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© 2014, John Bird
779
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) [ ]5 51.5 15 1.5 5 15∠ ° = ∠ × ° = 7.594∠75° (b) 1 + j2 = 5 63.435∠ °
Hence, (1 + j2) 6 = ( ) ( )6 65 63.435 5 6 63.435 125 380.61∠ ° = ∠ × ° = ∠ ° = 125∠20.61°
2. Determine in polar and Cartesian forms: (a) [3∠41°] 4 (b) (–2 – j) 5
(a) [ ]4 43 41 3 4 41∠ ° = ∠ × ° = 81∠164° = 8 cos 164° + j8 sin 164° = –77.86 + j22.33
(b) ( ) ( )5552 5 153.435 5 5 153.435j − − = ∠− ° = ∠ ×− °
= 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.
(3 – j) = 2 2 1 13 1 tan3
− + ∠ −
= 10 18.43∠− °
Hence, ( ) ( )7773 10 18.43 10 7 18.43j − = ∠− ° = ∠ ×− ° = 3162∠–129°
4. Express in both polar and rectangular forms: (6 + j5) 3
( ) ( )3336 5 61 39.806 61 3 39.806j + = ∠ ° = ∠ × °
= 476.4∠119.42°
© 2014, John Bird
780
= 476.4 cos 119.42° + j476.4 sin 119.42°
= –234 + j415 5. Express in both polar and rectangular forms: (3 – j8) 5
( ) ( )5553 8 73 69.444 73 5 69.444j − = ∠− ° = ∠ ×− ° = 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 = 2 22 7 53+ = and 1 7tan 74.0542
α − = = °
and 180 74.054 105.945θ = °− ° = °
Hence, ( ) ( )4442 7 53 105.945 53 4 105.945j − + = ∠ ° = ∠ × °
= 2809 423.78∠ ° = 2809 63.78∠ °
( )2809 63.78 2809cos 63.78 2809sin 63.78j∠ = °+ °
= 1241 2520j+
7. Express in both polar and rectangular forms: (–16 – j9) 6
From the diagram below, r = 2 216 9 337+ = and 1 9tan 29.35816
α − = = °
and 180 29.358 209.358θ = °+ ° = °
Hence, ( ) ( )66616 9 337 209.358 337 6 209.358j − − = ∠ ° = ∠ × °
© 2014, John Bird
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= 638.27 10 1256.148× ∠ ° = 6(38.27 10 ) 176.15× ∠ °
( )6 6(38.27 10 ) 176 9 ' 10 38.27cos176.15 38.27sin176.15j× ∠ ° = °+ °
= 610 ( 38.18 2.570)j− +
© 2014, John Bird
782
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 (b) j
(a) 121 2 45 2 45j + = ∠ ° = ∠ °
The first root is: ( )12 12 45 1.1892 22.5 (1.099 0.455)
2j∠ × ° = ∠ ° = +
and the second root is: 1.1892 (22.5 180 ) ( 1.099 0.455)j∠ °+ ° = − −
Hence, (1 ) (1.099 0.455)j j+ = ± + as shown in the Argand diagram below.
