complex analysis - METU | Department of Mechanical Engineering

Transkript

ME 210 Applied Mathematics for Mechanical Engineers
COMPLEX ANALYSIS
Complex Numbers
Complex numbers are useful in the fields such as:
Solutions of some types of linear differential equations
Electrical circuit analyses
Inverse transformations
Solutions of field problems
etc.
Prof. Dr. Bülent E. Platin
Spring 2014 – Sections 02 & 03
1/39
The general form of a complex number z, is z = x + i y or z = a + i b
where
i = -1
x = Re (z) is a real number and called as the real part of z.
y = Im (z) is a real number and called as the imaginary part of z.
Another representation of a complex number is its polar form which is obtained by
applying usual rectangular to polar coordinate transformation as
x = r cosθ
&
y = r sinθ
Using Euler’s identity eiθ = cosθ
cos + i sinθ
sin
z = r [cosθ + i sinθ] = r ei θ = r
θ
r = |z| = x 2 + y 2 = mod(z)
is called as the modulus (magnitude) of z
y
θ = arctan   = arg(z)
x
is called as the argument (angle) of z
2/39
The figure below denotes the geometric representation of a complex number in a
plane called complex z-plane, or Argand diagram, defined by a Cartesian
coordinate system whose abscissa is used to represent the real part of z, and
ordinate is used to represent the imaginary part of z.
y
z–plane
z=x+iy
z
θ = tan−1  
x
y
θ
x
z = r [ cos(θ ) + i sin(θ )]
x
z = r eiθ
z=r
x = r cos(θ )
y
y = r sin(θ )
 
z = r cos(θ ) + i r sin(θ )
r
0
x 2 + y2
r=
θ
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Note the following:
The modulus r, as the distance of the point z to the origin, is a non-negative quantity;
i.e., r ≥ 0
Geometrically, the argument θ is the directed angle measured in radians from
positive x–axis in counterclockwise direction. For z = 0, this angle is undefined.
For a given z ≠ 0, this angle is determined only up to integer multiples of 2π.
The value of θ that lies in the interval –π < θ ≤ π is called as the principle value
of the argument of z and is denoted by Arg(z), with capital A. Thus, – π < Arg(z) ≤ + π
It becomes extremely important to consider the quadrant of the z–plane in which
the point z lies when determining the value of the argument by using the above
equation.
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Example:
Determine the modulus and argument of the following complex numbers.
z=-1+i
⇒
r = 1+ 1 = 2
 1  3π
θ = arctan  =
≅ 2.356 rad (135o )
 − 1 4
z=3-4i
⇒
r = 9 + 16 = 5
−4
o
θ = arctan
 = −0.927 rad ( −53.1 )
 3 
z = - 5 - 12 i
⇒
r = 25 + 144 = 13
 − 12 
o
θ = arctan
 = −1.965 rad ( −112.6 )
 −5 
5/39
There are two fundamental rules for the manipulation of complex numbers:
1. A complex number z = x + i y is zero iff both its real and imaginary parts are zero;
i.e., z = 0
iff
x=0 & y=0
it follows that two complex numbers z1 = x1 + i y1 and z2 = x2 + i y2 are equal
iff both their real and imaginary parts are equal;
i.e., z1 = z2
iff
x1 = x2 & y1 = y2
Example:
What are the x and y values that satisfy the equation
(x2y – 2) + i (x + 2xy – 5) = 0
x2y – 2 = 0 and x + 2xy – 5 = 0
→
x = 1 and y = 2
x = 4 and y = 1/8
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2. Complex numbers obey the ordinary rules of algebra
(a + i b) ± (c + i d) = (a ± c) + i (b ± d)
(a + i b) (c + i d) = (a c – b d) + i (a d + b c)
(a + i b)2 = a2 – b2 + i (2 a b)
with the addition that
i2 = – 1 , i3 = – i , i4 = +1 , i5 = i
, i6 = – 1 , i7 = – i , ..... modularity
Addition and Subtraction
z = z1 ± z2 = (x1 ± x2) + i (y1 ± y2)
Easier to perform in Cartesian form
Similar to addition and subtraction of vectors in a plane
Triangle Inequality:
Length of any one side of a triangle is less than or equal to the sum of the lengths
of other two sides. That is r ≤ r1 + r2
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Minus, Conjugate, Addition, and Subtraction in Polar Coordinates
y
y
z1 + z2
z2
z=a+bi
z1
x
x
_
-z=-a-bi
z =a-ib
z1 - z2
conjugate
- z2
8/39
Multiplication
z = z1 z2 = (x1 x2 – y1 y2) + i (x1 y2 + y1 x2) = r1r2 θ1 + θ2
r = r1 r2
θ = θ1 + θ2
Easier to perform in polar form
Not similar to multiplication of vectors in plane
z1 = r1 ( cos(θ1) + i sin(θ1) )
z 2 = r2 ( cos(θ 2 ) + i sin(θ 2 ) )
z1 z2 = r1 r2 ( cos(θ1 + θ 2 ) + i sin(θ1 + θ 2 ) )
9/39
Division
z=
 y1 x 2 - x1 y 2  r1
z1  x1 x 2 + y1 y 2 
=
+
i


