CHAPTER 46 DE MOIVRE’S THEOREM
EXERCISE 192 Page 522
1. Determine in polar form: (a) [1.5∠15°] 5
(b) (1 + j2) 6
(a) [1.5∠15°] = 1.55 ∠5 ×15° = 7.594∠75°
5
5 ∠63.435°
(b) 1 + j2 =
Hence, (1 + j2) 6 =
(
) ( 5 ) ∠6 × 63.435=°
6
6
5 ∠63.435° =
2. Determine in polar and Cartesian forms: (a) [3∠41°] 4
125∠380.61° = 125∠20.61°
(b) (–2 – j) 5
(a) [3∠41°] = 34 ∠4 × 41° = 81∠164° = 8 cos 164° + j8 sin 164° = –77.86 + j22.33
4
(b)
( −2 − j=
)
5
5
 5∠ − 153.435°=


( 5 ) ∠5 × −153.435°
5
= 55.90∠–767.175° = 55.90∠–47.18°
= 55.90 cos – 47.18° + j55.90 sin –47.18°
= 38 – j41
3. Convert (3 – j) into polar form and hence evaluate (3 – j) 7 , giving the answer in polar form.
 1
32 + 12 ∠ tan −1  −  = 10∠ − 18.43°
 3
(3 – j) =
7

Hence, ( 3 − =
j )  10∠ − 18.43°=

7
( 10 ) ∠7 × −18.43° = 3162∠–129°
7
4. Express in both polar and rectangular forms: (6 + j5) 3
( 6 + j5=
)
3
3
 61 ∠39.806°=



( 61 ) ∠3 × 39.806°
3
= 476.4∠119.42°
779
© 2014, John Bird
= 476.4 cos 119.42° + j476.4 sin 119.42°
= –234 + j415
5. Express in both polar and rectangular forms: (3 – j8) 5
8)
( 3 − j=
5
5
 73∠ − 69.444=
°

(
)
5
73 ∠5 × −69.444° = 45530∠–347.22° = 45 530∠12.78°
= 45 530 cos 12.78° + j45 530 sin 12.78°
= 44 400 + j10 070
6. Express in both polar and rectangular forms: (–2 + j7)4
From the diagram below, r =
=
° 105.945°
θ 180° − 74.054=
and
Hence,
7
1
α tan −=
and =
  74.054°
2
22 + 7 2 =
53
( −2 + j 7=
)
4
4
 53∠105.945°=


(
)
4
53 ∠4 ×105.945°
= 2809∠423.78° = 2809∠63.78°
=
2809∠63.78
( 2809 cos 63.78° + j 2809sin 63.78° )
= 1241 + j 2520
7. Express in both polar and rectangular forms: (–16 – j9) 6
From the diagram below, r = 162 + 92 =
337
=
θ 180° + 29.358=
° 209.358°
and
Hence,
9
and =
α tan −1=
  29.358°
 16 
9)
( −16 − j=
6
6
 337∠209.358=
°

(
) ∠6 × 209.358°
6
337
780
© 2014, John Bird
= 38.27 ×106 ∠1256.148° = (38.27 ×106 )∠176.15°
(38.27
=
×106 )∠176°9 ' 106 ( 38.27 cos176.15° + j 38.27 sin176.15° )
= 106 (−38.18 + j 2.570)
781
© 2014, John Bird
EXERCISE 193 Page 524
1. Determine the two square roots of the given complex numbers in Cartesian form and show the
results on an Argand diagram: (a) 1 + j
(a) 1 + j =
 2 ∠45° =  2∠45°

 