(b) [ ] [ ]120 1 90 1 90j j= + = ∠ ° = ∠ °
The first root is: ( )12
11 90 1 45 (0.707 0.707)2
j∠ × ° = ∠ ° = +
and the second root is: 1 (45 180 ) ( 0.707 0.707)j∠ °+ ° = − −
Hence, (0.707 0.707)j j= ± + 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
© 2014, John Bird
783
(a) [ ] [ ]123 4 5 53.13 5 53.13j− = ∠− ° = ∠− °
The first root is: 12
15 53.13 2.236 26.57 (2 1)2
j∠ ×− ° = ∠− ° = −
and the second root is: 52.236 ( 26.57 180 ) ( 2 1)j∠ − °+ ° = − +
Hence, (3 4) (2 )j j− = ± − as shown in the Argand diagram of Figure (a) below
(a) (b)
(b) 121 2 5 243.435 5 243.435j − − = ∠ ° = ∠ °
The first root is: ( )12 15 243.435 1.495 121.72 ( 0.786 1.272)
2j∠ × ° = ∠ ° = − +
and the second root is: 1.495 (121.72 180 ) 1.495 58.28 (0.786 1.272)j∠ °− ° = ∠− = −
Hence, ( 1 2) (0.786 1.272)j j− − = ± − 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° (b) 12∠ 32π
(a) The first root of 7∠60° is: ( )12
17 60 7 30 (2.291 1.323)2
j∠ × ° = ∠ ° = +
and the second root is: 7 (30 180 ) 7 210 ( 2.291 1.323)j∠ °+ ° = ∠ = − −
Hence, 7 60 (2.291 1.323)j∠ ° = ± + as shown in the Argand diagram below
© 2014, John Bird
784
(b) The first root of 12∠ 32π is: ( )
12
1 3 312 12 ( 2.449 2.449)2 2 4
jπ π∠ × ° = ∠ = − +
and the second root is: 312 ( ) 12 (2.291 2.449)4 4
jπ ππ∠ − = ∠− = −
Hence, 312 ( 2.449 2.449)4
jπ∠ = ± − + as shown in the Argand diagram below
4. Determine the modulus and argument of: (3 + j4)1/3
( ) ( )1 1 1
33 3 313 4 5 53.13 5 53.13 5 17.71 1.710 17.713
j+ = ∠ ° = ∠ × ° = ∠ ° = ∠ °
Hence, the modulus is 1.710, and the arguments are 17.71°, 17.71° + 3603° = 137.71°,
and 137.71° + 3603° = 257.71° since the three roots are equally displaced by 120°
5. Determine the modulus and argument of: (–2 + j)1/4
( ) ( )111444
12 5 153.435 5 153.435 1.223 38.364
j − + = ∠ ° = ∠ × ° = ∠ °
There are four roots equally displaced 3604° , i.e. 90° apart
Hence, the modulus is 1.223, and the arguments are: 38.36°, 38.36° + 90° = 128.36°,
© 2014, John Bird
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128.36° + 90° = 218.36° and 218.36° + 90° = 308.36° 6. Determine the modulus and argument of: (–6 – j5)1/2
( ) ( )111222
16 5 61 219.806 61 219.806 2.795 109.902
j − − = ∠ ° = ∠ × ° = ∠ °
There are two roots equally displaced 3602° 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: (4 – j3) 2/3−
( ) [ ] ( )2 2 23 3 3
24 3 5 36.87 5 36.87 0.3420 24.583
j − − −− = ∠− ° = ∠− ×− ° = ∠ °
There are three roots equally displaced 3603° , i.e. 120° apart.
Hence, the modulus is 0.3420, and the arguments are: 24.58°, 24.58° + 120° = 144.58°,
and 144.58° + 120° = 264.58° 8. For a transmission line, the characteristic impedance Z0 and the propagation coefficient γ are given
by: Z0 = R j LG j C
ωω
+ +
and γ = ( )( )R j L G j Cω ω + +
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, Z0 and γ.
( )325 (2000 ) 5 10 25 31.416 40.15 51.49R j L j jω π −+ = + × = + = ∠ °
( )6 6 6 680 10 (2000 ) 0.04 10 10 (80 251.33) 263.755 10 72.34G j C j jω π− − − −+ = × + × = + = × ∠ °
Hence, ( )06
40.15 51.49 152 224.6 20.85263.755 10 72.34
R j LZG j C
ωω −
+ ∠ ° = = = ∠− ° + × ∠ °
= ( )1152 224.6 20.852
∠ − ° = 390.2∠–10.43° Ω
© 2014, John Bird
786
( )( ) ( )( )640.15 51.49 263.755 10 72.34R j L G j Cγ ω ω − = + + = ∠ ° × ∠ °
= ( ) 10.01058976 123.83 0.01058976 123.832
∠ ° = ∠ × °
= 0.1029∠61.92°
© 2014, John Bird
787
EXERCISE 194 Page 526
1. Change (5 + j3) into exponential form.
(5 + j3) = 5.83 30.96 or 5.83 0.54 rad∠ ° ∠
and 5.83∠0.54 ≡ 5.83 0.54e j
2. Convert (–2.5 + j4.2) into exponential form.
(–2.5 + j4.2) = 4.89 120.76 or 4.89 2.11rad∠ ° ∠
and 4.89∠2.11 ≡ 4.89 2.11e j
3. Change 3.6e 2j into Cartesian form.
3.6e 2j = 3.6∠2 rad = 3.6 cos 2 + j 3.6 sin 2 = –1.50 + j3.27 4. Express 2e 3 /6jπ+ in (a + jb) form.
