complex review

8/8/2019 Complex Review

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Introduction to Complex Numbers in Physics/Engineering

Reference: Mary L. Boas, Mathematical Methods in the Physical SciencesChapter 2 & 14

George Arfken, Mathematical Methods for PhysicistsChapter 6

The real numbers (denoted R ) are incomplete (not closed) in the sense that standard op-erations applied to some real numbers do not yield a real number result (e.g., square root:1). It is surprisingly easy to enlarge the set of real numbers producing a set of numbersthat is closed under standard operations: one simply needs to include 1 (and linearcombinations of it). This enlarged eld of numbers, called the complex numbers (denotedC ), consists of numbers of the form: z = a + b1 where a and b are real numbers. Thereare lots of notations for theses numbers. In mathematics, 1 is called i (so z = a + bi),whereas in electrical engineering i is frequently used for current, so

1 is called j (so

z = a + bj). In Mathematica complex numbers are constructed using I for i. Since com-plex numbers require two real numbers to specify them they can also be represented as anordered pair: z = ( a, b). In any case a is called the real part of z: a = Re( z) and b iscalled the imaginary part of z: b = Im( z). Note that the imaginary part of any complexnumber is real and the imaginary part of any real number is zero. Finally there is a polarnotation which reports the radius (a.k.a. absolute value or magnitude) and angle (a.k.a.phase or argument) of the complex number in the form: r. The polar notation canbe converted to an algebraic expression because of a surprising relationship between theexponential function and the trigonometric functions:

e j = cos + j sin (1)

Thus there is a simple formula for the complex number z1 in terms of its magnitude andangle:

|z1| a2 + b2 = r (2)a = r cos = |z1|cos (3)b = r sin = |z1|sin (4)

z1 = a + bj = |z1|(cos + j sin ) = |z1|e j (5)For example, we have the following notations for the complex number 1 + i:

1 + i = 1 + j = 1 + I = (1 , 1) = 245 = 2e j/ 4 (6)

Note that Equation 1 can be used to express the usual trigonometric functions in terms of complex exponentials:

cos = Re( e j ) =e j + e j

2sin = Im( e j ) =

e j e j

2 j(7)

Since complex numbers are closed under the standard operations, we can dene things whichpreviously made no sense: log( 1), arccos(2), ( 1) , sin(i), . . . . The complex numbers arelarge enough to dene every function value you might want. Note that addition, subtraction,


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Re

Im z 1

b

a Re

Im

b

a

r

z 1

a 2 +b 2 = r

Figure 1: Complex numbers can be displayed on the complex plane. A complex numberz1 = a + bi may be displayed as an ordered pair: ( a, b), with the real axis the usual x-axis and the imaginary axis the usual y-axis. Complex numbers are also often displayedas vectors pointing from the origin to ( a, b). The angle can be found from the usualtrigonometric functions; |z1| = r is the length of the vector.

multiplication, and division of complex numbers proceeds as usual, just using the symbolfor 1 (lets use j ): z1 = a + bj z2 = c + dj (8)

z1 + z2 = ( a + bj ) + ( c + dj ) = ( a + c) + ( b + d) jz1 z2 = ( a + bj ) (c + dj ) = ( a c) + ( bd) jz1 z2 = ( a + bj ) (c + dj ) = ac + adj + bcj + bdj 2 = ( ac bd) + ( ad + bc) j

1z1

=1

a + bj=

1a + bj

a bja bj

=a bja2 + b2

=a

a2 + b2+ b

a2 + b2j

Note in calculating 1 /z 1 we made use of the complex number a bj ; a bj is called thecomplex conjugate of z1 and it is denoted by z1 or sometimes z1 . See that zz = |z|2.Note that, in terms of the ordered pair representation of C , complex number addition andsubtraction looks just like component-by-component vector addition:

(a, b) + ( c, d) = ( a + b, c + d) (9)

Re

Im z 1

z 1 *

Re

Im z 1 + z 2

z 1

z 2

Figure 2: The complex conjugate is obtained by reecting the vector in the real axis.Complex number addition works just like vector addition.


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Thus there is a tendency to denote complex numbers as vectors rather than points in thecomplex plane.

