Chapter 31

Electromagnetic oscillations and AC circuit

Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton

(Date: July 25, 2019)

The phasor diagram is very useful in discussing the AC circuit formed of series

connection of R, L, and C. The detail of this concept will be discussed below. We note

that this method is not appropriate for the AC circuits which are formed of various kind

of combinations with connections of R, L, and C elements. In this case we need to use the

concept of impedance such as R, i L , and 1/ ( )i C in the frequency domain. We also

use the transformations [ ( ) Re( )i ti t Ie for the current and ( ) Re( )i tv t Ve for the

voltage] between the frequency domain and time domain, The quantities I and V are

complex numbers. In the frequency domain, the AC circuit can be simply solved by using the KCL and KVL laws. In general physics, in spite of convenience, this method is not

adopted.

1. LC oscillations

In a resistanceless LC circuit, the total energy U is given by

22

2

1

2

1CL CvLiU

where q = CvC.

From the energy conservation, U remains constant with time.

0dt

dvCvi

dt

dLi

dt

dU CCLL

Since dt

dqiL and

C

qvC , we have

02

2

2

qdt

qd (LC oscillation, simple harmonics).

with

LC

1

The total energy U is rewritten as

2

2

2

1

2

1q

Cdt

dqLU

We assume the initial condition as

At t = 0,

q = Q and dq/dt = 0.

The solution of the second order differential equation is given by

)cos( tQq

)sin( tQdt

dqiL

Out[11]=

0.2 0.4 0.6 0.8 1.0 1.2 1.4têT

-1.0

-0.5

0.5

1.0

qêQ, iLêHQwL

Fig. Qq / (red) and )/( QiL (blue) as a function of t/T.

Energy of UE and UB

C

QUUU

tC

QtLQU

tC

Q

C

qU

BE

B

E

2

)(sin2

)(sin2

1

)(cos22

2

22

222

222

Fig. )2//( 2 CQUE (red) and )2//( 2 CQUB (blue) as a function of t/T.

2. Damped oscillations in a RLC circuit

A circuit containing resistance, inductance, and capacitance is called an RLC circuit.

dt

dqi

Cvq

Riv

dt

diLv

L

C

LR

LL

From the KVL, we have

0 CRL vvv

or

the second order differential equation

02

2

LC

q

dt

dq

L

R

dt

qd

We assume that

q = x, 2 = R/L, and 02 = 1/(LC)

Then we have

0)()('2)("2

0 txtxtx

with the initial conditions

0)0(' vx and 0)0( xx

The solution of this differential equation depends is classed into three types,

(1) underdamping: 02

0

2

(2) critical damping 02

0

2

(3) overdamping 02

0

2

((Note))

We assume that x(t) is expressed by a solution given by exp(pt). Then we have

0)2()()('2)("2

0

22

0 ptepptxtxtx

The solution of the quadratic equation 2 2

02 0p p is

1

22

0

22

0

22

0

22

)(

)(2

iip

ip

pp

with

22

01

The solution is given by

)]sin()cos([)()( 1211 tCtCetxtq t

From the initial condition, we have

01 xC

1

002

xvC

The final form is given by

)]sin()cos([)( 1

1

0010 t

xvtxetq

t

3. AC circuit theory

3.1 phasor diagram

(a)

)sin(max tIi

)2

sin()cos(

)2

sin()cos(1

)sin(

maxmax

maxmax

max

tLItLIdt

diLv

tC

It

C

Iidt

Cv

tRIv

L

C

R

(b)

)cos(max tIi

)2

cos()sin(

)2

cos()sin(1

)cos(

maxmax

maxmax

max

tLItLIdt

diLv

tC

It

C

Iidt

Cv

tRIv

L

C

R

The phasor diagram is given by the following figure.

For more general case (more complicated AC circuits), we need to use the method of

complex number. This will be discussed in the Appendix.

3.2. Impedance of single elements in AC circuit

3.2.1 Inductance L

maxmax LIV

The relation between Vmax and Imax for the inductance is described in the following phasor

diagram.

We say that V leads to I by 90° (or I lags V by 90°).

3.2.2 Capacitance C

C

IV

max

max

The relation between Vmax and Imax for the capacitance is described in the following

phasor diagram.

