Transcript Chap. 2 - Sun Yat

Chapter 2. Analytic Functions
Weiqi Luo (骆伟祺)
School of Software
Sun Yat-Sen University
Email：[email protected] Office：# A313
Chapter 2: Analytic Functions






Functions of a Complex Variable; Mappings
Mappings by the Exponential Function
Limits; Theorems on Limits
Limits Involving the Point at Infinity
Continuity; Derivatives; Differentiation Formulas
Cauchy-Riemann Equations; Sufficient Conditions for
Differentiability; Polar Coordinates;
 Analytic Functions; Harmonic Functions; Uniquely
Determined Analytic Functions; Reflection Principle
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12. Functions of a Complex Variable
 Function of a complex variable
Let s be a set complex numbers. A function f defined on
S is a rule that assigns to each z in S a complex number w.
Complex
numbers
Complex
numbers
f
z
w
S
S’
The domain of definition of f
The range of f
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12. Functions of a Complex Variable
Suppose that w=u+iv is the value of a function f at z=x+iy,
so that
u  iv  f ( x  iy)
Thus each of real number u and v depends on the real
variables x and y, meaning that
f ( z)  u( x, y)  iv( x, y)
Similarly if the polar coordinates r and θ, instead of x and
y, are used, we get
f ( z)  u(r, )  iv(r, )
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12. Functions of a Complex Variable
 Example 2
If f(z)=z2, then
case #1: z  x  iy
When v=0, f is a real-valued function.
f ( z)  ( x  iy)2  x2  y 2  i 2xy
u( x, y)  x2  y 2 ; v( x, y)  2 xy
case #2: z  rei
f ( z)  (rei )2  r 2ei 2  r 2 cos 2  ir 2 sin 2
u(r, )  r 2 cos 2 ; v(r, )  r 2 sin 2
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12. Functions of a Complex Variable
 Example 3
A real-valued function is used to illustrate some important
concepts later in this chapter is
f ( z) | z |2  x2  y 2  i0
 Polynomial function
P( z)  a0  a1z  a2 z 2  ...  an z n
where n is zero or a positive integer and a0, a1, …an are complex
The domain of definition is the entire z plane
constants, an is not 0;
 Rational function
the quotients P(z)/Q(z) of polynomials
The domain of definition is Q(z)≠0
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12. Functions of a Complex Variable
 Multiple-valued function
A generalization of the concept of function is a rule that
assigns more than one value to a point z in the domain of
definition.
Complex
numbers
Complex
numbers
f
w2
z
w1
wn
S’
S
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12. Functions of a Complex Variable
 Example 4
Let z denote any nonzero complex number, then z1/2 has
the two values
z
1/2

  r exp(i )
2
Multiple-valued function
If we just choose only the positive value of  r
z
1/2

 r exp(i ), r  0
2
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Single-valued function
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12. Homework
pp. 37-38
Ex. 1, Ex. 2, Ex. 4
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13. Mappings
 Graphs of Real-value functions
f=tan(x)
f=ex
Note that both x and f(x) are real values.
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13. Mappings
 Complex-value functions
f ( z)  f ( x  yi)  u( x, y)  iv( x, y)
mapping
v
y
w(u(x,y),v(x,y))
z(x,y)
u
x
Note that here x, y, u(x,y) and v(x,y) are all real values.
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13. Mappings
 Examples
y
w  z  1  ( x  1)  iy
v
z(x,y)
w(x+1,y)
Translation Mapping
x
v
y
w  z  x  yi
u
z(x,y)
Reflection Mapping
u
w(x,-y)
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13. Mappings
 Example

