Chapter 2: Complex Numbers

I. Background

• Complex number (z): z = A + iB where (note: engineers use )

“real part”: Re{z} = A“imaginary part”: Im{z} = B(both A and B are real numbers)

• Graphical Representation:

ex: 2+3i 3-4i

1i 1j

Ch. 2- Complex Numbers>Background

• Polar vs. rectangular coordinates:Rectangular coordinates: z = A + iBPolar coordinates: z = reiθ

(note: |z| ≡ r , z = θ)

To move between the two: A + iB = (r cosθ) + i (r sinθ) = reiθ

because eiθ = cosθ + sinθ (see section 7-9 for proof)so:

θ is in radians, r ≥ 0

ex: z=3-i4 (write in polar coordinates)

AB

BArrB

rA1

22

tanor

sin

cos

Ch. 2- Complex Numbers>Background

• Beware when using

Your calculator can’t tell the difference between θ and θ+π.It should always find . You may need to add π to get to the correct quadrant!

ex: z = -3 + i4

In practice, polar form reiθ is often easier to use because it’s easier to differentiate and integrate reiθ.

)(tan 1AB

)(tan)(tan 11AB

AB

22 0

Ch. 2- Complex Numbers>Background

II. Algebra with complex numbers

1) Complex conjugate:Let z = A + iBThen is the complex conjugate.

In polar coordinates: z = reiθ z* = re-iθ

Why? Recall: z = reiθ = r (cosθ + isinθ) z* = r (cosθ - isinθ) = r (cos(-θ) + isin(-θ))

= re-iθ

ex:1) z = 2 + i32) z = 3 – i43) z = 5ei(/3)

Note: z z* = |z|2 (Proof in a minute…)

iBA (or z*)z

Ch. 2- Complex Numbers>Algebra>Complex Conjugate

For the remaining examples:Let z1 = 2 + i3 r = 3.6, (θ) = 0.98 rad z1 = 3.6ei(0.98)

z2 = 4 – i2 r = 4.5, (θ) = -0.46 rad z1 = 4.5ei(-0.46)

2) Addition: (A + iB) + (C + iD) = (A+C) + i(B+D) there is no easy way to do this in polar coordinates:

ex: z1 + z2 (in rectangular coordinates)

ex: z1 + z2 (in polar coordinates)

?2121 ii erer

Ch. 2- Complex Numbers>Algebra>Addition

3) Multiplication: (A + iB) (C + iD) = AC + i(AB) + i(BC) – BD (term by term)

ex: z1•z2 (rectangular)

ex: z1•z2 (polar)

)(2121

2121 )())(( iii errerer

Ch. 2- Complex Numbers>Algebra>Multiplication

4) Multiplying by complex conjugate:

ex: z1z1*

|z1|2

This is true for any complex number: |z1|2 = z1z1*Proof: Let z = A + iB

z z*= (A + iB) (A – iB) = A2 – iAB + iBA + B2

= A2 + B2

= r2

= |z|2

Ch. 2- Complex Numbers>Algebra>Multiplying by Complex Conjugate

5) Division:

(more difficult in rectangular form)

ex: (polar)

ex: (rectangular)

)(

2

1 21

2

1

2

1

i

rr

i

i

eer

er

2

1

zz

2

1

zz

Ch. 2- Complex Numbers>Algebra>Division

6) More complex conjugates:

Say I want the complex conjugate of a messy equation:

Change all i -i

7) Complex equations: (A + iB) = (C + iD) iff A = C and B = D

ex: z1 + z2 = x + i(3x + y)

42

32

3 9

32

ixx

ixxiz

42

32

3 9

32*

ixx

ixxiz

Ch. 2- Complex Numbers>Algebra>More Complex Conjugates & Complex Equations

8) Powers: do these in polar form: rectangular form switch to polar first

ex: z12 (polar)

ex: z12 (rectangular)

1111 )( innni erer

Ch. 2- Complex Numbers>Algebra>Powers

9) Roots: Polar coordinates:

ex:

Check by changing back to rectangular coordinates.