We say that V lags I by 90° (or I leads V by 90°).

3.2.3 Resistance

For resistance, one can write down a relation

maxmax RIV

The relation between Vmax and Imax for the resistance is described in the following phasor

diagram.

3.3. Power dissipated in single elements

3.3.1 Definition

The time average of the power P(t) over a period time T (= 2/) is given by

cos

cos2

1

)]2cos([cos2

1

)sin()sin(1

maxmax

0

maxmax

0

maxmax

rmsrms

T

T

avg

VI

VI

tdtVIT

tVtdtIT

P

where cos is the power factor, Imax and Vmax are the amplitudes of current and voltage.

Note that the definition of the root-mean squares Irms and Vrms will be given later (Section

3.4), where

max

1

2rmsI I , max

1

2rmsV V

3.3.2 inductance

0avgP

since =/2. No power is dissipated in an inductor.

3.3.3 Capacitance

0avgP

since = -/2. No power is dissipated in a capacitor.

3.3.4 Resistance

2

max2

1RIPavg

Power is dissipated only in a resistor.

3.4. Root-mean square value of current and voltage

The root-mean square value of the current is defined as

2

)]2cos(1[2

1

)(sin1

max

0

2

max

0

22

max

I

dttIT

dttIT

i

T

T

rms

The root-mean square value of the voltage is defined as

2

maxVvrms

((Note))

Domestic electricity is provided at a frequency of 60 Hz in the United States, and a

residential outlet provides 120 Vac. This means the rms voltage is max 120 2V = 170 V.

The power dissipated in the resistor is rewritten as

R

VIRP rms

rmsavg

22

3.5. The series RLC circuit

Phasor diagram of the RLC circuits is given in the following way, where we use I and V

instead of Imax and Vmax (or Irms and Vrms)

where

IC

V

LIV

RIV

IZV

C

L

R

ss

1

where I is the amplitude of the current, Vs is the amplitude of the voltage source, Zs is the

total impedance defined by

22)

1(

CLRZ s

and the angle between Vs and I is defined as

)1

(1

tanC

LR

The average power dissipated in the system is

2coscos

2

1rmsrmsrmsss RIIVIVP

3.6. Example (RLC circuit)

Problem 31-46 (SP-31)

31-46

Figure shows a driven RLC circuit that contains two identical capacitors and two

switches. The emf amplitude is set at 12.0 V, and the driving frequency is set at 60.0 Hz.

With both switches open, the current leads the emf by 30.9°. With switches S1 closed and

switch S2 still open, the emf leads the current by 15.0°. With both switches closed, the

current amplitude is 447 mA. What are (a) R, (b) C, and (c) L?

((Solution))

E0 = 12 V

f = 60 Hz

= 2f

(1) S1 and S2 open

R

LC

ILC

REs

)2

(

tan

)2

(

1

1

22

Es lags I1 by 1. The average power dissipated in the system is

2

11111 )(cos)(cos2

1rmsrmsrmsss IRIEIEP

where 11cos RIEs

(2) S1 closed and S2 open

R

CL

IC

LREs

)1

(

tan

)1

(

2

2

22

Es leads I2 by 2. The average power dissipated in the system is

2

22222 )(cos)(cos2

1rmsrmsrmsss IRIEIEP

where 22cos RIE s

(3) S1 closed and S2 closed

CL

EI s

13

3.7. Example (RC circuit)

Problem 31-44 (HW-31)) Hint

An alternating emf source with a variable frequency fd is connected in series with a

50.0 resistor and 20.0 F capacitor. The emf amplitude is 12.0 V. (a) Draw a phasor

diagram for phasor VR (the potential across the resistor) and phasor VC (the potential

across the capacitor). (b) At what driving frequency fd do the two phasors have the same

length? At that driving frequency, what are (c) the phase angle in degrees, (d) the angular

speed at which the phasors rotate, and (e) the current amplitude?