=
2
2
z 2  x 22 + y 22 
x
+
y
2
2

 r2
θ1 − θ2
r = r1 / r2
θ = θ1 – θ2
Easier to perform in polar form
Not similar to any operation of vectors in plane
z1 = r1 [ cos(θ1) + i sin(θ1)]
z2 = r2 [ cos(θ 2 ) + i sin(θ 2 )]
z1
r1
=
z2
r2
[cos(θ1 - θ2 ) + i sin(θ1 - θ2 )]
,
r2 ≠ 0
10/39
Complex conjugate
_
The complex conjugate of a complex number z = x + i y is defined as z = x - i y
Another practical way to obtain division of two complex numbers is to use the
conjugate of the denominator:
Example:
a+ib
a + i b  c - i d 
a c + b d 
b c - a d 
= 
=
+
i
 

 c 2 + d2 
 c 2 + d2 
c+id
 c+id  c-id




c 2 + d2 ≠ 0
11/39
Note the following:
z + z = 2x
2
→
2
z+z
x = Re(z) =
2
2
z z = x + y =| z |
z − z = i2y
(z1 ± z 2 ) = z1 ± z2
z−z
→ y = Im(z) =
i2
(z1z 2 ) = z1z2
 z1  z1
  =
 z 2  z2
Example:
Solve the following (find z): z 2 + 2z = − 1 + i6
Let
z = x + iy → z = x – iy
then
(x + iy)2 + 2(x – iy) = x2 – y2 + i2xy + 2x – i2y = –1 + i6
which gives the following two real equations
x2 – y2 + 2x = –1 → y = ± (x+1)
2xy – 2y = 6 → y(x–1) = 3
whose solutions are found as
x=2&y=3
and
x = –2 & y = –1
Hence, the required solution for z is found as
z1 = 2 + i3
and
z2 = – 2 – i
12/39
Integer powers of complex numbers (de Moivre’s formula)
The idea of product of two complex numbers can be extended to the product of
n complex numbers
n
∏z
k
= z1z 2 … zn = r1r2 … rne
i(θ1 +θ2 + ... +θn )
k =1

=


n
 i ∑ θk
rk e k =1

k =1 
n
∏
If all the complex numbers multiplied are the same, the above expression gives
an important result
zn = rn einθ = rn [cosθ + i sinθ]n = rn [cos(nθ) + i sin(nθ)]
where n is either an integer or a rational number (that is, n=p/q where p and q are
integers) with the condition that z ≠ 0 for n = –1.
For r = 1, the expression reduces the form
(cosθ + i sinθ)n = cos(nθ) + i sin(nθ)
called as the de Moivre’s formula named after Abraham de Moivre (1667-1754)
13/39
Example: Find z5 for z = 1 + i
3
Solution 1: Taking the power of a binomial
(
5
z = 1+i 3
)
5
= 1 + 5 ( i 3) + 10 ( i 3)2 + 10 ( i 3)3 + 5 ( i 3)4 + ( i 3)5
= 1 + i 5 3 - 30 - i 30 3 + 45 + i 9 3
= 16 - i 16 3
Solution 2: Using de Moivre’s formula
5
(
z = 1+i 3
)
5

