Chapter 2: Complex numbers - UCF Physicsschellin/teaching/phz3113/lec1-2.pdfChapter 2: Complex...

Chapter 2: Complex numbers

Complex numbers are commonplace in physics and engineering. Inparticular, complex numbers enable us to simplify equations and/ormore easily find solutions to equations. We will explore thedamped, driven simple-harmonic oscillator as an example of theuse of complex numbers.By the end of this chapter you should be able to...

I Represent complex numbers in various ways

I Use complex algebraI Complex infinite seriesI Determine functions of complex numbersI Use Euler’s formulaI Use exponential and trigonometric functionsI Define and use hyperbolic functionsI Use logarithmsI Do all of the above with complex numbers!I Solve harmonic oscillator and driven-damped oscillator

Patrick K. Schelling Introduction to Theoretical Methods



I Represent complex numbers in various waysI Use complex algebra

I Complex infinite seriesI Determine functions of complex numbersI Use Euler’s formulaI Use exponential and trigonometric functionsI Define and use hyperbolic functionsI Use logarithmsI Do all of the above with complex numbers!I Solve harmonic oscillator and driven-damped oscillator




I Represent complex numbers in various waysI Use complex algebraI Complex infinite series

I Determine functions of complex numbersI Use Euler’s formulaI Use exponential and trigonometric functionsI Define and use hyperbolic functionsI Use logarithmsI Do all of the above with complex numbers!I Solve harmonic oscillator and driven-damped oscillator




I Represent complex numbers in various waysI Use complex algebraI Complex infinite seriesI Determine functions of complex numbers

I Use Euler’s formulaI Use exponential and trigonometric functionsI Define and use hyperbolic functionsI Use logarithmsI Do all of the above with complex numbers!I Solve harmonic oscillator and driven-damped oscillator




I Represent complex numbers in various waysI Use complex algebraI Complex infinite seriesI Determine functions of complex numbersI Use Euler’s formula

I Use exponential and trigonometric functionsI Define and use hyperbolic functionsI Use logarithmsI Do all of the above with complex numbers!I Solve harmonic oscillator and driven-damped oscillator




I Represent complex numbers in various waysI Use complex algebraI Complex infinite seriesI Determine functions of complex numbersI Use Euler’s formulaI Use exponential and trigonometric functions

I Define and use hyperbolic functionsI Use logarithmsI Do all of the above with complex numbers!I Solve harmonic oscillator and driven-damped oscillator




I Represent complex numbers in various waysI Use complex algebraI Complex infinite seriesI Determine functions of complex numbersI Use Euler’s formulaI Use exponential and trigonometric functionsI Define and use hyperbolic functions

I Use logarithmsI Do all of the above with complex numbers!I Solve harmonic oscillator and driven-damped oscillator




I Represent complex numbers in various waysI Use complex algebraI Complex infinite seriesI Determine functions of complex numbersI Use Euler’s formulaI Use exponential and trigonometric functionsI Define and use hyperbolic functionsI Use logarithms

I Do all of the above with complex numbers!I Solve harmonic oscillator and driven-damped oscillator




I Represent complex numbers in various waysI Use complex algebraI Complex infinite seriesI Determine functions of complex numbersI Use Euler’s formulaI Use exponential and trigonometric functionsI Define and use hyperbolic functionsI Use logarithmsI Do all of the above with complex numbers!

I Solve harmonic oscillator and driven-damped oscillator




I Represent complex numbers in various waysI Use complex algebraI Complex infinite seriesI Determine functions of complex numbersI Use Euler’s formulaI Use exponential and trigonometric functionsI Define and use hyperbolic functionsI Use logarithmsI Do all of the above with complex numbers!I Solve harmonic oscillator and driven-damped oscillator


Real and imaginary parts of a complex number

Consider a complex number written as z = x + iy

I Re z = x

I Im z = y

What if it is not presented as z = x + iy? We can always simplifyto this form with complex algebra.

(2 + i)2 = 4 + 4i − 1 = 3 + 4i (1)

What if there is a complex denominator? We can multiply in thenumerator and denominator by the complex conjugate or conjugateof the denominator.

i + 1

i − 1=

(i + 1

i − 1

)(−i − 1

−i − 1

)= −i (2)


The complex plane

I Representing a complex number as z = x + iy suggests wecan plot them in a complex plane.

I We can also represent in polar form z = re iθ

I Equivalently z = r (cos θ + i sin θ)

I The modulus or magnitude is r

I The angle or phase is θ


Complex algebra

We need to work with complex representationsz = x + iy = re iθ = r (cos θ + i sin θ). It should become easy to gofrom one representation to another!