R

C

IC

REs

)/1(tan

)1

(22

Es lags I by the angle . The average power dissipated in the system is

2coscos

2

1rmsrmsrmsss RIIEIEP

where

RIEs cos

3.8. Example (RL circuit)

R

L

ILREs

tan

)( 22

Es leads I by the angle . The average power dissipated in the system is

2coscos

2

1rmsrmsrmsss RIIEIEP

where

RIEs cos

3.9 Resonance of the RLC circuit

IC

LREs

22)

1(

or

22 )1

(C

LR

E

Z

EI s

s

s

where

22)

1(

CLRZs

(impedance)

The amplitude of the current has a maximum when

LC

1

We define the quality factor of the circuit by

R

LQ 0

Then the amplitude of the current can be rewritten as

20

0

2)(1

1

QR

EI s

We now consider the frequency dependence of I2, instead of I.

20

0

2

2

2

)(1

1

Q

R

EI s

We note that this is a scaling function of /0. In the figures we show the plot of

)/( 22

2

REs

Ivs

0

, where the quality factor Q is changes as a parameter.

Q=100 - 1000

0.6 0.8 1.0 1.2 1.4wêw0

0.2

0.4

0.6

0.8

1.0Es2êR2

Q = 10 - 100

0.6 0.8 1.0 1.2 1.4wêw0

0.2

0.4

0.6

0.8

1.0Es

2êR2

Fig. )/( 22

2

REs

Ivs

0

, where the quality factor Q is changes as a parameter.

_______________________________________________________________________

Suppose that

10

(<<1).

1)1( 10

Then we have

22

2

max

22

2

22

2

2

4141

1

)]1(1[1

1

Q

I

QR

E

QR

EI

s

s

When

12 Q , or Q2

1

I2 is equal to 2

2

maxI. Then the width is estimated as

000 2)1()1(2

or

Q

12

2

0

2

0Q (definition of the quality factor)

or

Q

02

(full-width at half-maximum)

4. Transformer

dt

dNV

dt

dNV

Bss

Bpp

Then we have

p

s

p

s

N

N

V

V (transformation of voltage)

If Ns>Np, the transformer is called a step-up transformer. If Np>Ns, the transformer is

called a step-down transformer

From the energy conservation, we have

ppss VIVI

or

s

p

s

p

p

s

N

N

V

V

I

I (transformation of current)

The equivalent resistance Req is the value of the load resistance as seen by the generator.

RN

N

I

V

I

I

V

V

I

VR

s

p

s

s

p

s

s

p

p

p

eq

2)(

5. Typical examples

5.1 Problem 31-17 (SP-31)

In an oscillating LC circuit with C = 64.0 F, the current is given by i = 1.60 sin(2500

t + 0.680), where t is in seconds, i in amperes, and the phase constant in radians. (a) How

soon after t = 0 will the current reach its maximum value? What are (b) the inductance L

and (c) the total energy?

((Solution))

C = 64.0 F )680.02500sin(60.1 ti

Imax = 1.60 A

= 2500 rad/s

(a) ,1)680.02500sin( t 2

680.02500

t

or

t = 356.30 s.

(b)

HC

L

LC

0025.01

1

2

(c)

The total energy is conserved.

JCvLiCvLi 322

max

2

max

22 102.36.10025.02

1

2

1

2

1

2

1

2

1

5.2 Problem 31-61 (SP-31)

In Fig. R = 15.0 , C = 4.70 F, and L = 25.0 mH. The generator provides an emf

with rms voltage 75.0 V and frequency 550 Hz. (a) What is the rms current? What is the

rms voltage across (b) R, (c) C, (d) L, (e) C, and L together, and (f) R, C, and L together?

At what average rate is energy dissipated by (g) R, (h) C, and (i) L?

((Solution))

R = 15

C = 4.70 F

L = 25.0 mH

= 75.0 V

f = 550 Hz

Phasor diagram

rmsrmsLC

rmsrmsC

rmsrmsR

rmsrmsL

rmsrms

IC

LV

IC

V

RIV

LIV

IC

LR

)1

()(

1

)1

(22

22

22

)1

(

)1

(

CLR

I

IC

LR

rmsrms

rmsrms

(a) Irms = 2.58576 A

(b) rmsrmsR RIV = 38.7865 V

(c) rmsrmsC IC

V1

= 159.202 V

(d) rmsrmsL LIV = 223.394 V

(e) rmsrmsLC IC

LV )1

()(

= 64.192 V

(f) rms = 75.0 V

(g) 2

rmsR RIP = 100.293 W

(h) 0CP

(i) 0LP

5.3 Problem 31-60 (SP-31) A typical light dinner used to dim the stage lights in a theater consists of a variable

inductor L (whose inductance is adjustable between 0 and Lmax) connected in series with a lightbulb B as shown in Fig. The electrical supply is 120 V (rms) at 60.0 Hz; the light-