= 2

5
1

3 
+
i
=


2
 2

2  


i
5π
5π 

5
= 2  cos
+ i sin
=2 e

3
3 

5
1
3
= 32  - i
 = 16 - i 16
2
2 

5

i
π
π 

 cos 3 + i sin 3   =  2 e



π
3




5
5π
3
3
14/39
De Moivre’s formula may also be employed to evaluate the nth root of a complex
number z = r (cos θ + i sin θ)
Let
z = wn where
w = R eiφ = ?
then
z = Rn (cos nφ + i sin nφ) = Rn einφ
Hence
r (cos θ + i sin θ) = Rn (cos nφ + i sin nφ)
or
r eiθ = Rn einφ
Therefore
r = Rn
R = r1/n = n r
or
nφ = θ + 2kπ or
1/n
w= z
=r
1/n
and cos θ = cos nφ
&
sin θ = sin nφ
,
k = 0, 1, ..., n-1
θ + 2kπ
φ=
, k = 0, 1, … , n − 1
n

θ + 2 k π 
θ+ 2k π
 + i sin 
 cos 
n
n







15/39
Example: Find
Let
3
8
z=8+i0

0 + 2 k π 
 0 + 2 k π 
z1/3 = 81/3  cos 
+
i
sin



3
3





,
k = 0, 1, 2
z1 = 81/3 cos(0) + i sin(0) = 2

0 + 2 π 
0 + 2 π
z2 = 81/3  cos 
+
i
sin


3
3




1/3
z3 = 8

 = 2


0 + 4 π 
 0 + 4 π 
cos
+
i
sin
=2






3
3





 1
3
+
i

 = - 1 + i 3
2 
 2
 1
3
-i
  = - 1 - i 3
2 
 2
16/39
Note that all solutions have a common modulus of 2, but their arguments differ
from each other, and they are located on a circle of radius of 2 about the origin of
the z–plane, equally spaced around it with an incremental angle of 2π/3 as
illustrated
(81/3)2
Im
i2
z–plane
2π/3
(81/3)1
4π/3
0
2
Re
(81/3)3
17/39
Example: Find
Let
i
z = 0 + i = 1 ei π /2

