x¡,Ji) + (x,y)(X2,Y2) = ZZ¡ + ZZ2-
2
10. Toe problem here is to solve the equation z + z + 1 = O for z = (x,y) by writing
(x,y)(x,y) + (x,y) + (1,0) = (0,0).
Since
2
(x - y2 +X+ 1, 2.xy +y)= (0,0),
it follows that
x2 -y2 + x + 1 = O and 2xy + y = O.
By writing the second of these equations as (2x + l)y = O, we see that either 2x + 1 = O or
2
y= O. If y= O, the first equation becomes x + x + 1 = O, which has no real roots
(according to the quadratic formula). Hence 2x + 1 = O, or x = -1/2. In that case, the first
equation reveals that y2 = 3/4, or y= ±,['j/2. Thus
z =(x,y)=(-~. ± ~)-
SECTION3
l. (a) 1 + 2i + 2 - i = (1 + 2i)(3 + 4i) + (2 - i)(-5i) = -5 + IOi + -5 - IOi = _ 2.
3-4i Si (3-4i)(3 + 4i) (5i)(-5i) 25 25 5'
Si Si Si 1
(b) (1-i)(2-i)(3-i) = (1-3i)(3-i) = --1-0i = - 2;
(e) (I-i)4 =[(1-i)(l-i)]2 =(-2i)2 =-4.
2. (a) (-I)z=-z since z+(-l)z=z[l+(-l)]=z·O=O;
1 1 z z
(b) -=-·-=-=z (z;tO).
1
l / z z- z 1
3
6.
=(~J(~)=(~Jz(.!.)=(~)(zz- )=(~)·1 =~
7. 1
Z1Z
Z2Z Z2 Z Zz Z Z2 Z2 Z2
SECTION 4
2 .
1. (a) z = 2i, Z =--¡
1 2 3
y
X
y
o
X
4
(e) Z = (-3,1), z = (1,4)
1 2
X
(d ) Z¡ = X¡ + iy¡, Z2 = X¡ - iy¡
y
X
2. Inequalities (3), Sec. 4, are
Rez S IRezl S lzl and Imz S llmzl S lzl.
These are obvious if we write them as
3. In order to verify the inequality "V21zl ~ IRezl + llmzl, we rewrite it in the following ways:
"'12~x2 + y2 ~ lxl + lyl,
2(x2 + y2) ~ lxl2 + 21xllyl + lyl2,
lxl2-21xllyl + lyl2 ~ O,
(lxl-lyl)2 ~ O.
This last form of the inequality to be verified is obviously true since the left-hand side is a
perfect square.
5
4. (a) Rewrite lz -1 + il= 1 as ~ -(1-i)I = l. This is the circle centered at 1-i with radius l.
It is shown below.
o
5. (a) Write lz-4il+lz+ 4il= 10 as lz-4il+lz-(-4i)I= 10 to see that this is the locus of all
points z such that the sum of the distances from z to 4i and -4i is a constant. Such a
curve is an ellipse with foci ±4i.
(b) Write lz-ll=lz + il as lz - ll=lz -(-i)I to see that this is the locus of all points z such
that the distance from z to 1 is always the same as the distan ce to -i. Toe curve is,
tlten, the perpendicular bisector of the line segment from 1 to -i.
SECTION5
1. (a) z+3i=z+3i=z-3i;
= =
(b) iz iz -iz;
(e) (2+i)2 =(2+i)2 =(2-i)2 =4-4i+i2 =4-4i-1=3-4i;
(d) 1(2.z + 5)( "'2 - i)l=l2Z + 511"'2 - il=l2z + 5 h/2 + 1 = "1312z + 51.
2. (a) Rewrite Re{z - i) = 2 as Re[x + i(-y -1)] = 2, or x = 2. This is the vertical line
through the point z = 2, shown below.
y
o
2 X
6
i
lz-
~ ~
(b) Rewrite 12z-il= 4 as 21z- 1= 4, or 1= 2. This is the circle centered at with
radius 2, shown below.