The first root is:
( )
2
(b)
1
∠ ×=
45° 1.1892∠22.5
=
° (1.099 + j 0.455)
2
(1 + j ) =
±(1.099 + j 0.455) as shown in the Argand diagram below.
j = 0+ j =
1
[ 1∠90°] = [1∠90°] 2
1
The first root is:
(1) 2 ∠
and the second root is:
Hence,
1
2
1.1892∠(22.5° + 180°) = (−1.099 − j 0.455)
and the second root is:
Hence,
1
2
(b) j
1
× 90° = 1∠45° = (0.707 + j 0.707)
2
1∠(45° + 180°) = (−0.707 − j 0.707)
j=
±(0.707 + j 0.707) as shown in the Argand diagram below.
2. Determine the two square roots of the given complex numbers in Cartesian form and show the
results on an Argand diagram: (a) 3 – j4
(b) –1 – j2
782
© 2014, John Bird
(a)
3 − j 4=
1
[5∠ − 53.13°]= [5∠ − 53.13°] 2
The first root is:
1
1
5 2 ∠ × −53.13°= 2.236∠ − 26.57°= (2 − j1)
2
52.236∠(−26.57° + 180°) = (−2 + j1)
and the second root is:
(3 − j 4) =
±(2 − j ) as shown in the Argand diagram of Figure (a) below
Hence,
(a)
(b)
1
(b)
 5 ∠243.435° 2
−1 − j 2 = 5 ∠243.435° =


The first root is:
( )
and the second root is:
Hence,
5
1
2
1
∠ × 243.435° = 1.495∠121.72° = (−0.786 + j1.272)
2
1.495∠(121.72° − 180
=
°) 1.495∠ − 58.28
= (0.786 − j1.272)
(−1 − j 2) =±(0.786 − j1.272) as shown in the Argand diagram of Figure (b)
above
3. Determine the two square roots of the given complex numbers in Cartesian form and show the
results on an Argand diagram:
(a) 7∠60°
1
(a) The first root of 7∠60° is:
and the second root is:
Hence,
(b) 12∠
3π
2
1
× 60=
°
2
7 ∠30=
° (2.291 + j1.323)
7∠(30° + 180°) =
7∠210 = (−2.291 − j1.323)
(7)2 ∠
7∠60° = ±(2.291 + j1.323) as shown in the Argand diagram below
783
© 2014, John Bird
(b) The first root of 12∠
3π
is:
2
12∠(
and the second root is:
Hence,
12∠
1 3π
× °=
2 2
1
(12 ) 2 ∠
3π
)
− π=
4
12 ∠
3π
= (−2.449 + j 2.449)
4
π
12∠ − = (2.291 − j 2.449)
4
3π
=±(−2.449 + j 2.449) as shown in the Argand diagram below
4
4. Determine the modulus and argument of: (3 + j4) 1/3
1
1
( 3 + j 4 ) 3 = ( 5∠53.13° ) 3 =
1
1
5 3 ∠ × 53.13°=
3
3
5∠17.71°= 1.710∠17.71°
Hence, the modulus is 1.710, and the arguments are 17.71°,
and
137.71° +
360°
= 257.71°
3
( −2 + j )=
1
4
 5∠153.435°=


( 5)
360°
= 137.71°,
3
since the three roots are equally displaced by 120°
5. Determine the modulus and argument of:
1
4
17.71° +
1
4
There are four roots equally displaced
(–2 + j) 1/4
1
∠ ×153.435=
° 1.223∠38.36°
4
360°
, i.e. 90° apart
4
Hence, the modulus is 1.223, and the arguments are: 38.36°,
784
38.36° + 90° = 128.36°,
© 2014, John Bird
128.36° + 90° = 218.36°
and
6. Determine the modulus and argument of:
( 61 )
1
1
2
2
( −6 − j5=
)  61∠219.806°=

There are two roots equally displaced
1
2
218.36° + 90° = 308.36°
(–6 – j5) 1/2
1
∠ × 219.806
=
° 2.795∠109.90°
2
360°
2
i.e. 180° apart.
Hence, the modulus is 2.795, and the arguments are: 109.90° and 109.90° + 180° = 289.90°
7. Determine the modulus and argument of:
2
( 4 − j 3) 3 =
−
2
[5∠ − 36.87°] 3 =
−
( 5)
−
2
3
(4 – j3) −2/3
2
∠ − × −36.87°= 0.3420∠24.58°
3
There are three roots equally displaced
360°
, i.e. 120° apart.
3
Hence, the modulus is 0.3420, and the arguments are: 24.58°,
and
24.58° + 120° = 144.58°,
144.58° + 120° = 264.58°
8. For a transmission line, the characteristic impedance Z 0 and the propagation coefficient γ are given
by:
Z0 =
 R + jω L 


