Ch. 2- Complex Numbers>Background
Chapter 2: Complex Numbers
I. Background
•
Complex number (z):
z = A + iB where
i
1
(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
j
1
)
Ch. 2- Complex Numbers>Background
•
Polar vs. rectangular coordinates:
Rectangular coordinates: z = A + iB
Polar 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:
2
2
A r cos
r A B
or
1 B
B
r
sin
tan A
θ is in radians, r ≥ 0
ex: z=3-i4 (write in polar coordinates)
Ch. 2- Complex Numbers>Background
•
Beware when using tan
1
( BA )
tan
1
( BA ) tan
1
( BA )
Your calculator can’t tell the
difference between θ and θ+π.
It should always find 2 0 2 .
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.
Ch. 2- Complex Numbers>Algebra>Complex Conjugate
II. Algebra with complex numbers
1) Complex conjugate:
Let z = A + iB
Then z (or z*) A iB 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 + i3
2) z = 3 – i4
3) z = 5ei(/3)
Note: z z* = |z|2 (Proof in a minute…)
Ch. 2- Complex Numbers>Algebra>Addition
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)
i
i
there is no easy way to do this in polar coordinates: r1 e r2 e ?
1
ex: z1 + z2 (in rectangular coordinates)
ex: z1 + z2 (in polar coordinates)
2
Ch. 2- Complex Numbers>Algebra>Multiplication
3) Multiplication: (A + iB) (C + iD) = AC + i(AB) + i(BC) – BD (term by term)
( r1e
ex: z1•z2 (rectangular)
ex: z1•z2 (polar)
i 1
)( r2 e
i 2
) ( r1 r2 ) e
i ( 1 2 )
Ch. 2- Complex Numbers>Algebra>Multiplying by Complex Conjugate
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>Division
5) Division:
r1 e
r2 e
i 1
i 2
r1
r2
e
i ( 1 2 )
(more difficult in rectangular form)
ex:
ex:
z1
z2
z1
z2
(polar)
(rectangular)
Ch. 2- Complex Numbers>Algebra>More Complex Conjugates & Complex Equations
6) More complex conjugates:
Say I want the complex conjugate of a messy equation:
2 i 3 x ix
2
z3
9 x ix
2
3
4
Change all i -i
2 i 3 x ix
2
z3*
9 x ix
2
3
4
7) Complex equations: (A + iB) = (C + iD)
ex: z1 + z2 = x + i(3x + y)
iff A = C and B = D
Ch. 2- Complex Numbers>Algebra>Powers
8) Powers: do these in polar form: ( r1 e
i 1
n
) r1 e
n
rectangular form switch to polar first
ex: z12 (polar)
ex: z12 (rectangular)
in 1
9) Roots: Polar coordinates:
ex:
( re
i
1
n
1
n
) r e
i ( n )
z1
Check by changing back to rectangular coordinates.
Find another root (add 2 to ).
Convert back to rectangular coordinates:
Ch. 2- Complex Numbers>Algebra>Roots
In general: z
1
n
ex: z 4
ex: z 27
Ch. 2- Complex Numbers>Algebra>Roots
has n possible roots!
z 2
find
3
z
Ch. 2- Complex Numbers>Algebra>Complex Exponentials
10) Complex Exponentials:
let z = x + iy
then ez = ex+iy = exeiy = ex(cosy + isiny)
ex:
e
e
e
e
3 i
3 i 2
i
i 2
Ch. 2- Complex Numbers>Algebra>Trig Functions
11) Trig Functions:
e
e
e
e
(e
e
i
cos i sin
i
i
cos i sin
e
i
2 cos
cos
i
e
i
2
cos i sin
i
i
i
e
cos i sin )
e
i
2 i sin
sin
e
i
e
2i
i
Ch. 2- Complex Numbers>Algebra>Trig Functions
This is very useful for derivatives and integrals:
ex:
d
dz
(cos z )
Good Trick:
1
i
i
i
1
i
ex:
cos( 2 x ) cos( 3 x ) dx
Ch. 2- Complex Numbers>Algebra>Hyperbolic Functions
12) Hyperbolic Functions:
iz
iz
sin z
e e
2i
cos z
e e
2
iz
iz
Usual sin/cos functions
Likewise: tanh z
sinh z
cosh z
sech z
1
cosh z
etc…
↔
sinh z
cosh z
z
e e
2
z
z
e e
2
z
Hyperbolic functions
(entirely real)
Ch. 2- Complex Numbers>Algebra>Natural Logarithm
13) Natural logarithm:
i
i
ln z ln( re ) ln( r ) ln( e ) ln( r ) i
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 0 , 2 .
Ch. 2- Complex Numbers>Example: RLC Circuit
Physics Example: RLC Circuit
applied emf
V V sin ωt
Then I I sin (ωω )
Find I0 & Φ and Z (the complex impedance).
From Physics 216:
V R IR RI 0 sin( t )
VL L
dI
dt
VC
1
C
LI 0 cos( t )
Idt
1
C
I 0 cos( t )
What is the impedance?
Z
V
I
V R VC V L
I
(yuck!)
Ch. 2- Complex Numbers>Example: RLC Circuit
It’s easier to do this:
V V0e
i t
Ch. 2- Complex Numbers>Example: RLC Circuit
What’s the physically real I?
i ( t )
I I 0e
where
V0
I0
R ( L
2
tan
1
1
C
1
C
)
2
L
R
Recall our actual driving voltage:
We wrote this as V 0 e
Only Im V 0 e
i t
i t
.
has any physical reality.
So, same goes for current:
The physically real current is Im I 0 e
i ( t )
I
0
sin( t ) with same I o & Φ as above.
Ch. 2- Complex Numbers>Example: RLC Circuit
Resonance
Defn: The frequency at which Z is entirely real:
Z R i L
1
C
So, at resonance ωR:
RL
1
RC
0 R
1
LC
(resonance frequency)
Note that for this circuit, I0 is max when ω=ωR:
I0
V0
R ( L
2
See plot of I0 vs. ω.
1
C
)
2
V0
R
(at resonance)
Ch. 2- Complex Numbers>Example: RLC Circuit
Complex Impedances
We found:
V R RI
XR R
V L i LI V L X L I
X L i L
VC
1
i C
I VC X C I
where
XC
1
i C
“reactance” or
“complex impedance”
Complex impedances behave just like resistors in series and parallel.
Ch. 2- Complex Numbers>Example: RLC Circuit
From Physics 216:
Wave : Asin( ω - kx φ)
Wavelength
: k
Period : ω
2π
T
2π
λ
λ
T
2π
k
2π
ω
Phase : φ
Velocity : v
Amplitude
ω
k
(to the right)
: A
Instead, it’s easier to use complex notation: Ae
i( ω( kx φ)