Superposition of Oscillation

While the closure property of the complex numbers is dear to the hearts of mathematicians,the main use of complex numbers in science is to represent sinusoidally varying quantitiesin a simple wayallowing them to be combined with relative ease. (Remember that thesuperposition of sinusoidal quantities is itself sinusoidal, but with a new amplitude andphase.) For example, in a series RC circuit the voltage across the resistor might be givenby V R (t) = A cos t whereas the voltage across the capacitor might be given by V C (t) =B sin t, and the voltage across the combination (according to Kirchhoff) is the sum:

V R (t) + V C (t) = A cos t + B sin t where: A, B R

= A2 + B 2A

A2 + B 2 cos t +B

A2 + B 2 sin t

= A2 + B 2 (cos cos t + sin sin t) where: cos = AA2 + B 2= A2 + B 2 cos(t )Yuck! Thats a lot of work just to add two sinusoidal functions; we seek a simpler method(which might not seem overly simple at rst glance). Note that V R can be written asRe(Ae jt ) and V C can be written as Re( jBe jt ) so:

V R (t) + V C (t) = Re (A jB )e jt (10)Now using the polar form of the complex number A jB :

A jB = A2 + B 2 e

j where: tan = B/A (11)

we have:

V R (t) + V C (t) = Re (A jB )e jt= Re A2 + B 2 e

j e jt

= A2 + B 2 Re e j (t )

= A2 + B 2 cos(t )If we have more sinusoidal waves to add up, the problem is not much more difficult. Considerthe case of adding four waves, which for ease of notation we assume have unit amplitudeand constant phase difference ( ). [Im also going to switch to 1 = i; you should getused to both notations.]

f (t) = cos( t) + cos( t + ) + cos( t + 2 ) + cos( t + 3 ) (12)

We can write this as the real part of a complex expression:

f (t) = Re 1 + ei + ei2 + ei3 eit (13)


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1 2 3 4

-3

-2

-1

1

2

3

Figure 3: Consider the problem of adding four sinusoidal functions with the same frequencyand amplitude, but with different offsets (see left). The result (right) is a sinusoidal functionwith the same frequency we seek to easily determine the resulting amplitude and offset.

Well use some tricks below to add these four complex numbers, but for now the main pointis that they add to some complex number which can be expressed in polar form:

1 + ei + ei2 + ei3 = A ei (14)

sof (t) = Re A ei(t + ) = A cos(t + ) (15)

Thus adding sinusoidal waves is as simple as adding the corresponding (complex) ampli-tudes.

Now for the trick that applies to this particular problem. Recall the formula for the sum of the geometric series:

1 + r + r 2 + + r N 1 =

1 r N 1 r

(16)

(The proof of this result is easy: just multiply both sides by (1 r ), and notice that therhs terms telescope down to 1

r N .) Here:

r = ei (17)

So the sum is:

A ei =1 ei41 ei

=ei2

ei/ 2ei2 e

i2

ei/ 2 e i/ 2= ei3/ 2

sin(2)sin(/ 2)

(18)

Re

Im

2

3

Figure 4: Finding the sum of four cosines amounts to adding four complex amplitudes:1 + ei + ei2 + ei3


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1 2 3 4

5

10

15

Figure 5: A2 is plotted as a function of . Note that if = 0 , 2, 4 , . . . , the four wavesare in phase so the amplitude is 4 (so A2 = 16). The rst zero of A2 occurs when the fouramplitude vectors form a (closed) square for = / 2.

and hence:A =

sin(2)sin(/ 2)

= 3 / 2 (19)

In physics we are usually most interested in the value of A2 which is plotted in Figure 5 asfunction of .

In a more general case we might need to add N cosines with possibly different real amplitudesak and offsets k :

h(t) = a1 cos(t + 1) + a2 cos(t + 2) + + aN cos(t + N ) (20)=

N

k=1

ak cos(t + k ) (21)

= Re a1ei 1 + a2ei1 + + aN eiN eit (22)

= ReN

k=1akei k eit (23)

Once again all that is required is to express the sum of the complex amplitudes in polarfashion:

N

k=1

akei k = Aei (24)

thenh(t) = A cos(t + ) (25)

Differential Equations and e it

Complex exponentials provide a fast and easy solution for many differential equations.Consider the damped harmonic oscillator:

F net = kx bv = ma (26)


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or:0 =

d2xdt2

+bm

dxdt

+km

x (27)

Seeking a more compact notation, we redene the constants in this expression:bm 2

km

20 (28)

So:d2xdt2

+ 2 dxdt

+ 20 x = 0 (29)

If we guess a solution of the form: x = Aert , we nd:

r 2 + 2 r + 20 Aert = 0 (30)

