Post on 26-Dec-2015
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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)
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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
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2
1
2
1
i
rr
i
i
eer
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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