I For z = x + iy , the complex conjugate z∗ = x − iy

I For z = re iθ, z∗ = re−iθ

I Given z = x + iy , we can find r fromr = |z | =

√z∗z =

√x2 + y2

I Given z = x + iy , we see cos θ = x/r and sin θ = y/r

I We can find θ using θ = cos−1 x/r = sin−1 y/r

I Also tan θ = yx , or θ = tan−1 y

x


Complex power series

Previously we had real power series

∞∑n=1

anxn (3)

We can also consider a power series in z = x + iy

∞∑n=1

anzn (4)

We need to determine whether a series converges using the ratiotest:

ρ = limn→∞

|an+1zn+1

anzn| (5)

When ρ < 1, the series converges.


Example of convergence test

Consider the series,

1 + iz +(iz)2

2!+

(iz)3

3!+

(iz)4

4!+ ... (6)

We see from this that,

ρ = limn→∞

|an+1zn+1

anzn| = lim

n→∞| iz

n + 1| = 0 (7)

So indeed this series converges. In fact we will see that itconverges to e iz .


Euler’s formula

We previously had the infinite-series (i.e. Mclaurin series) for cos θand sin θ,

cos θ = 1− θ2

2!+θ4

4!− ... (8)

sin θ = θ − θ3

3!+θ5

5!− ... (9)

Using dez

dz = ez , we can also represent ez by the series,

ez = 1 + z +z2

2!+

z3

3!+

z4

4!+

z5

5!+ .... (10)

If we take z = iθ, then comparison of these series proves Euler’sformula,

e iθ = cos θ + i sin θ


Examples using Euler’s formula

• Express z = 2eπi4 in the form z = x + iy

From Euler’s formula,

2eiπ4 = 2 cos

π

4+ 2i sin

π

4=√

2 + i√

2

• Express z =(

i√

21+i

)6in the form z = x + iy

Using Euler’s formula, i = eiπ2 and 1 + i =

√2e

πi4 , then we see(

i√

2

1 + i

)6

=

(e

iπ2

eiπ4

)6

=(

eiπ4

)6= e

3πi2

Then we use Euler’s formula,

e3πi2 = cos

3π

2+ i sin

3π

2= −i


Powers and roots of complex numbers

We can start with the form z = re iθ, then to take to the nth power,

zn =(

re iθ)n

= rne inθ

Likewise, if we want the nth root of z,

z1/n =(

re iθ)1/n

= r1/ne iθ/n

We used this in the second example in the last slide


Examples of roots of complex numbers

• For z = −8, determine z1/3 = (−8)1/3 in the form x + iyFor z = −8, we can see r = 8 and θ = π, so z = 8e iπ, and then

z1/3 = (−8)1/3 =(8e iπ

)1/3= 2e iπ/3

Then we use Euler’s formula,

2e iπ/3 = 2 cosπ/3 + 2i sinπ/3 = 1 + i√

3

This can be easily checked without invoking Euler’s formula,(1 + i

√3)3

= −8


Exponential and trigonometric functions

Euler’s formula can be used to find representations of cos θ andsin θ

sin θ =e iθ − e−iθ

2i(11)

cos θ =e iθ + e−iθ

2(12)

Instead of just real θ, this also applies for complex z ,

sin z =e iz − e−iz

2i(13)

cos z =e1z + e−iz

2(14)


Example of complex exponentials for integration

Complex exponentials are useful for integrating products of sin andcos functions. For example• Solve the integral

∫ π−π cos 2x cos 3xdx .

First we make note that cos2x = e2ix+e−2ix

2 and cos3x = e3ix+e−3ix

2

∫ π

−πcos 2x cos 3xdx =

1

4

∫ π

−π

(e5ix + e ix + e−ix + e−5ix

)dx

This integral is easy, and we get

1

4

[(e5ix − e−5ix

5i

)+

(e ix − e−ix

i

)]π−π

=

[1

10sin 5x +

1

2sin x

]π−π

= 0

The complex exponential form is also useful in differentialequations.


Hyperbolic functions

If we start with our representations of cos and sin as complexexponentials, then consider pure imaginary argument (e.g. z = iy)

sin iy = iey − e−y

2(15)

cos iy =ey + e−y

2(16)

This provides us with definitions for the hyperbolic functions,sinh y = −i sin iy and cosh y = cos iy . More generally for any z ,

sinh z =ez − e−z

2(17)

cosh z =ez + e−z

2(18)

Also tanh z = sinh zcosh z , coth z = cosh z

sinh z ,etc.