( )( )3 /6 3 /6 32e 2e e 2e / 6 radj jπ π π+ = = ∠ = 3 32e cos( / 6) 2e sin( / 6)jπ π+
= 34.79 + j20.09
5. Convert 1.7e1.2 2.5j− into rectangular form.
( )( )1.2 2.5 1.2 2.5 1.21.7e 1.7e e 1.7e 2.5radj j− −= = ∠− = 1.2 1.21.7e cos( 2.5) 1.7e sin( 2.5)j− + −
= –4.52 – j3.38
6. If z = 7e 2.1j , determine ln z (a) in Cartesian form, and (b) in polar form.
(a) If 2.17 e jz = then ln z = ( )2.1 2.1ln 7e ln 7 ln ej j= + = ln 7 + j2.1 in Cartesian form (b) ln 7 + j2.1 = 2.86∠47.18° or 2.86∠0.82 rad
© 2014, John Bird
788
7. Given z = 4e1.5 2j− , determine ln z in polar form.
If z = 4e1.5 2j− then ln z = ln ( )1.5 24e j− = ln 4 + 1.5 2ln e j− = 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) (b) ln(– 4 – j3).
(a) ln(2 + j5) = ( ) ( )1.19 1.19ln 29 1.19 ln 29 e ln 29 ln ej j∠ = = +
= ln 29 1.19 1.6836 1.19j j+ = +
= 2.06∠35.25° or 2.06∠0.615 rad
(b) ln(–4 – j3) = ( ) ( ) ( )3.785ln 5 216.87 ln 5 3.785 ln 5e j∠ ° = ∠ =
= ln 5 3.785 1.6094 3.785j j+ = +
= 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
the equation: ( )24
2 2em f hht j
m m ax A − − + − =
Determine the real part of x in terms of t, assuming ( )24mf h− is positive.
( )22
44 2
2 22 2 2
4e e e e
2
mf hht j t mf hht htj tm m am m a m
mf hx A A A t
m a
− − + − − −− − − = = = ∠ −
= 2 2
2 24 4
e cos e sin2 2
ht htm m
mf h mf hA t jA t
m a m a− − − −
+ − −
Hence, the real part is: 2
24
e cos2
htm
mf hA t
m a− −
−
© 2014, John Bird
789
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,
2 2z x y= +
In this case, 2 2 2x y+ = from which, 2 2 22x y+ =
From Chapter 28, 2 2 4x y+ = 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, 2 2z x y= +
In this case, 2 2 5x y+ = from which, 2 2 25x y+ =
From Chapter 28, 2 2 25x y+ = is a circle, with centre at the origin and with radius 5
(b) The locus (or path) of z = 5 is shown below
© 2014, John Bird
790
3. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
arg (z – 2) = 3π
(a) If arg (z – 2) = 3π , then arg (x + jy – 2) =
3π
i.e.
arg[( 2) ]
3x jy π− + =
From the Argand diagram in Problem 1 above, 1tan yx
θ − =
i.e. arg z = 1tan yx
−
Hence, in this example, 1tan2
yx
− −
= 3π i.e.
2y
x − = tan
3π = tan 60° = 3
Thus, if 2
yx −
= 3 , then y = 3 (x – 2)
(b) Hence, the locus of arg (z – 2) =3π 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) = 6π
© 2014, John Bird
791
(a) If arg (z + 1) =6π , then arg (x + jy +1) =
6π
i.e.
arg[( 1) ]
6x jy π+ + =
From the Argand diagram in Problem 1 above, 1tan yx
θ − =
i.e. arg z = 1tan yx
−
Hence, in this example, 1tan1
yx
− +
= 6π i.e.