Since ert is never zero, r must be a root of the quadratic equation in square brackets.

r = 2 4 2 4202 = i 20 2 (31)where we have assumed 0 > . Dening the free oscillation frequency

= 20 2, wehave a solution:x = Re A e t ei

t =

|A

|e t cos( t + ) (32)

In the driven, damped harmonic oscillator, we have a driving force: F 0 cos t in addition tothe other forces:

F net = F 0 cos t kx bv = ma (33)Dening A0 = F 0/m yields the differential equation:

d2xdt2

+ 2 dxdt

+ 20 x = A0 cos(t) (34)

If we seek a solution that oscillates at the driving frequency: x = Re Aeit , our differentialequation becomes:

2 + 2 i + 20 Ae

it = A0eit (35)So:

A =A0

(20 2) + 2 i(36)

From which we can extract the amplitude and phase of the oscillation, for example:

|A| =A0

(20 2)2 + 4 22(37)

One can show that the amplitude is largest for

= 20 2 2 (38)Finally it is customary to describe driven oscillators in terms of a dimensionless

quality factor , Q,

Q =02

(39)

If we then let x be the dimensionless frequency ratio / 0, we can write the oscillationamplitude in a particularly simple form:

|A| =A0/ 20

(1 x2)2 + x2/Q 2(40)

Notice that the case of small damping (small ) corresponds to large Q.


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0.6 0.8 1 1.2 1.4

5

10

15

20

Figure 6: Resonance: the amplitude factor: 1(1 x 2 )2 + x 2 /Q 2 is plotted as a function of thedimensionless frequency ratio: x = / 0 for the case Q = 20. Clearly the largest amplitudeoccurs when x 1, i.e., 0

Homework

1. Prove that when you multiply complex numbers z1 and z2, the magnitude of the result

is the product of the magnitudes of z1 and z2, and the phase of the product is thesum of the phases of z1 and z2 .

Re

Im

1

z 1

r 1

z 2 r 22

z 3

z 3 = z 1 z 2

3 = 1 + 2r 3 = r 1 r 2

2. Express the following in the r format (I bet your calculator can do this automati-cally):

(a)1

1 + i(b)

3 + i1 + 3 i

(c) 25e2i (d) (1 / (1 + i)) (e)1

(1 + i)

3. Find the following in ( a, b) format (I bet your calculator can do this automatically):

(a)3i 7i + 4

(b) ( .64 + .77i)4 (c) 3 + 4 i (d) 25e2i (e) ln(1)4. Three cosine functions with amplitudes and offsets: a1 = 1 .32, 1 = .253 rad; a2 =

3.21, 2 = .532 rad; and a3 = 2 .13, 3 = .325 rad are to be added together. Find theresult.

5. Find the formula for the amplitude A that results from adding 5 unit-amplitude cosinewaves with constant phase difference:

g(t) = cos( t) + cos( t + ) + cos( t + 2 ) + cos( t + 3 ) + cos( t + 4 )= A cos(t + )

Use your favorite graphing program to make a hardcopy plot of A2 vs. for (0, 4).


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6. Consider a driven RC circuit (see below left). According to Kirchhoffs law the voltagedrop across the resistor ( IR , for current I ) plus the voltage drop across the capacitor(Q/C , for charge Q) must equal the voltage applied by the a.c. generator ( V 0 cos(t))(where the generator is operating at an angular frequency ). Thus:

QC

+ R I = V 0 cos(t) (41)

Since the the current owing must accumulate as charge on the capacitor we have:

I =dQdt

(42)

Thus we have the differential equation:

QC

+ RdQdt

= V 0 cos(t) (43)

Using complex variable methods and guessing a solution of the form:

Q = Q0 eit (44)

Determine the amplitude and phase of Q0 so you can express your nal answer in realform:

Q = A cos(t + ) (45)

Q

V o

c o s

( t )

I

V o

c o s

( t )

I

7. Consider a driven RL circuit (see above right). According to Kirchhoffs law thevoltage drop across the resistor ( IR , for current I ) plus the voltage drop across theinductor ( L dI/dt ) must equal the voltage applied by the a.c. generator ( V 0 cos(t))(where the generator is operating at an angular frequency ). Thus:

LdI dt

+ R I = V 0 cos(t) (46)

Using complex variable methods and guessing a solution of the form:

I = I 0 eit (47)

Determine the amplitude and phase of I 0 so you can express your nal answer in realform:

I = A cos(t + ) (48)

complex review

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