i
w  iz  i (re )  r exp(i (  ))
2
y
Rotation Mapping
v
w
z
r
θ

2 r
θ
u
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13. Mappings
 Example 1
w  z2
u  x2  y 2 , v  2xy
Let u=c1>0 in the w plane, then x2-y2=c1 in the z plane
Let v=c2>0 in the w plane, then 2xy=c2 in the z plane
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13. Mappings
 Example 2
The domain x>0, y>0, xy<1 consists of all points lying on
the upper branches of hyperbolas
u  x2  y 2 ;
v  2 xy  2  xy  1
x=0,y>0
x>0,y=0
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13. Mappings
 Example 3
2 i 2
w z r e
2
In polar coordinates
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14. Mappings by the Exponential Function
 The exponential function
we e
z
x  iy
 e e , z  x  iy
x iy
ρeiθ
ρ=ex, θ=y
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14. Mappings by the Exponential Function
 Example 2
w=exp(z)
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14. Mappings by the Exponential Function
 Example 3
w=exp(z)=ex+yi
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14. Homework
pp. 44-45
Ex. 2, Ex. 3, Ex. 7, Ex. 8
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15. Limits
 For a given positive value ε, there exists a positive value δ
(depends on ε) such that
when 0 < |z-z0| < δ, we have |f(z)-w0|< ε
meaning the point w=f(z) can be made arbitrarily chose to w0
if we choose the point z close enough to z0 but distinct from it.
lim f ( z )  w0
z  z0
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15. Limits
 The uniqueness of limit
If a limit of a function f(z) exists at a point z0, it is unique.
f ( z )  w0 & lim f ( z )  w1
Proof: suppose that zlim
z
zz
then  / 2  0, 0  0, 1  0
when 0 | z  z0 | 0
| f ( z)  w0 |  / 2;
0 | z  z0 | 1
| f ( z)  w1 |  / 2;
Let   min(0 , 1 ) , when 0<|z-z0|<δ, we have
0
0
| w1  w0 || ( f ( z)  w0 )  ( f ( z)  w1 ) |
| f ( z )  w0 |  | f ( z )  w1 |
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
2


2

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15. Limits
 Example 1
Show that f ( z)  iz / 2 in the open disk |z|<1, then
lim f ( z ) 
Proof:
z 1
i
2
i
iz i | i || z  1| | z  1|
| f ( z )  ||  |

2
2 2
2
2
  0,   2 , s.t.
when 0 | z  1|  ( 2 )
0
| z  1|
i
  | f ( z )  | 
2
2
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15. Limits
 Example 2
If
f ( z) 
z  ( x, 0)
z
z
then the limit
lim f ( z )
z 0
does not exist.
x  i0
lim
1
x 0 x  i 0
≠
z  (0, y)
0  iy
lim
 1
y 0 0  iy
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16. Theorems on Limits
 Theorem 1
Let f ( z)  u( x, y)  iv( x, y) z  x  iy
and z0  x0  iy0 ; w0  u0  iv0
then
lim f ( z )  w0
(a)
z  z0
if and only if
lim
( x , y ) ( x0 , y0 )
u ( x , y )  u0
and
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lim
( x , y ) ( x0 , y0 )
v( x, y )  v0
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(b)
16. Theorems on Limits
 Proof: (b)(a)
lim
( x , y ) ( x0 , y0
u ( x , y )  u0 &
)
lim
( x , y ) ( x0 , y0 )
v( x, y )  v0
lim f ( z )  w0
z  z0
 / 2  0, 1  0,  2  0s.t.
When
| u ( x, y )  u0 |
0  ( x  x0 )  ( y  y0 )  1
2
2
| v( x, y )  v0 |
0  ( x  x0 )  ( y  y0 )   2
2
Let
  min(1,  2 )
2

2

2
When 0  ( x  x0 ) 2  ( y  y0 ) 2   , i.e.0 | z  z0 | 
| f ( z)  w0 || (u( x, y)  iv( x, y))  (u0  iv0 ) | | u( x, y)  u0  i(v( x, y)  v0 ) |
| u ( x, y )  u0 |  | v( x, y )  v0 |
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
2