 π /2 + 2 k π 
 π /2 + 2 k π  
z1/2 = 11/2  cos 
+
i
sin



2
2





,
k = 0, 1

2
2
π 
 π 
z1 = 1 cos   + i sin    = ei π /4 =
+i
2
2
4
 4 


2
2
5 π 
 5 π 
i 5π /4
z2 = 1  cos 
+
i
sin
=
e
=
i

 4 
2
2
 4 



18/39
Complex Functions
Let z and w are two complex variables defined as z = x + i y and w = u + i v
where x, y, u, and v are real variables.
If, for each value of z in some portion of the complex z–plane, one or more value(s)
of w are defined, then w is said to be a complex function of z.
w = f(z) = u + i v = f(x + i y) = u(x,y) + i v(x,y)
This complex functional relationship between z and w may be regarded as a
complex mapping or complex transformation of points P within a region in
the z–plane (called as the Domain) to corresponding image point(s) Q within a
region in the w–plane (called as the Range).
19/39
Complex Functions as Complex Mapping
y
v
z–plane
P
w = f(z)
mapping
Q
domain
range
x
w–plane
u
20/39
Example:
Find the ranges, R, in the complex w–plane of the following complex functions
corresponding to their domains, D, in z–plane. Plot D and R regions.
Use the standard notation, z = x + i y = r eiθ & w = u + i v
(a) w = f(z) = i z , D: Re(z) ≥ 0
w = i z = i (x + i y) = – y + i x
u(x,y) = – y & v(x,y) = x
Re(z) = x ≥ 0
=>
v(x,y) ≥ 0
21/39
(b) w = f(z) = 3 z – π , D: – π ≤ Re(z) ≤ π
w = 3 z – π = (3x – π) + i 3 y
u(x,y) = 3 x – π
&
–π ≤ x ≤ π
– 4 π ≤ u(x,y) ≤ 2 π
=>
v(x,y) = 3 y
(c) w = f(z) = z2 , D: |z| ≤ 1 and 0 ≤ Arg(z) ≤ π /4
w = z2 = (r ei θ)2 = r2 ei 2θ
|w| = r2 = |z|2 & Arg(w) = 2 Arg(z)
|z| ≤ 1 => |w| ≤ 1
0 ≤ Arg(z) ≤ π /4 => 0 ≤ Arg(w) ≤ π /2
22/39
In dealing with complex functions, it is possible to distinguish the following two cases.
Complex valued functions of a real variable
Complex valued functions of a complex variable
Complex valued functions of a real variable
In this case, to each value of a real variable, t (a ≤ t ≤ b), one or more complex
value(s) of z are assigned, which can be shown as z(t) = x(t) + i y(t) = f(t)
Some examples for this type of complex function are:
z = ei t (0 ≤ t ≤ 2 π) => z = cos(t) + i sin(t) => x(t) = cos(t) & y(t) = sin(t)
z = t + i t2 (for all t) => x(t) = t & y(t) = t2
z = t (1 – i t ) (1 + i 2 t ) (t ≥ 0) => x(t) = t (1 + 2 t) & y(t) = t
23/39
This form can conveniently be used to represent the parametric equations of planar
curves in complex z–plane.
Example:
For the slider-crank mechanism shown in the figure, it is desired to represent the
position of the point Q as a (complex valued) function of the horizontal position t
(a real variable) of the slider P as z(t) = x(t) + i y(t)
(h – r ≤ t ≤ h + r)
Q
h
r
θ
P
y
β
x
t
24/39
Q
h
r
θ
P
y
β
x
t
Note that as P moves on a straight line, Q moves on a circle.
For every position of P denoted by t, it is possible to find two positions for Q
denoted by z.
Therefore, the relationship between the positions of P and Q can be considered as
a mapping of points lying on a straight line to points lying on a curve (circle).
By using either θ or β angle, the coordinates x and y of Q can be related to the
position t of P.
25/39
Q
h
r
θ
P
y
β
x
t
x(t) = r cos(θ) = t – h cos(β)
y(t) = r sin(θ) = h sin(β)
Since x2 + y2 = r2
=> [t – h cos(β)]2 + [h sin(β)]2 = r2
t2 + h2 - r 2
cos(β) =
2ht
 t 2 + h2 - r 2 
sin( β ) = ± 1 - 

2
h
t


2
26/39
Q
h
r
θ
P
y
β
x
t
The parametric representation of the position of Q in the complex z–plane becomes
2

2
2
2
 t +h −r
t +h −r  

 
z ( t ) = t −
 ± i h 1 − 

2t
2ht

 

 


