Table Of ContentStudent 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·