2

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16. Theorems on Limits
 Proof: (a)(b)
lim f ( z )  w0
lim
( x , y ) ( x0 , y0 )
z  z0
  0,   0s.t.
When
u ( x , y )  u0 &
lim
( x , y ) ( x0 , y0 )
v( x, y )  v0
| f ( z)  w0 | 
0 | z  z0 | 
| f ( z)  w0 || u( x, y)  iv( x, y)  (u0  iv0 ) |
| (u( x, y)  u0 )  i(v( x, y)  v0 ) | 
| u( x, y)  u0 || (u( x, y)  u0 )  i(v( x, y)  v0 ) | 
| v( x, y)  v0 || (u( x, y)  u0 )  i(v( x, y)  v0 ) | 
Thus
| u( x, y)  u0 |  ;| v( x, y)  v0 | 
When (x,y)(x0,y0)
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16. Theorems on Limits
 Theorem 2
f ( z )  w0 and
Let zlim
z
lim F ( z )  W0
z  z0
0
then
lim[ f ( z )  F ( z )]  w0  W0
z  z0
lim[ f ( z ) F ( z )]  w0W0
z  z0
w0
f ( z)
lim[
]
,W0  0
z  z0 F ( z )
W0
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16. Theorems on Limits
lim f ( z )  w0
z  z0
&
lim F ( z )  W0
z  z0
lim[ f ( z ) F ( z )]  w0W0
z  z0
f ( z )  u( x, y)  iv( x, y), F ( z)  U ( x, y)  iV ( x, y)
Let
z0  x0  iy0 ; w0  u0  iv0 ;W0  U0  iV0
f ( z) F ( z)  (uU  vV )  i(vU  uV )
lim f ( z )  w0
z  z0
lim F ( z )  W0
When (x,y)(x0,y0);
u(x,y)u0; v(x,y)v0; & U(x,y)U0; V(x,y)V0;
z  z0
Re(f(z)F(z)):
Im(f(z)F(z)):
(u0U0  v0V0 )
(v0U0  u0V0 )
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16. Theorems on Limits
It is easy to verify the limits
lim z  z0
lim c  c
z  z0
z  z0
lim z n  z0n (n  1, 2,...)
z  z0
For the polynomial
P( z)  a0  a1z  a2 z 2  ...  an z n
We have that
lim P( z )  P( z0 )
z  z0
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17. Limits Involving the Point at Infinity
 Riemannsphere & Stereographic Projection
N: the north pole
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17. Limits Involving the Point at Infinity
 The ε Neighborhood of Infinity
y
When the radius R is large enough
R1
O
i.e. for each small positive number ε
R2
R=1/ε
x
The region of |z|>R=1/ε is called the
ε Neighborhood of Infinity(∞)
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17. Limits Involving the Point at Infinity
 Theorem
If z0 and w0 are points in the z and w planes,
respectively, then
lim f ( z )  
z  z0
lim f ( z )  w0
z 
lim f ( z )  
z 
1
lim
0
z  z0 f ( z )
iff
iff
1
lim f ( )  w0
z 0
z
1
lim
0
z 0
1
f( )
z
iff
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17. Limits Involving the Point at Infinity
 Examples
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18. Continuity
 Continuity
A function is continuous at a point z0 if
lim f ( z )  f ( z0 )
z  z0
meaning that
1. the function f has a limit at point z0 and
2. the limit is equal to the value of f(z0)
For a given positive number ε, there exists a positive number δ, s.t.
When
| z  z0 | 
| f ( z)  f ( z0 ) | 
0 | z  z0 |  ?
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18. Continuity
 Theorem 1
A composition of continuous functions is itself continuous.
Suppose w=f(z) is a continuous at the point z0;
g=g(f(z)) is continuous at the point f(z0)
Then the composition g(f(z)) is continuous at the point z0
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18. Continuity
 Theorem 2
If a function f (z) is continuous and nonzero at a point z0,
then f (z) ≠ 0 throughout some neighborhood of that point.
lim f ( z )  f ( z0 )  0
Proof
When
z  z0
| f ( z0 ) |Why?
 