Example with hyperbolic functions

• Write sinh(ln 2 + iπ

3

)in x + iy form

We use the representation of sinh in terms of exponentials,

sinh

(ln 2 +

iπ

3

)=

e(ln 2+iπ/3) − e−(ln 2+iπ/3)

2=

2e iπ/3 − (1/2)e−iπ/3

2

Then using Euler’s formula for the complex exponentials, we get

sinh

(ln 2 +

iπ

3

)=

3

8+

5√

3

8i


Logarithms of complex numbers

I It is possible to take logarithms of negative or even complexnumbers

I If z = ew then w = ln z where z is complex

I w = ln z = ln(re iθ) = ln r + iθ

I Since we can add 2nπ to θ, n integer, and get same result, wehave most generally:

ln (re iθ) = ln r + i(θ ± 2nπ) (19)


Complex roots and complex powers

We can take a complex number a to a complex power b! We canoften evaluate using,

ab = eb ln a (20)

• For example, evaluate i i in the form x + iy

i i = e i ln i

Then using i = e iπ/2e±2nπi (from Euler’s formula), we seei ln i = −π/2± 2nπ, and finally,

i i = e−π/2±2nπ

While there are an infinite number of answers, note that they areall real!


Example of a complex number and a real root

• Evaluate i1/2 in the form x + iy .

i1/2 = e(1/2) ln i = e(1/2) ln (e iπ/2±2nπi ) = e iπ/4±inπ

Since e inπ = 1 for even n and e inπ = −1 for odd n, we have twoanswers,

i1/2 = ±e iπ/4 = ±1 + i√2

Not surprising that the square root gives two possible results, as itdoes for real numbers.• Check directly this result,

i1/2i1/2 =

[1 + i√

2

] [1 + i√

2

]= i


Inverse trigonometric and hyperbolic functions

I For w = cos z , we have z = arccos w = cos−1 w

I Likewise w = sin z , we have z = arcsin w = sin−1 w

I If z is real, w is always between −1 and +1

I If z is complex, w does not have to be between −1 and +1

It is convenient to use the forms,

w = cos z =e iz + e−iz

2

w = sin z =e iz − e−iz

2i


Example of cos z , sin z with complex z

• Find z = arccos (i√

8) in the form x + iyWe start with z = arccos (i

√8) and write equivalently cos z = i

√8

cos z =e iz + e−iz

2= i√

8

For simplicity take u = e iz then we can write,

u + u−1

2= i√

8

Which gives the quadratic equation,

u2 − 4i√

2u + 1 = 0

This has the roots u = (2√

2± 3)i , so iz = ln[(

2√

2± 3)

i]...

complete in the homework!


Simple-harmonic oscillator

For a pendulum oscillating with small angle θ, or a mass-springsystem obeying Hooke’s law, we get simple harmonic motion

d2θ

dt2+

g

lθ = 0

d2y

dt2+

k

my = 0

y(t) = A cos(ω0t) + B sin(ω0t) = Ce iω0t . Note that C is complex,but we usually take A and B to be real. Also ω0 =

√k/m.

I A and B (or C ) are determined by conditions at t = 0(position and velocity)

I When using y(t) = Ce iω0t , we assume Re[y(t)] gives actualdisplacement

I The A and B can be directly related to C (homework)


Damped, driven simple-harmonic oscillator

Complex exponentials useful in the damped, driven oscillator

d2y

dt2+ 2b

dy

dt+ ω2

0y = FD cos(ωt)

We will assumeY (t) = Ce iωt = Ae i(ωt+φ)

with the driving force FDe iωt . The A and δ are real and C iscomplex. Substitute and solve for C ,

C =FD

ω20 − ω2 + 2ibω

We would like to get C into the form Ae iφ, so we multiplynumerator and denominator by complex conjugate of denominator


Damped, driven simple-harmonic oscillator, cont.

C =FD

ω20 − ω2 + 2ibω

[ω2

0 − ω2 − 2ibω

ω20 − ω2 − 2ibω

]=

FD

[ω2

0 − ω2 − 2ibω]

(ω20 − ω2)2 + 4b2ω2

We now need A = |C | =√

C ∗C and also an expression for φ. Wefind

A = |C | =FD√

(ω20 − ω2)2 + 4b2ω2

and for φ,

tanφ =2bω

ω20 − ω2


Damped, driven simple-harmonic oscillator, cont.

Then we get Y (t) = Ae i(ωt−φ) with the A and φ given on lastslide. Then finally, for a driving force FD cosωt, we take the realpart of Y (t),

y(t) = A cos (ωt − φ)

Using complex exponentials was somewhat simpler, becauseotherwise we would have both sin and cos terms to keep track of,and also solve for A and B (see previously).


LCR circuit

I LCR circuit completely analogous to damped, drivensimple-harmonic oscillator

Consider an alternating current source with potentialVD = V0 cosωt

VL + VR + VC = VD

Ld2q

dt2+ R

dq

dt+

q

C= V0 cosωt

Q(t) = Q0e i(ωt−φ)

Solve as a homework problem, but given damped-driven oscillatorsolution, this should be easy!


Damped, driven simple-harmonic oscillator

Is that everything for this problem?

I No! We only considered particular solutions

I Particular solution valid for long-time behavior

I We also need homogeneous solutions

I Homogeneous solutions relevant for no driving force andbehavior right after driving force turned on

I This is a subject for more study in Chapter 8!


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