1y
x + = tan
6π = tan 30° = 1
3
Thus, if 1
yx +
= 13
, then y = 13
(x +1)
(b) Hence, the locus of arg (z + 1) =6π is a straight line y = 1
3x + 1
3 or y = 1
3(x + 1)
as shown below.
5. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
2 4z − = (a) If z = x + jy, then 2 2 ( 2) 4z x jy x jy− = + − = − + =
On the Argand diagram shown in Problem 1, 2 2z x y= +
Hence, in this case, 2z − = 2 2( 2) 4x y− + = from which, 2 2 2( 2) 4x y− + =
or 2 24 4 16 0x x y− + + − = i.e. 2 24 12 0x x y− − + =
From Chapter 28, 2 2 2( ) ( )x a y b r− + − = is a circle, with centre (a, b) and radius r
Hence, 2 2 2( 2) 4x y− + = is a circle, with centre (2, 0) and radius 4
(b) The locus of 2 4z − = is shown below
© 2014, John Bird
792
6. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
3 5z + =
(a) If z = x + jy, then 3 3 ( 3) 5z x jy x jy+ = + + = + + =
On the Argand diagram shown in Problem 1, 2 2z x y= +
Hence, in this case, 3z + = 2 2( 3) 5x y+ + = from which, 2 2 2( 3) 5x y+ + =
or 2 26 9 25 0x x y+ + + − = i.e. 2 26 16 0x x y+ − + =
From Chapter 28, 2 2 2( ) ( )x a y b r− + − = is a circle, with centre (a, b) and radius r
Hence, 2 2 2( 3) 5x y+ + = is a circle, with centre (–3, 0) and radius 5
(b) The locus of 3 5z + = is shown below
7. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
1 31
zz+
=−
(a) z + 1 = x + jy + 1 = (x + 1) + jy z – 1 = x + jy –1 = (x – 1) + jy
© 2014, John Bird
793
Hence, 2 2
2 2
( 1)1 ( 1) 31 ( 1) ( 1)
x yz x jyz x jy x y
+ ++ + += = =
− − + − +
and squaring both sides gives:
2 2
2 2
( 1) 9( 1)x yx y+ +
=− +
from which,
2 2 2 2( 1) 9[( 1) ]x y x y+ + = − +
2 2 2 22 1 9[ 2 1 ]x x y x x y+ + + = − + +
2 2 2 22 1 9 18 9 9x x y x x y+ + + = − + +
2 20 8 20 8 8x x y= − + +
i.e.
2 28 20 8 8 0x x y− + + =
and dividing by 4 gives:
2 22 5 2 2 0 which is theequation of the locusx x y− + + =
Rearranging gives: 2 25 12
x x y− + = −
Completing the square (see Chapter 14) gives:
2
25 25 14 16
x y − − + = −
i.e. 2
25 2514 16
x y − + = − +
i.e. 2
25 94 16
x y − + =
i.e. 2 2
25 34 4
x y − + =
which is the equation of a circle
(b) Hence the locus defined by 1 31
zz+
=−
is a circle of centre 5 , 04
and radius 34
© 2014, John Bird
794
8. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
1 2zz−
=
(a) z – 1 = x + jy –1 = (x – 1) + jy
Hence, 2 2
2 2
( 1)1 ( 1) 2x yz x jy
z x jy x y− +− − + −
= = =+ +
and squaring both sides gives:
2 2
2 2
( 1) 2x yx y− +
=+
from which,
2 2 2 2( 1) 2[ ]x y x y− + = +
2 2 2 22 1 2 2x x y x y− + + = +
2 20 2 1x x y= + − + i.e.