2
2
2
(
 t 2 − h 2 + r 2   4h 2 t 2 − t 2 + h 2 − r 2
=
 ± i
2t
2t

 
)
2




27/39
For r = 1 and h = 4, x(t) and y(t) are shown for 3 ≤ t ≤ 5 in the Figure.
1.00
0.75
y(t)
0.50
x(t)
0.25
0.00
-0.25
-0.50
-0.75
-1.00
3.0
3.5
4.0
Real variable t
4.5
5.0
Position of Q as a function of Slider Position
28/39
Complex valued functions of a complex variable
In this case, to each value of a complex variable z in a domain D of the z-plane,
one or more complex values w are assigned, which can be shown as w = f(z)
For a given complex variable z, a complex value w may be obtained regardless
of the function, f, itself being a real function or a complex function.
Some examples for this type of complex function are:
w=iz →
→
i (x + i y) = – y + i x
u=–y & v=x
w = z2 → (x + i y)2 = (x2 – y2) + i 2 x y →
u = x2 – y2 & v = 2 x y
w = z1/4 → (r eiθ)1/4 = R eiΦ
→
R = r1/4 & Φ = (θ + 2 k π) / 4
w = ez → ex + iy = ex (cos y + i sin y)
→
u = ex cos y & v = ex sin y
29/39
Note that once w = f(z) is known, it is a straightforward algebra to obtain u & v (or R
& f) in terms of x & y (or r & q).
However, if u(x,y) & v(x,y) are given in turn, to express w as a function of z is not a
trivial problem at all.
In fact, it may even not have a solution.
If there is a solution, then a suitable manipulation must be carried out in order
to come up with the correct w = f(z) expression.
Example:
Given w = u(x,y) + i v(x,y) = (x2 + x – y2 + 1) + i y (2x + 1) express w as a function
of z = x + i y
Rearranging gives
w = x2 + x – y2 + 1 + i 2 x y + i y
= x2 + i 2 x y – y2 + x + i y + 1
= (x + i y)2 + (x + i y) + 1
= z2 + z + 1
30/39
Some elementary functions of z
Exponential function: Using the basic definition of exponential function for real
variables, one gets
∞
zn
z2
z3
z4
f (z) = e = ∑
=1+z+
+
+
+…
2!
3!
4!
k =1 n!
z
u(x,y) = ex cos(y) & v(x,y) = ex sin(y)
Note that
f ( z ) = e- z
f ( z ) = ei z
f (z) = e
-iz
Mod(ez) = |ez| = ex & Arg(ez) = y
∞
(-z)n
z 2 z3
z4
=∑
=1-z+
+
-…
2! 3!
4!
k =1 n!
∞
(i z)n
z 2 i z3
z4
=∑
=1+iz+
-…
2!
3!
4!
k =1 n!
∞
(- i z)n
z2
i z3
z4
=∑
=1-iz+
+
-…
n!
2!
3!
4!
k =1
31/39
Hyperbolic functions:
cosh(z) = (ez + e–z)/2 = cosh(x) cos(y) + i sinh(x) sin(y)
sinh(z) = (ez – e–z)/2 = sinh(x) cos(y) + i cosh(x) sin(y)
If z = 0 + iy
=>
cosh(iy) = cos(y)
sinh(iy) = i sin(y)
Familiar laws for the hyperbolic functions:
cosh2(z) – sinh2(z) = 1
cosh(z1 + z2) = cosh(z1) cosh(z2) + sinh(z1) sinh(z2)
sinh(z1 + z2) = sinh(z1) cosh(z2) + cosh(z1) sinh(z2)
cosh(2 z) = cosh2(z) + sinh2(z) = 1 + 2 sinh2(z) = 2 cosh2(z) – 1
sinh(2 z) = 2 sinh(z) cosh(z)
32/39
Trigonometric functions:
cos(z) = (eiz + e–iz)/2 = cos(x) cosh(y) – i sin(x) sinh(y)
sin(z) = (eiz – e–iz)/2i = sin(x) cosh(y) + i cos(x) sinh(y)
If z = 0 + iy
=>
cos(iy) = cosh(y)
sin(iy) = i sinh(y)
Familiar laws for the trigonometric functions:
cos2(z) + sin2(z) = 1
cos(z1 + z2) = cos(z1) cos(z2) – sin(z1) sin(z2)
sin(z1 + z2) = sin(z1) cos(z2) + cos(z1) sin(z2)
cos(2 z) = cos2(z) – sin2(z) = 1 – 2 sin2(z) = 2 cos2(z) – 1
sin(2 z) = 2 sin(z) cos(z)
33/39
Logarithmic function:
The logarithm of z = r eiθ that is defined implicitly as the function w = ln(z) which
satisfies the equation z = ew
Hence, eu = r
or
u = ln(r),
or
r eiθ = eu + i v = eu eiv
and
v=θ
Thus, w = u + i v = ln(r) + i θ = ln |z| + i arg(z)
If the principal argument of z is denoted by Arg(z), then this equation can
be rewritten as ln(z) = ln |z| + i [Arg(z) + 2 k π]
k = 0, ±1, ±2, …
This indicates that complex logarithmic function is infinitely multi–valued.
For k = 0, the part (branch) of the logarithmic function is called as the principal
value.
34/39
Familiar laws for the logarithms of real quantities all hold for the logarithms of
complex quantities in the following sense:
Im(w)
i23π/4
ln(z1z2) = ln(z1) + ln (z2)
w-plane
ln(z1/z2) = ln(z1) – ln (z2)
ln(zk)
= k ln(z)
i15π/4
k = 0, ±1, ±2, …
Example:
i7π/4
Compute w(z) = ln(1 – i)
ln( 2 )
ln(z) = ln |z| + i [Arg(z) + 2 k π]
k = 0, ±1, ±2, …
–iπ/4
Re(w)
( )
 π