bulb is rated at 120 V, 1000 W. (a) What Lmax is required if the rate of energy dissipation in the lightbulb is to be varied by a factor of 5 from its upper limit of 1000 W? Assume

that the resistance of the lighbulb is indepenent of its temperature. (b) Could one use a variable resistor (adjustable between zero and Rmax) instead of an inductor? (c) If so, what

Rmax is required? (d) Why isn’t this done?

\ ((Solution))

L = 0 – Lmax 120 V (rms) at 60 Hz

Light bulb 1000 W at 120 V

rms = 120 V

f = 60 Hz

22

22

22

2

22

22

)()(

)(

)(

LR

R

LRRRIP

LRI

ILR

rmsrmsrmsR

rmsrms

rmsrms

(a) When L = 0,

RW

RP rms

R

22120

1000

or R = 14.4

5

1000

)(1

1000

)(1

1

)(2

2

2

2

2

22

2W

R

L

W

R

LRLR

RP rmsrms

R

or

5)(

12

2

R

L

or

2R

L mH

RL 39.762

(b) and (c) Yes.

The inductance is replaced by a variable resistance R0.

20

2

2

0

2

)1(

1

)(

R

RRRR

RP rmsrms

R

2361.115

5)1(

0

0

R

R

R

R

or

799.174.142361.10R

(d) This is not done because we lose the energy in the variable resistance.

5.4 Phasor diagram of RLC circuit

G. Gladding, M. Selen, and T. Steltzer, SmartPhysics; Electricity and

Magnetism (II) (W.H. Freeman, 2011)

Consider the driven LCR circuit. The source voltage has a maximum voltage of 10 V

and oscillates at a frequency of 60 Hz, which converts to an angular frequency of 377

rad/s. The values for the circuit components are 20 , 20 mH, and 150 F, respectively.

Our goal is to determine maxI , the maximum current, and , the phase angle between

the current and the source voltage. Once we know these two quantities at any time. We

can determine both of these quantities by simply constructing the phasor diagram.

Fig. RLC circuit. f 60 Hz. L = 20 mH. C = 150 F. 2 f 376.991 rad/s.

10mE V.

((Solution))

Suppose that max sin( )I I t . The voltages across the capacitance, inductance, and

resistance are drawn in the x-y plane, as well as the current.

Fig. Phasor diagram for voltages of the circuit. We assume that maxRI is directed along

the x axis. The voltage source is actually expressed by m sin( )t . So it is

necessary to rotate the phasor diagram by appropriate angle at the last stage.

Note that

max max20 RV RI I

max

1( )CV I

C 17.6839 Imax

max( )LV L I 7.53982 Imax

The source voltage;

2 2

max max

1( ) 22.4255 ZV R L I I

C

Since 0 max10.0 V 22.4255 ZV E I , we have the maximum current as

0max 0.445921

z

EI

V [A]

max 8.91843RI [V]

max 3.36217LI [V]

max

17.88562I

C V

The phase factor:

arctan[ ] 26.89C L

R

V V

V

.

Since 0 sin( )sV E t , we need to rotate the phasor diagram (in the x-y plane) around the

origin by the angle in counter clockwise. So we get the final phasor diagram

The average power dissipated in the circuit;

2

max

2

max

max max

max max

1

2

1cos

2

1cos

2

1cos

2

avP RI

Z I

ZI I

I V

where cos is the power factor.

RImaxLImax

1

CImax

1

CL Imax

Vmax ZImax

Fig. Phasor diagram of the RLC circuit. 26.89 .

For the voltage source 10 sin( )t [V]

For the voltage across the resistance: 8.91834 sin( )t [V]

For the voltage across the inductance 3.36217 sin( )2

t

[V]

For the voltage across the capacitance 7.88562 sin( )2

t

[V]

For the current 0.445921 sin( )t [A]

________________________________________________________________________

Appendix A AC circuit theory based on the complex plane

Here we discuss the AC circuit theory based on the complex number. This theory is mathematically correct.