 0,   0, s.t.
2
| z  z0 | 
| f ( z )  f ( z0 ) |  
If f(z)=0, then
| f ( z0 ) |
f(z)
f(z0)
| f ( z0 ) |
2
| f ( z0 ) |
2
 | f ( z0 ) |
Contradiction!
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18. Continuity
 Theorem 3
If a function f is continuous throughout a region R that is
both closed and bounded, there exists a nonnegative real
number M such that
| f ( z) | M
for all points z in R
where equality holds for at least one such z.
Note:
| f ( z ) | u 2 ( x, y )  v 2 ( x, y )
where u(x,y) and v(x,y) are continuous real functions
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18. Homework
pp. 55-56
Ex. 2, Ex. 3, Ex. 6, Ex. 9, Ex. 11, Ex. 12
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19. Derivatives
 Derivative
Let f be a function whose domain of definition contains
a neighborhood |z-z0|<ε of a point z0. The derivative of f at
z0 is the limit
f ( z )  f ( z0 )
f '( z0 )  lim
z  z0
z  z0
And the function f is said to be differentiable at z0 when
f’(z0) exists.
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19. Derivatives
 Illustration of Derivative
f '( z0 )  lim
z  z0
Any position
f ( z )  f ( z0 )
z  z0
f ( z0  z )  f ( z0 )
f '( z0 )  lim
z  0
z
f(z0+Δz)
z  z0  z
v
w  f ( z0  z)  f ( z0 )
Δw
dw
w
 lim
dz z 0 z
f(z0)
O
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19. Derivatives
 Example 1
Suppose that f(z)=z2. At any point z
w
( z  z )2  z 2
lim
 lim
 lim(2 z  z )  2 z
z 0 z
z 0
z 0
z
since 2z + Δz is a polynomial in Δz. Hence dw/dz=2z or f’(z)=2z.
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19. Derivatives
 Example 2
If f(z)=z, then
w z  z  z z  z  z z



z
z
z
z
z  (x, y)  (0,0)
In any direction
Δy
Case #2
Case #1: Δx0, Δy=0
Case #1
z x  i 0

1
x 0 z
x  i 0
O
lim
Case #2: Δx=0, Δy0
z 0  iy

 1
x 0 z
0  iy
lim
Since the limit is unique, this function does not exist anywhere
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Δx
19. Derivatives
 Example 3
Consider the real-valued function f(z)=|z|2. Here
w | z  z |2  | z |2 ( z  z )( z  z )  z z
z


 z  z  z
z
z
z
z
Case #1: Δx0, Δy=0
lim ( z  z  z
x 0
z
x  i0
)  lim ( z  x  z
) zz
z x0
x  i 0
Case #2: Δx=0, Δy0
lim ( z  z  z
y 0
zz  zzz 0
z
0  iy
)  lim ( z  iy  z
) zz
z y0
0  iy
dw/dz can not exist when z is not 0
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19. Derivatives
 Continuity & Derivative
Continuity
Derivative
For instance,
f(z)=|z|2 is continuous at each point, however, dw/dz does not exists when z is not 0
Derivative
Continuity
lim[ f ( z )  f ( z0 )]  lim
z  z0
z  z0
f ( z )  f ( z0 )
lim( z  z0 )  f '( z0 )0  0
z
 z0
z  z0
Note: The existence of the derivative of a function at a point implies the continuity
of the function at that point.
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20. Differentiation Formulas
 Differentiation Formulas
d
d
d
c  0; z  1; [cf ( z )]  cf '( z )
dz
dz
dz
F ( z )  g ( f ( z ))
d n
[ z ]  nz n 1
dz
Refer to pp.7 (13)
d
[ f ( z )  g ( z )]  f '( z )  g '( z )
dz
d
[ f ( z )  g ( z )]  f ( z )  g '( z )  f '( z )  g ( z )
dz
F '( z0 )  g '( f ( z0 )) f '( z0 )
dW dW dw