Find another root (add 2 to ).

Convert back to rectangular coordinates:

1z

)(11

)( nnn ii erre

Ch. 2- Complex Numbers>Algebra>Roots

In general: has n possible roots!

ex:

ex:

nz1

24 zz

3 find 27 zz

Ch. 2- Complex Numbers>Algebra>Roots

10) Complex Exponentials:

let z = x + iythen ez = ex+iy = exeiy = ex(cosy + isiny)

ex:

2

23

3

i

i

i

e

e

e

ei

Ch. 2- Complex Numbers>Algebra>Complex Exponentials

11) Trig Functions:

cos2

sincos

sincos

ii

i

i

ee

ie

ie

sin2

)sincos(

sincos

iee

ie

ie

ii

i

i

Ch. 2- Complex Numbers>Algebra>Trig Functions

2cos

ii ee

i

ee ii

2sin

This is very useful for derivatives and integrals:

ex:

ex:

)(cos zdzd

Good Trick:

ii

ii

1

1

dxxx )3cos()2cos(

Ch. 2- Complex Numbers>Algebra>Trig Functions

12) Hyperbolic Functions:

↔ Usual sin/cos functions Hyperbolic functions

(entirely real)

Likewise:

etc…

2

2

cos

siniziz

iziz

ee

iee

z

z

2

2

cosh

sinhzz

zz

ee

ee

z

z

z

zz

z

z

cosh1

coshsinh

sech

tanh

Ch. 2- Complex Numbers>Algebra>Hyperbolic Functions

13) Natural logarithm:

Note: i = i(+2) = i(+4) = … So ln(z) has infinitely many solutions: ln(z) = ln(r) + i = ln(r) + i(+2) = …

‘Principal solution’ has .

irerrez ii )ln()ln()ln()ln(ln

2,0

Ch. 2- Complex Numbers>Algebra>Natural Logarithm

Physics Example: RLC Circuit

applied emf

Then

Find I0 & Φ and Z (the complex impedance).

Ch. 2- Complex Numbers>Example: RLC Circuit

)(ωωII

ωtVV

sin

sin

From Physics 216:

What is the impedance?(yuck!)

IVVV

IV

CCC

dtdI

L

R

LCRZ

tIIdtV

tLILV

tRIIRV

)cos(

)cos(

)sin(

011

0

0

Ch. 2- Complex Numbers>Example: RLC Circuit

It’s easier to do this:

tieVV 0

Ch. 2- Complex Numbers>Example: RLC Circuit

What’s the physically real I?

where

Recall our actual driving voltage:

We wrote this as .

Only has any physical reality.

So, same goes for current:

The physically real current is with same Io & Φ as above.

R

L

LR

VI

tieII

C

C

11

212

00

0

tan

)(

)(

tieV 0

tieV 0Im

)sin(Im 0)(

0 tIeI ti

Ch. 2- Complex Numbers>Example: RLC Circuit

Resonance

Defn: The frequency at which Z is entirely real:

So, at resonance ωR:

(resonance frequency)

Note that for this circuit, I0 is max when ω=ωR:

(at resonance)

See plot of I0 vs. ω.

CLiRZ 1

LCRCR RL 11 0

R

V

LR

VI

C

0

212

00

)(

Ch. 2- Complex Numbers>Example: RLC Circuit

Complex Impedances

We found:

“reactance” or where “complex impedance”

Complex impedances behave just like resistors in series and parallel.

IXVIV

IXVLIiV

RIV

CCCiC

LLL

R

1

CiC

L

R

X

LiX

RX

1

From Physics 216:

Instead, it’s easier to use complex notation:

A:Amplitude

right) the (to v :Velocity

φ :Phase

Tω :Period

λk :Wavelength

φ)kx- Asin(ω :Wave

kω

ω2π

T2π

k2π

λ2π

φ)kxi(ω(Ae

Ch. 2- Complex Numbers>Example: RLC Circuit