3. w rite Z¡ = X¡ + iy¡ and Z2 = X2 + iy2. Then
Z¡ -z2 = (x1 + iy1)-(x2 + iy2) = (x1 - X2) + i(y¡ - y2)
=(X¡ -x2)-i(y¡-y2)=(x¡-iy¡)-(X2-iY2)=z1 -z2
and
z1z2 = (x1 + iy¡)(x2 + iy2) = (X1X2 - y¡y2) + i(y1x2 + X1y2)
= (X¡X2 - YiY2) - i(Y1X2 + X¡Ji) = (X¡ - iy¡ )(X2 - ÍY2) = Z1Z2.
(zz)(zz) zzzz
(b) z4 = z2z2 = z2 z2 = zzzz = = = Z4 •
6. (a)
~ =~= lz 1
(b) 1
zz liiz 1 lz 11z 1
2 3 3 2 3
8. In this problem, we shall use the inequalities (see Sec. 4)
Specifically, when lzl:S; l,
z z
1Re(2 + + z3)1 :S; 12 + + z31 :S; 2+1.zl +lz31 = 2+1zl+lzl3 :S; 2 + 1 + 1 = 4.
7
10. First write z4 - 4z2 + 3 = (z2 l)(z2 3). Then observe that when lzl= 2,
- -
and
Thus, when lzl = 2,
lz4 -4z2 + 31=1z2 -ll·lz2 -31 ~ 3· 1 = 3.
Consequently, when z líes on the circle lzl= 2,
I= <.!.
1 1
1z4 -4z2 +3 lz4 -4z2 +31- 3·
z
11. (a) Prove that z is real ~ = z.
z
( <=) Suppose that = z, so that x - iy = x + iy. This means that i2 y = O, or y = O.
Thus z = x + iO = x, or z is real.
z
(=>) Suppose that z is real, so that z = x + iO. Toen = x -iO = x + iO = z.
(b) Prove that z is either real or pure imaginary ~ z2 = z2.
(<=) Suppose that z2 =z2. Toen (x-iy)2 =(x+iy)2, or i4xy=O. But this can be
only if either x = O or y = O, or possibly x = y = O. Thus z is either real or pure
imaginary.
( =>) Suppose that z is either real or pure imaginary. If z is real, so that z = x, then
z2 = x2 = z2. If Z is pure imaginary, so that Z = iy, then z2 = ( -iy) 2 = (iy )2 = z2•
12. (a) We shall use mathematical induction to show that
(n = 2,3, ... ).
This is known when n = 2 (Sec. 5). Assuming now that it is true when n = m, we may
write
Z1 + Z2 +···+ Zm + Zm+l = (Z1 + Z2 +···+ Zm) + Zm+l
= (Z1 +z2 +···+zm)+zm+l
= Z1 + z'2+···+zm) + Zm+1
= z'1 + Z2+··-+zm + Zm+1·
8
(b) In the same way, we can show that
(n = 2,3, ... ).
This is true when n = 2 (Sec. 5). Assuming that it is true when n = m, we write
Z¡Zz ···ZmZm+l = (Z1Z2 ···zm)Zm+l = (Z1Z2 ···zm) Zm+l
= =
(z'iz2 ·•· zm)zm+i Z1Zz ·•· ZmZm+i ·
zz z
14. Toe identities (Sec. 5) =lzl2 and Rez = z + enable us to write lz-z 1= Ras
2 0
(z-z )(z-z = R2
0 0) ,
ZZ - (Z Zo + ZZo) + ZoZo = R2
,
zz
1z l2 - 2 Re( + 1z / = R2
0) •
;i
15. Since x = z; z and y = z z, the hyperbola x2 -y2 = 1 can be written in the following
ways:
2 2
2z + 2z = 1,
4
z2 +z2 = 2.
SECTION7
1. (a) Since
arg( i .)=argi-arg(-2-2i),
-2-2z
3
one value of arg( i .) is 1C -(- ,c), or Stc. Consequently, the principal value is
-2-2z 2 4 4
51C 31C
--2,c or --
4 ' 4·