# chapter2

```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 )
difference between θ and θ+π.


It should always find 2  0  2 .
You may need to add π to get to
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…)
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)
(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)
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