A1. Impedance of single elements in AC circuit

We represent all voltages and current (sinusoidally change with time) by complex numbers, using the exponential notation. The time-varying current is written by

2]Re[)(

* tititi eIIe

Ieti

where (= 2f) is the angular frequency, I represents a complex number that is

independent of t, and I* is the complex conjugate. The complex number I is described by ieI0 where I0 is the amplitude and is the phase. For convenience, we use the phasor

diagram in the complex plane, where for I is plotted as

)cos(]Re[]Re[)( 00 tIeeIIeti tititi

The voltage is also described by

]Re[)( tiVetv

((Note))

The absolute values V and I are the amplitudes of the real voltage and current.

(a) Inductance L

For inductance one can write down the relation

]Re[]Re[]Re[)( tititi LIeiIedt

dL

dt

diLVetv

or

ILeLIiV i 2/

The relation between V and I for the inductance is described in the following phasor diagram.

We say that V leads to I by 90° (or I lags V by 90°). The impedance of the inductance is given by

LiI

VZL

ɺ

ɺ

(b) Capacitance C

For capacitance, one can write down a relation

]]Re[]Re[)(

]Re[)( tititi CVeiVedt

dC

dt

tdvCIeti

CViI

or

IeCCi

IV i 2/1

The relation between V and I for the capacitance is described in the following phasor diagram.

We say that V lags I by 90° (or I lags V by 90°). The impedance of the capacitance is

given by

CiI

VZC

1

(c) Resistance

For resistance, one can write down a relation

]]Re[]Re[)(]Re[)( tititi RIeIeRtRiVetv

RIV

The relation between V and I for the resistance is described in the following phasor diagram.

The impedance of the resistance is given by

RZR

In summary

((Note)) The analysis of the AC theory without the concept of the complex number will be given

in the Appendix B.

A2. Power dissipated in single elements

The power is given by

)(4

1

22

)()()(

**2**2

**

VIIVeIVVIe

eIIeeVVe

titvtP

titi

titititi

The time average of the power P(t) over a period time T (= 2/) is given by

]Re[2

1]Re[

2

1)(

4

1)(

1 ****

0

VIIVVIIVdttPT

P

T

avg

When 0II and ieVV 0 (I0 and V0 are real numbers), then the power is rewritten as

2

00 coscos2

1rmsrmsrmsavg RiviVIP

where cos is the power factor, and the definition of the root-mean squares irms and vrms will given later.

(a) inductance

For the capacitance, we have ** LIiV . Then Pavg is calculated as

0)Re(2

1)Re(

2

1]Re[

2

1 2** ILiILIiIVPavg

No power is dissipated in an inductor.

(b) Capacitance

For the capacitance, we have Ci

IV

** . Then Pavg is calculated as

0]1

Re[2

1]

1Re[

2

1])Re[(

2

1]Re[

2

1 2**

* IC

iIIC

iICi

IIVPavg

No power is dissipated in a capacitor.

(c) Resistance

For the resistance, we have ** RIV . Then Pavg is calculated as

R

VIRIRIRIIVPavg

2

22**

2

1

2

1]Re[

2

1]Re[

2

1]Re[

2

1

Power is dissipated in a resistor.

A3. Root-mean square value of current and voltage

What is the root-mean square of the current and voltage?

)cos(]Re[]Re[)( 00 tIeeIIeti tititi

The root-mean square value of the current is defined as

2)]22cos(1[

2

)(cos)]([1

0

0

2

0

0

2

2

0

0

2

Idtt

T

I

dttT

Idtti

Ti

T

TT

rms

or

22

maxIIirms

where Imax (=|I| = |I0|) is the maximum of the current.

Similarly, we have the root-mean value of the voltage as

22

maxVVvrms

The power dissipated in the resistor is rewritten as

R

viRP rmsrmsavg

22

((Note))

The rms value of any quantity that varies as sin(t) or cos(t) is always 1/ 2 times the

maximum value, so

vrms = 0.707vmax.