2 22 1 0x x y+ − + = which is the equation of the locus
Rearranging gives: 2 22 1x x y+ + = Completing the square (see Chapter 14) gives: ( )2 21 1 1x y+ − + = i.e. ( )2 21 2x y+ + =
i.e. ( ) ( )22 21 2x y+ + = which is the equation of a circle
(b) Hence the locus defined by 1 2zz−
= is a circle of centre ( )1, 0− and radius 2
© 2014, John Bird
795
9. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
arg 14
zz
π−=
(a) ( )( )( )
2
2 2
[( 1) ]1 ( 1) ( 1) ( 1)x jy x jyz x jy x x jy x jxy yz x jy x jy x jy x y
− + − − − + − − − + + = = = + + − +
2 2 2 2
2 2 2 2
x x jxy jy jxy y x x jy yx y x y
− − + + + − + += =
+ +
= ( )2 2 2 2
2 2 2 2 2 2
x x y jy x x y jyx y x y x y
− + + − += +
+ + +
Since arg 14
zz
π−= then
2 21
2 2
2 2
tan4
yx y
x x yx y
π−
+ =
− + +
i.e.
1
2 2tan
4y
x x yπ
−
= − + from which,
2 2tan 1
4y
x x yπ
= =− +
Hence,
2 2y x x y= − +
Hence, the locus defined by arg 14
zz
π− =
is: 2 2 0x x y y− − + =
Completing the square gives:
2 21 1 12 2 2
x y − + − =
1 1 1which is a circle,centre , and radius2 2 2
(b) A sketch of the locus is shown below
© 2014, John Bird
796
10. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
arg 24
zz
π+=
(a) ( )( )( )
2
2 2
[( 2) ]2 ( 2) ( 2) ( 2)x jy x jyz x jy x x jy x jxy yz x jy x jy x jy x y
+ + − + + + + − + + + = = = + + − +
2 2 2 2
2 2 2 2
2 2 2 2x x jxy j y jxy y x x j y yx y x y
+ − − + + + − += =
+ +
= ( )2 2 2 2
2 2 2 2 2 2
2 2 2 2x x y j y x x y j yx y x y x y
+ + − + += −
+ + +
Since arg 24
zz
π+= then
2 21
2 2
2 2
2
tan 2 4
yx y
x x yx y
π−
− + =
+ + +
i.e.
1
2 2
2tan2 4
yx x y
π− −
= + + from which,
2 2
2 tan 12 4
yx x y
π−= =
+ + Hence,
2 22 2y x x y− = + +
Hence, the locus defined by arg 24
zz
π+ =
is: 2 22 2 0x x y y+ + + =
Completing the square gives:
( ) ( )2 21 1 2x y+ + + =
( )which is a circle,centre 1, 1 and radius 2− −
(b) A sketch of the locus is shown below.
© 2014, John Bird
797
11. If z = x + jy, (a) determine the equation of the locus (b) sketch the locus of the following:
2z j z+ = + Since 2z j z+ = + then ( 1) ( 2)x j y x jy+ + = + + and 2 2 2 2( 1) ( 2)x y x y+ + = + +
Squaring both sides gives: 2 2 2 2( 1) ( 2)x y x y+ + = + + i.e. 2 2 2 22 1 4 4x y y x x y+ + + = + + + from which, 2y + 1 = 4x + 4 i.e. 2y = 4x + 3 or y = 2x + 1.5 Hence, the locus defined by 2z j z+ = + 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:
4 2z z j− = − Since 4 2z z j− = − then ( 4) ( 2)x jy x j y− + = + − and 2 2 2 2( 4) ( 2)x y x y− + = + −
Squaring both sides gives: 2 2 2 2( 4) ( 2)x y x y− + = + − i.e. 2 2 2 28 16 4 4x x y x y y− + + = + − +
© 2014, John Bird
798
from which, –8x + 16 = –4y + 4 i.e. 4y = 8x – 12 or y = 2x – 3 Hence, the locus defined by 4 2z z 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:
1z z− = Since 1z z− = then ( 1)x jy x jy− + = + and 2 2 2 2( 1)x y x y− + = +
Squaring both sides gives: 2 2 2 2( 1)x y x y− + = + i.e. 2 2 2 22 1x x y x y− + + = + from which, –2x + 1 = 0 i.e. 1 = 2x
or x = 12
Hence, the locus defined by 1z z− = is a straight line: x = 12
© 2014, John Bird
799
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