 G + jωC 
and γ = ( R + jω L )( G + jωC ) 
Given R = 25 ohms, L = 5 × 10–3 henry, G = 80 × 10–6 siemens, C = 0.04 × 10–6 farads and
ω = 2000π rad/s, determine, in polar form, Z 0 and γ.
R + jω L =
25 + j (2000π ) ( 5 ×10−3 ) =
25 + j 31.416 =
40.15∠51.49°
G + jωC =80 ×10−6 + j (2000π ) ( 0.04 ×10−6 ) =10−6 (80 + j 251.33) =263.755 ×10−6 ∠72.34°
Hence,
=
Z0
 R + jω L 
=

 G + jωC 
40.15∠51.49°


=


−
6
 263.755 ×10 ∠72.34° 
= 152 224.6 ∠
(152 224.6∠ − 20.85° )
1
( −20.85° ) = 390.2∠–10.43° Ω
2
785
© 2014, John Bird
γ=
( R + jω L )( G + jωC )  =
=
( 40.15∠51.49° )( 263.755 ×10−6 ∠72.34° ) 
∠123.83° )
( 0.01058976=
1
0.01058976 ∠ × 123.83°
2
= 0.1029∠61.92°
786
© 2014, John Bird
EXERCISE 194 Page 526
1. Change (5 + j3) into exponential form.
(5 + j3) = 5.83∠30.96° or 5.83∠0.54 rad
5.83∠0.54 ≡ 5.83 e j 0.54
and
2. Convert (–2.5 + j4.2) into exponential form.
(–2.5 + j4.2) = 4.89∠120.76° or 4.89∠2.11rad
4.89∠2.11 ≡ 4.89 e j 2.11
and
3. Change 3.6e j 2 into Cartesian form.
3.6e j 2 = 3.6∠2 rad = 3.6 cos 2 + j 3.6 sin 2
= –1.50 + j3.27
4. Express 2e 3+ j π /6 in (a + jb) form.
/6
2 e3 + j π=
/6
( 2 e3 )( e j π =
)
2 e3 ∠π / 6 rad = 2 e3 cos(π / 6) + j 2 e3 sin(π / 6)
= 34.79 + j20.09
5. Convert 1.7e 1.2− j 2.5 into rectangular form.
j 2.5
1.7e1.2 −=
j 2.5
(1.7e1.2 )( e− =
)
1.7e1.2 ∠ − 2.5 rad = 1.7 e1.2 cos(−2.5) + j 1.7 e1.2 sin(−2.5)
= –4.52 – j3.38
6. If z = 7e j 2.1 , determine ln z (a) in Cartesian form, and (b) in polar form.
(a) If z = 7 e j 2.1 then ln z = ln ( 7 e j 2.1
=
) ln 7 + ln e j 2.1 = ln 7 + j2.1 in Cartesian form
(b) ln 7 + j2.1 = 2.86∠47.18° or 2.86∠0.82 rad
787
© 2014, John Bird
7. Given z = 4e 1.5− j 2 , determine ln z in polar form.
If z = 4e 1.5− j 2 then ln z = ln ( 4e1.5− j 2 ) = ln 4 + ln e1.5− j 2
= ln 4 + 1.5 – j2 = (ln 4 + 1.5) – j2
= 2.886 – j2
= 3.51∠–0.61 or 3.51∠–34.72°
8. Determine in polar form (a) ln(2 + j5)
(a) ln(2 + j5) = ln
(
)
29∠1.19 = ln
(
(b) ln(– 4 – j3).
)
29 e j 1.19 = ln 29 + ln e j 1.19
= ln 29 + j 1.19 = 1.6836 + j 1.19
= 2.06∠35.25° or 2.06∠0.615 rad
(b) ln(–4 – j3) = ln ( 5∠216.87°=
) ln ( 5∠3.785=) ln ( 5e j 3.785 )
= ln 5 + j 3.785 =1.6094 + j 3.785
= 4.11∠66.96° or 4.11∠1.17 rad
9. When displaced electrons oscillate about an equilibrium position the displacement x is given by
x = Ae
the equation:
 ht
+j
−
2m