dz
dw dz
d f ( z)
f '( z )  g ( z )  f ( z )  g '( z )
[
]
dz g ( z )
[ g ( z )]2
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20. Differentiation Formulas
 Example
To find the derivative of (2z2+i)5, write w=2z2+i and
W=w5. Then
d
(2 z 2  i)5  (5w4 ) w '  5(2 z 2  i) 4 (4 z )  20 z (2 z 2  i) 4
dz
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20. Homework
pp. 62-63
Ex. 1, Ex. 4, Ex. 8, Ex. 9
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21. Cauchy-Riemann Equations
 Theorem
Suppose that f ( z)  u( x, y)  iv( x, y)
and that f’(z) exists at a point z0=x0+iy0. Then the first-order
partial derivatives of u and v must exist at (x0,y0), and they must
satisfy the Cauchy-Riemann equations
ux  vy ; u y  vx
then we have
f '( z0 )  ux ( x0 , y0 )  ivx ( x0 , y0 )
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21. Cauchy-Riemann Equations
 Proof:
Let
z0  x0  iy0 ; z  x  iy
w  f ( z0  z )  f ( z0 )
 [u( x0  x, y0  y)  u( x0 , y0 )]  i[v( x0  x, y0  y)  v( x0 , y0 )]
w
z 0 z
u ( x0  x, y0  y )  u ( x0 , y0 )]  i[v( x0  x, y0  y )  v( x0 , y0 )
 lim
( x , y ) (0,0)
x  iy
f '( z0 )  lim
Note that (Δx, Δy) can be tend to (0,0) in any manner .
Consider the horizontally and vertically directions
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21. Cauchy-Riemann Equations
 Horizontally direction (Δy=0)
u ( x0  x, y0  y )  u ( x0 , y0 )  i[v( x0  x, y0  y )  v( x0 , y0 )]
x 0
x  i 0
u ( x0  x, y0 )  u ( x0 , y0 )  i[v( x0  x, y0 )  v( x0 , y0 )]
 lim
x 0
x
 ux ( x0 , y0 )  ivx ( x0 , y0 )
Cauchy-Riemann equations
f '( z0 )  lim
ux  vy ; u y  vx
 Vertically direction (Δx=0)
u ( x0 , y0  y )  u ( x0 , y0 )  i[v( x0 , y0  y )  v( x0 , y0 )]
y 0
0  iy
f '( z0 )  lim
i{[u( x0 , y0  y)  u( x0 , y0 )]  i 2 [v( x0 , y0  y)  v( x0 , y0 )]}
 lim
y 0
i(iy)
 vy ( x0 , y0 )  iuy ( x0 , y0 )
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21. Cauchy-Riemann Equations
 Example 1
f ( z)  z 2  x2  y 2  i 2xy
is differentiable everywhere and that f’(z)=2z. To verify that
the Cauchy-Riemann equations are satisfied everywhere, write
u( x, y)  x  y
2
v( x, y)  2 xy
2
ux  2 x  v y
uy  2 y  vx
f '( z)  2 x  i 2 y  2( x  iy)  2 z
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21. Cauchy-Riemann Equations
 Example 2
f ( z ) | z |2
u( x, y)  x2  y 2
v( x, y )  0
If the C-R equations are to hold at a point (x,y), then
ux  uy  vy  vx  0
x y0
Therefore, f’(z) does not exist at any nonzero point.
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22. Sufficient Conditions for Differentiability
 Theorem
f ( z)  u( x, y)  iv( x, y)
be defined throughout some ε neighborhood of a point z0 = x0
+ iy0, and suppose that
a.
b.
the first-order partial derivatives of the functions u and v with
respect to x and y exist everywhere in the neighborhood;
those partial derivatives are continuous at (x0, y0) and satisfy the
Cauchy–Riemann equations
ux  vy ; u y  vx
at (x0,y0)
Then f ’(z0) exists, its value being f ’ (z0) = ux + ivx where the righthand side is to be evaluated at (x0, y0).
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22. Sufficient Conditions for Differentiability
 Proof
Let
z  x  iy
w  f ( z0  z)  f ( z0 )  u  iv
u  u( x0  x, y0  y)  u( x0 , y0 )
v  v( x0  x, y0  y)  v( x0 , y0 )
Note : (a) and (b) assume that the first-order partial derivatives of u and v are continuous
at the point (x0,y0) and exist everywhere in the neighborhood
u  ux ( x0 , y0 )x  uy ( x0 , y0 )y  1x   2y
v  vx ( x0 , y0 )x  vy ( x0 , y0 )y  3x   4y
Where ε1, ε2, ε3 and ε4 tend to 0 as (Δx, Δy) approaches (0,0) in the Δz plane.
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22. Sufficient Conditions for Differentiability
-vx(x0,y0)
u  ux ( x0 , y0 )x  uy ( x0 , y0 )y  1x   2y
v  vx ( x0 , y0 )x  vy ( x0 , y0 )y  3x   4y
ux(x0,y0)
Note : The assumption (b) that those partial derivatives are continuous at (x0, y0) and
satisfy the Cauchy–Riemann equations
u  ux ( x0 , y0 )x  vx ( x0 , y0 )y  1x   2y
v  vx ( x0 , y0 )x  ux ( x0 , y0 )y   3x   4y
w u  iv

