chapter 46 de moivre’s theorem - s3-eu-west … · exercise 195 page 528 . 1. if. z = x + jy, (a)...

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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°

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= 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 − − = ∠ ° = ∠ × °

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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

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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.


results on an Argand diagram: (a) 3 – j4 (b) –1 – j2

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(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


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

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(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

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( )( ) ( )( )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

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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

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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

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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


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

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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.


arg (z + 1) = 6π

© 2014, John Bird

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(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.


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

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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


1 31

zz+

=−

(a) z + 1 = x + jy + 1 = (x + 1) + jy z – 1 = x + jy –1 = (x – 1) + jy

© 2014, John Bird

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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

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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

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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

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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

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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


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

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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


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

chapter 46 de moivre’s theorem - s3-eu-west … · exercise 195 page 528 . 1. if. z = x + jy, (a)...

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