( 4 m f − h2 ) 
2 m−a


Determine the real part of x in terms of t, assuming ( 4mf − h 2 ) is positive.
=
x Ae
 ht
+j
−
2m

( 4 mf − h2 )

t

2
 − ht   j 42mfm −−ah t   − 2htm   4mf − h 2
=  Ae 2 m   e =
  Ae
 ∠ 


  2m − a
 
2m−a
= Ae
Hence, the real part is:
−
Ae
ht
2m
−
 4mf − h 2
cos 
 2m − a

ht
2m

 t

ht

 4mf − h 2
−
 t + jAe 2 m sin 

 2m − a
 4mf − h 2
cos 
 2m − a


 t


 t

788
© 2014, John Bird
EXERCISE 195 Page 528
1. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
z =2
(a) If z = x + jy, then on an Argand diagram as shown below, the modulus z,
=
z
In this case,
x2 + y 2
x2 + y 2 =
2 from which, x 2 + y 2 =
22
From Chapter 28, x 2 + y 2 =
4 is a circle, with centre at the origin and with radius 2
(b) The locus (or path) of z = 2 is shown below
2. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
z =5
(a) Modulus z, =
z
In this case,
x2 + y 2
x2 + y 2 =
5 from which, x 2 + y 2 =
52
From Chapter 28, x 2 + y 2 =
25 is a circle, with centre at the origin and with radius 5
(b) The locus (or path) of z = 5 is shown below
789
© 2014, John Bird
3. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
arg (z – 2) =
(a) If arg (z – 2) =
π
3
, then arg (x + jy – 2) =
i.e.
π
3
π
3
π
arg[( x − 2) + jy ] =
3
 y
From the Argand diagram in Problem 1 above, θ = tan −1  
x
 y  π
tan −1 
=
 x−2 3
Hence, in this example,
Thus,
if
i.e.
y
= 3 , then y =
x−2
(b) Hence, the locus of arg (z – 2) =
π
3
i.e.
 y
arg z = tan −1  
x
π
y
= tan
= tan 60° =
3
x−2
3
3 (x – 2)
is a straight line y = 3 x – 2 3 or y =
3 (x – 2)
as shown below.
4. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
arg (z + 1) =
790
π
6
© 2014, John Bird
(a) If arg (z + 1) =
π
6
, then arg (x + jy +1) =
i.e.
π
6
π
arg[( x + 1) + jy ] =
6
 y
From the Argand diagram in Problem 1 above, θ = tan −1  
x
 y  π
tan −1 
=
 x +1  6
Hence, in this example,
Thus,
if
i.e.
 y
arg z = tan −1  
x
1
y
π
= tan
= tan 30° =
x +1
6
3
i.e.
1
1
y
=
, then y =
(x +1)
x +1
3
3
(b) Hence, the locus of arg (z + 1) =
π
6
is a straight line y =
1
1
x+
3
3
or y =
1
(x + 1)
3
as shown below.
5. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
z−2 =
4
(a) If z = x + jy, then
z − 2 = x + jy − 2 = ( x − 2) + jy = 4
On the Argand diagram shown in Problem 1,
Hence, in this case,
or
=
z
x2 + y 2
z − 2 =( x − 2) 2 + y 2 =
4 from which, ( x − 2) 2 + y 2 =
42
x 2 − 4 x + 4 + y 2 − 16 =
0
i.e. x 2 − 4 x − 12 + y 2 =
0
From Chapter 28, ( x − a ) 2 + ( y − b) 2 =
r 2 is a circle, with centre (a, b) and radius r
Hence,
( x − 2) 2 + y 2 =
42 is a circle, with centre (2, 0) and radius 4
(b) The locus of z − 2 =
4 is shown below
791
© 2014, John Bird
6. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
z +3 =
5
(a) If z = x + jy, then
z + 3 = x + jy + 3 = ( x + 3) + jy = 5
On the Argand diagram shown in Problem 1,
Hence, in this case,
or
=
z
x2 + y 2
z + 3 =( x + 3) 2 + y 2 =
52
5 from which, ( x + 3) 2 + y 2 =
x 2 + 6 x + 9 + y 2 − 25 =
0
i.e. x 2 + 6 x − 16 + y 2 =
0
From Chapter 28, ( x − a ) 2 + ( y − b) 2 =
r 2 is a circle, with centre (a, b) and radius r
Hence,
( x + 3) 2 + y 2 =
52 is a circle, with centre (–3, 0) and radius 5
(b) The locus of z + 3 =
5 is shown below
7. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
z +1
=3
z −1
(a) z + 1 = x + jy + 1 = (x + 1) + jy
z – 1 = x + jy –1 = (x – 1) + jy
792
© 2014, John Bird
Hence,
z +1
=
z −1
( x + 1) 2 + y 2
= 3
( x − 1) 2 + y 2
( x + 1) + jy
=
( x − 1) + jy
and squaring both sides gives:
( x + 1) 2 + y 2
=9
( x − 1) 2 + y 2
from which,
( x + 1) 2 + y 2 = 9[( x − 1) 2 + y 2 ]
2
x 2 + 2 x + 1 + y=
9[ x 2 − 2 x + 1 + y 2 ]
x 2 + 2 x + 1 + y 2= 9 x 2 − 18 x + 9 + 9 y 2
0= 8 x 2 − 20 x + 8 + 8 y 2
i.e.
8 x 2 − 20 x + 8 + 8 y 2 =
0
and dividing by 4 gives:
2 x2 − 5x + 2 + 2 y 2 =
0 which is the equation of the locus
x2 −
Rearranging gives:
5
x + y2 =
−1
2
Completing the square (see Chapter 14) gives:
2
5  25