ln(1− i) = ln 2 + i − + 2kπ  , k = 0, ± 1, ± 2,L
 4

( )
( )
( )
–i9π/4
π
7π
9π
ln(1− i) = ln 2 − i
, ln 2 + i
, ln 2 − i
,L
4
4
4
–i17π/4
35/39
General Powers of z
w(z) = zc , c is any number, real or complex
zc = ec ln(z) = e
If c = n:
n
z =e
n ln(z)
Because
If c = m/n:
c ln(r) + i (θ0 + 2 k π ) 
=e
n ln(r) + i (θ0 ) 
ei 2 n k π = 1
zm / n = r m / n e
=e
,
k = 0, ± 1, ± 2, ...
( ) ei n θ = r n ei n θ
ln r n
0
0
for all k’s
i m/n (θ0 + 2 k π )

m

m

= r m / n cos 
θ
+
2
k
π
+
i
sin
θ
+
2
k
π
(0
)
) 
n ( 0
n





k = 0, 1, 2, ... , n-1
36/39
If c is an irrational number, not expressible in the form m/n
zc = r c e
i c (θ0 + 2 k π )
= r c  cos ( c (θ0 + 2 k π ) ) + i sin ( c (θ0 + 2 k π ) )
k = 0, ± 1, ± 2, ...
If c is complex, i.e., c = a + i b
zc = e
=e
=e
(a + i b) ln(r) + i (θ0 + 2 k π ) 
a ln(r) - b (θ0 + 2 k π )
a ln(r) - b (θ0 + 2 k π )
e
i b ln(r) +4 a (θ0 + 2 k π )
 cos (b ln(r) + a (θ0 + 2 k π ) )



 + i sin (b ln(r) + a (θ0 + 2 k π ) )
k = 0, ± 1, ± 2, ...
37/39
Example:
Find all possible values of (1)i
→
r = 1 , θo = 0 , c = i
(1)i = ei ln(1) = ei[ln(1)+i(0+2kπ )] = e −2kπ , k = 0, ± 1, ± 2,...
Example:
→
Find all possible values of (i)i
π
i[ln(1)+i( + 2kπ )]
2
(i)i = ei ln(i) = e
r = 1 , θo = π/2 , c = i
= e −( 4k +1)π / 2 , k = 0, ± 1, ± 2,...
Summary: zc is
single valued if c is an integer
n-valued if c = 1/n
n-valued if c = m/n
infinite valued if c is real and irrational
infinite valued if Im(c) ≠ 0
38/39
END OF WEEK 13
39/39

complex analysis - METU | Department of Mechanical Engineering

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