A4. The series RLC circuit

Circuit in the time domain

AC circuit in the frequency domain

)()()()( tvtvtvte CLRs

The relation of the expressions in the time domain and the frequency domain is given by

)Re()(

)Re()(

)Re()(

)Re()(

)Re()(

ti

RR

ti

CC

ti

LL

ti

ti

ss

eVtv

eVtv

eVtv

Ieti

eEte

Es, I, VL, VC, and VR are complex numbers. The phasor diagram of these will be given

later.

ICi

V

LIiV

RIV

IZVVVE

C

L

R

sCLRs

1

or

s

sC

s

sL

s

sR

s

s

Z

CiEV

Z

LiEV

Z

REV

Z

EI

)/1(

where Zs is the total impedance defined by

i

s

i

s

eZ

eC

LR

iC

LR

CLiR

CiLiRZ

22

22

)1

(

)sin(cos)1

(

)1

(1

where is an angle and ie is the phase factor,

R

CL

1

tan

and

22)

1(

CLRZ s

or

A5. Phasor diagram of the RLC circuits

((Example))

A series RLC circuit has R = 150 , L = 0.25 H, and C = 16.0 F. It is driven by 60 Hz

voltage of peak value 170 V. Determine the time dependence of the voltages across each

element.

-0.01 0.01 0.02 0.03t HsecL

-1.0

-0.5

0.5

1.0

i HAL

vL

vR

vCvS

-0.01 0.01 0.02 0.03t HsecL

-200

-100

100

200v

A6. Infinite ladder network (from the book of Feynman)

What would happen if the following network we keep on adding a pair of the

impedances (z1 and z2) forever, as we indicate by the dashed lines. Can we solve such an

infinite network?

First, we notice that such an infinite network is unchanged even if we add one more

section at the front end. If we add one more section to an infinite network it is still the

same infinite network. Suppose we call the impedance between the two terminals a and b

of the infinite network z0. Then the impedance of all the stuff to the right of the two

terminals c and d is also z0. Taking into account of the similarity of the infinite network,

we get an equation to determine the value of z0 as

02

0210

zz

zzzz

We can solve for z0 to get

21

2

110

42zz

zzz

This impedance z0 is called the characteristic impedance of the infinite network.

Now we consider a specific example where an AC source is connected between the

terminals a and b, and Liz 1 and )/(12 Ciz . We define the complex current In and

voltage Vn.

In the above Fig. we get the following recursion relation,

0

111z

VzIzVV n

nnn

or

0

10

0

11 1z

zz

z

z

V

V

n

n

or

n

nV

We can call this ratio the propagation factor for one section of the ladder; we will call it .

It is the same for all sections. Note that

42

42

22

21

2

110

L

C

LLi

zzzz

z

Then we have

2

2

22

22

0

10

1

1

24

24

xix

xix

LiL

C

L

LiL

C

L

z

zz

where 0

x and LC

20

Out[23]=

1 2 3 4 5wêw0

0.2

0.4

0.6

0.8

1.0

»a»

Out[21]=

1 10 100 1000wêw0

10-5

0.001

0.1

»a»

Fig. as a function of 0

x .

ReHaL

ImHaL

1 2 3 4 5wêw0

-1.0

-0.5

0.0

0.5

1.0Re@aD, Im@aD

Fig. Re[] and Im[] as a function of 0

x .

In summary, the network propagates energy for <0 and blocks it for >0. We say

that the network passes low frequencies and reject or filters out the high frequencies. Any

network designed to have its characteristics vary in a prescribed way wit frequency is

called a filter. In this case, we have a low pass filter.

A7. Impedance matching

Given a circuit having a load impedance ( LLL iXRZ ), what impedance will result in

the maximum average power P being absorbed by the load?

L

sLL

L

s

ZZ

VZIZV

ZZ

VI

0

0

The power P (absorbed by the lad) is given by

]Re[2

1]Re[

2

1 ** IVVIP

where

L

sL

L

sL

L

s

ZZ

VZ

ZZ

VZ

ZZ

VVI

0

2

0

*

0

*

*

)(

and

)()( 000 LLL

LLL

XXiRRZZ

iXRZ

Then P is given by

])()[(2 2

0

2

0

2

LL

Ls

XXRR

RVP

First we assume that RL is constant. Only XL is a unknown parameter. P has a maximum

when

0XX L

Then P obtained as

2

0

2

)[(2 L

Ls

RR

RVP

has a maximum at RL = R0.