2
−1
x−  − + y =
4
16


2
i.e.
5
25

2
 x −  + y =−1 +
4
16

i.e.
5
9

2
x−  + y =
4
16

i.e.
5

3
2
x−  + y =
  which is the equation of a circle
4

4
(b) Hence the locus defined by
z +1
= 3 is a circle of centre
z −1
2
2
2
793
3
5 
 , 0  and radius
4
4 
© 2014, John Bird
8. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
z −1
= 2
z
(a) z – 1 = x + jy –1 = (x – 1) + jy
z −1
Hence, =
z
( x − 1) + jy −
=
x + jy
and squaring both sides gives:
( x − 1) 2 + y 2
=
x2 + y 2
2
( x − 1) 2 + y 2
=2
x2 + y 2
from which,
( x − 1) 2 + y 2 = 2[ x 2 + y 2 ]
2
x2 − 2 x + 1 + y=
2x2 + 2 y 2
0 = x2 + 2 x −1 + y 2
i.e.
x 2 + 2 x − 1 + y 2 =0 which is the equation of the locus
x2 + 2 x + y 2 =
1
Rearranging gives:
Completing the square (see Chapter 14) gives:
( x + 1)
−1+ y2 =
1
( x + 1)
i.e.
i.e.
2
( x + 1)
(b) Hence the locus defined by
2
2
+ y2 =
2
( )
+ y 2 =2
2
which is the equation of a circle
z −1
= 2 is a circle of centre ( −1, 0 ) and radius
z
794
2
© 2014, John Bird
9. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
arg
z −1
π
=
z
4
 z − 1   ( x − 1) + jy  [( x − 1) + jy ] ( x − jy ) x( x − 1) − jy ( x − 1) + jxy + y 2
(a) 
=
=

 =
x2 + y 2
( x + jy )( x − jy )
 z   x + jy 
x 2 − x − jxy + jy + jxy + y 2 x 2 − x + jy + y 2
= =
x2 + y 2
x2 + y 2
=
Since arg
z −1
π
=
z
4
then
y 2 ) + jy
( x 2 − x +=
x2 + y 2
y

 x2 + y 2
tan −1  2
2
 x −x+ y
 x2 + y 2

x2 − x + y 2
jy
+
2
2
2
x +y
x + y2

 π
=
 4


i.e.

