APPENDIX A
Exercises
EA.1
Given Z 1  2  j 3 and Z 2  8  j 6, we have:
Z 1  Z 2  10  j 3
Z 1  Z 2  6  j 9
Z 1 Z 2  16  j 24  j 12  j 2 18  34  j 12
2  j 3 8  j 6 16  j 12  j 24  j 2 18

Z1 / Z2 

 0.02  j 0.36
8 j6 8 j6
100
EA.2
Z 1  1545   15 cos( 45  )  j 15 sin( 45  )  10.6  j 10.6
Z 2  10  150   10 cos( 150  )  j 10 sin( 150  )  8.66  j 5
Z 3  590   5 cos(90  )  j 5 sin(90  )  j 5
EA.3
Notice that Z1 lies in the first quadrant of the complex plane.
Z 1  3  j 4  32  4 2  arctan( 4 / 3)  553.13
Notice that Z2 lies on the negative imaginary axis.
Z 2   j 10  10  90 
Notice that Z3 lies in the third quadrant of the complex plane.
Z 3  5  j 5  52  52 (180   arctan( 5 /  5))  7.07 225   7.07   135 
EA.4
Notice that Z1 lies in the first quadrant of the complex plane.
Z 1  10  j 10  10 2  10 2  arctan(10 / 10)  14.1445   14.14 exp( j 45  )
Notice that Z2 lies in the second quadrant of the complex plane.
Z 2  10  j 10  10 2  10 2 (180   arctan( 10 / 10))
 14.14135   14.14 exp( j 135  )
1
EA.5
Z 1Z 2  (1030  )(20135  )  (10  20)(30   135  )  200(165  )
Z 1 / Z 2  (1030  ) /(20135  )  (10 / 20)(30   135  )  0.5( 105  )
Z 1  Z 2  (1030  )  (20135  )  (8.66  j 5)  ( 14.14  j 14.14)
 22.8  j 9.14  24.6  21.8
Z 1  Z 2  (1030  )  (20135  )  (8.66  j 5)  ( 14.14  j 14.14)
 5.48  j 19.14  19.9106
Problems
PA.1
Given Z 1  2  j 3 and Z 2  4  j 3, we have:
Z1  Z2  6  j 0
Z 1  Z 2  2  j 6
Z 1 Z 2  8  j 6  j 12  j 2 9  17  j 6
Z1 / Z2 
PA.2
2  j 3 4  j 3  1  j 18


 0.04  j 0.72
4  j3 4  j3
25
Given that Z 1  1  j 2 and Z 2  2  j 3, we have:
Z1  Z2  3  j 1
Z 1  Z 2  1  j 5
Z1 Z2  2  j 3  j 4  j 2 6  8  j 1
Z1 / Z2 
1  j2 2  j3  4  j 7


 0.3077  j 0.5385
2  j3 2  j3
13
2
PA.3
Given that Z 1  10  j 5 and Z 2  20  j 20, we have:
Z 1  Z 2  30  j 15
Z 1  Z 2  10  j 25
Z 1 Z 2  200  j 200  j 100  j 2 100  300  j 100
Z1 / Z2 
PA.4
PA.5
PA.6
10  j 5 20  j 20 100  j 300


 0.125  j 0.375
20  j 20 20  j 20
800
(a)
Z a  5  j 5  7.071  45  7.071 exp j 45 
(b)
Z b  10  j 5  11.18153.43  11.18 expj 153.43 
(c)
Z c  3  j 4  5  126.87   5 exp j 126.87  
(d)
Z d   j 12  12  90   12 exp j 90  
(a)
Z a  545  5 expj 45   3.536  j 3.536
(b)
Z b  10120   10 expj 120    5  j 8.660
(c)
Z c  15  90   15 exp j 90     j 15
(d)
Z d  1060   10 exp j 120    5  j 8.660
(a)
Z a  5e j 30  530   4.330  j 2.5
(b)
Z b  10e  j 45  10  45  7.071  j 7.071
(c)
Z c  100e j 135  100135  70.71  j 70.71



3
PA.7
(d)
Z d  6e j 90  690   j 6
(a)
Z a  5  j 5  1030   13.66  j 10
(b)
Z b  545  j 10  3.536  j 6.464
(c)
Zc 
1045 
1045 
2  8.13  1.980  j 0.283

3  j4
553.13
(d)
Zd 
15
 3  90    j 3

590

4