In conclusion, in order to get maximum power to ZL, we select

*

000 ZiXRZL

_______________________________________________________________________

B. Complex number

B1. Imaginary unit

The imaginary unit is defined by 1i : 12 i .

ii

i

522020

55

These numbers are called a pure imaginary number.

B2. Properties of complex number

For real numbers a, b, c, and d,

dicbia → a = c and b = d.

0 bia → a = 0 and b = 0.

22

222222

)()(

))((

))((

)1()())((

)())((

)()()()(

)()()()(

dc

iadbcbdac

dicdic

dicbia

dic

bia

bababiabiabia

ibcadbdacdicbia

idbcadicbia

idbcadicbia

B3. Simplifying powers of i

The expression of ni , where n is a positive whole number, can be reduced to either ±1

or ±i.

i0 = 1, i1 = i i2 = -1, i3 = i2i=-i

i4 = 1, i5 = i4 i =i i6 = i4 i2=i2 = -1 i7 = i4i3=-i

In general, we have

i4n = 1, i4n+1 = i i4n+2 = -1, i4n+1 = -i

where n is an integer.

B4. Graphing complex number: complex plane

(a) Definition

z = x+iy

A complex number in x+iy form can be graphed in the complex plane by measuring x

units along the horizontal (real axis) and y units along the vertical (imaginary axis).

We define the absolute value of the complex number as

22

yxiyxz

which is the distance between the ponits of z (= x+i y) and the origin.

(b) Expression of the complex number in the real-imaginary plane

We make a plot of the complex numbers in the complex plane.

a = 3+2i, b = -2+i

a+b = 1+3i, a - b=5+i,

a

b

a+b

a-b

x

y

O

B5. Complex conjugate

The complex number z and z* are complex conjugate.

ibaz

ibaz

*

where a and b are real.

22* ))(( baibaibazz .

B6. Angles and polar coordinates of complex number

We find the real (horizontal) and imaginary (vertical) components in terms of r (the

length of the vector) and θ (the angle made with the real axis):

From Pythagoras, we have: r2 = x2 + y2 and basic trigonometry gives us:

x

ytan

x = r cosθ, y = r sinθ

Multiplying the last expression throughout by i gives us:

siniryi

So we can write the polar form of a complex number as:

)sin(cos iriyxz

where r is the absolute value (or modulus) of the complex number, and θ is the

argument of the complex number.

q

f

q +f

O

P

Q

R

Real

Imaginary

Points P, Q and R on the unit circle in the complex plane, are denoted by complex

numbers z1, z2, and z1z2.

)sin()cos(

)sincoscos(sin)sinsincos(cos

)sin)(cossin(cos

sincos

sincos

21

2

1

i

i

iizz

iz

iz

The complex conjugate of z1 is

)sin()cos(

sincos*

1

i

iz

((Example))

We make a plot of zn (n = 0, 1, 2, 3, …), where

).sin(cos irz

We choose r = 1.05 and = 5º.

B7. Euler’s formula

Euler's formula states that, for any real number ,

sincos iei

where e is the base of the natural logarithm, i is the imaginary unit. Richard Feynman

called Euler's formula "our jewel" and "the most remarkable formula in mathematics".

B8. de Moivre’s theorem

When ireiriyxz )sin(cos

)]sin()[cos( ninrerrez ninnnin .

where

x

y

yxr

tan

22

((Note))

(a) The expression of z-n

)]sin()[cos(

)]sin()[cos(

ninr

ninr

er

rez

n

n

inn

nin

(b) The expression of z* (complex conjugate of z)

ire

ir

irz

)]sin()[cos(

)sin(cos*

APPENDIX

Single-phase rms voltage and frequency over the world

Europe (Great Britain, France, Germany, Belgium, Netherland, Denmark, Sweden) 230 V 50 Hz

Australia 230 V 50 Hz U.S.A. 120 V 60 Hz

Equador 120 V 60 Hz China 220 V 50 Hz

Malawi 220 V 50 Hz

Japan: 100 V 50 (East)/60Hz (West); regions of west and east separated by Fossa Magna.