 π
y
tan −1 
=
2
2
 x −x+ y  4
from which,
π
y
= tan
= 1
x2 − x + y 2
4
Hence,
y = x2 − x + y 2
 z −1  π
2
2
Hence, the locus defined by arg 
0
 = is: x − x − y + y =
 z  4
Completing the square gives:
2
2
1 
1
1

x−  + y−  =
2 
2
2

1
1 1
which is a circle, centre  ,  and radius
2
2 2
(b) A sketch of the locus is shown below
795
© 2014, John Bird
10. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
arg
z+2
π
=
z
4
 z + 2   ( x + 2) + jy  [( x + 2) + jy ] ( x − jy ) x( x + 2) − jy ( x + 2) + jxy + y 2
(a) 
=
=

 =
x2 + y 2
( x + jy )( x − jy )
 z   x + jy 
=
=
Since arg
z+2
π
=
z
4
then
x 2 + 2 x − jxy − j 2 y + jxy + y 2 x 2 + 2 x − j 2 y + y 2
=
x2 + y 2
x2 + y 2
y2 ) − j2 y
( x 2 + 2 x +=
x2 + y 2
−2 y

 x2 + y 2
tan −1  2
2
 x + 2x + y
 x2 + y 2

x2 + 2x + y 2
j2 y
−
2
2
2
x +y
x + y2

 π
=
 4


i.e.

 π
−2 y
tan −1 
=
2
2
 x + 2x + y  4
from which,
π
−2 y
= tan
= 1
x2 + 2x + y 2
4
Hence,
−2 y = x 2 + 2 x + y 2
 z+2 π
2
2
Hence, the locus defined by arg 
0
 = is: x + 2 x + 2 y + y =
 z  4
Completing the square gives:
2
2
2
( x + 1) + ( y + 1) =
which is a circle, centre ( −1, − 1) and radius
2
(b) A sketch of the locus is shown below.
796
© 2014, John Bird
11. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
z+ j = z+2
Since z + j = z + 2
then
x + j ( y + 1) = ( x + 2) + jy
and
x 2 + ( y + 1) 2 =
( x + 2) 2 + y 2
Squaring both sides gives: x 2 + ( y + 1) 2 = ( x + 2) 2 + y 2
i.e.
from which,
i.e.
or
x2 + y 2 + 2 y + 1 = x2 + 4x + 4 + y 2
2y + 1 = 4x + 4
2y = 4x + 3
y = 2x + 1.5
Hence, the locus defined by z + j = z + 2 is a straight line: y = 2x + 1.5
12. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
z−4 = z−2j
Since z − 4 = z − 2 j
then
( x − 4) + jy = x + j ( y − 2)
and
( x − 4) 2 + y 2 =
x 2 + ( y − 2) 2
Squaring both sides gives: ( x − 4) 2 + y 2 = x 2 + ( y − 2) 2
i.e.
x 2 − 8 x + 16 + y 2 = x 2 + y 2 − 4 y + 4
797
© 2014, John Bird
from which,
–8x + 16 = –4y + 4
i.e.
4y = 8x – 12
or
y = 2x – 3
Hence, the locus defined by z − 4 = z − 2 j is a straight line: y = 2x – 3
13. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
z − 1 =z
Since z − 1 =z
then
( x − 1) + jy = x + jy
and
( x − 1) 2 + y 2 =
x2 + y 2
Squaring both sides gives: ( x − 1) 2 + y 2 = x 2 + y 2
i.e.
from which,
x2 − 2x + 1 + y 2 = x2 + y 2
–2x + 1 = 0
i.e.
1 = 2x
or
x=
1
2
Hence, the locus defined by z − 1 =z is a straight line: x =
798
1
2
© 2014, John Bird
799
© 2014, John Bird