2v (x ,y )
i
x 0 0
z
z
u ( x , y )x  vx ( x0 , y0 ) y  1x   2 y vx ( x0 , y0 )x  u x ( x0 , y0 ) y   3 x   4 y
 x 0 0
i
z
z
u ( x , y )x  iu x ( x0 , y0 )y vx ( x0 , y0 )x  ivx ( x0 , y0 )y (1  i 3 )x ( 2  i 4 )y
 x 0 0
i


z
z
z
z
x
y
 u x ( x0 , y0 )  ivx ( x0 , y0 )  (1  i 3 )
 ( 2  i 4 )
z
z
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22. Sufficient Conditions for Differentiability
w
x
y
 u x ( x0 , y0 )  ivx ( x0 , y0 )  (1  i 3 )
 ( 2  i 4 )
z
z
z
|
x
y
| 1;|
| 1
z
z
| (1  i 3 )
x
|| 1  i 3 || 1 |  |  3 |
z
| ( 2  i 4 )
y
||  2  i 4 ||  2 |  |  4 |
z
w
 u x ( x0 , y0 )  ivx ( x0 , y0 )
z  0 z
lim
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22. Sufficient Conditions for Differentiability
 Example 1
f ( z)  ez  exei y  e x cos y  ie x sin y
u( x, y)  e x cos y
v( x, y)  ex sin y
Both Assumptions (a) and (b) in the theorem are satisfied.
f '( z)  ux  ivx  ex cos y  iex sin y  f ( z)
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22. Sufficient Conditions for Differentiability
 Example 2
f ( z ) | z |2
v( x, y )  0
u( x, y)  x2  y 2
ux  vy  2x  0  x  0
uy  vx  2x  0  y  0
Therefore, f has a derivative at z=0, and cannot have a derivative at any nonzero points.
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23. Polar Coordinates
 Assuming that z0≠0
x  r cos , y  r sin 
u u x u y


r x r y r
Similarly
u u x u y


 x  y 
ur  ux cos  uy sin 
u  ux r sin   uy r cos
vr  vx cos  vy sin 
v  vx r sin   vy r cos
If the partial derivatives of u and v with respect to x and y satisfy the Cauchy-Riemann equations
ux  vy ; u y  vx
rur  v ; u  rvr
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23. Polar Coordinates
 Theorem
Let the function f(z)=u(r,θ)+iv(r,θ) be defined throughout
some ε neighborhood of a nonzero point z0=r0exp(iθ0) and
suppose that
(a) the first-order partial derivatives of the functions u and v
with respect to r and θ exist everywhere in the
neighborhood;
(b) those partial derivatives are continuous at (r0, θ0) and satisfy
the polar form rur = vθ, uθ = −rvr of the Cauchy-Riemann
equations at (r0, θ0)
Then f’(z0) exists, its value being
f '( z0 )  ei (ur (r0 ,0 )  ivr (r0 ,0 ))
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23. Polar Coordinates
 Example 1
Consider the function
f ( z) 
Then
1
1
1
1
 i  e i  (cos   i sin  ), z  0
z re
r
r
u (r , ) 
rur  
cos 
sin 
, v( r ,  )  
r
r
cos 
sin 
 v & u  
 rvr
r
r
 i
cos 
sin 
1
1
 i e
f '( z )  e ( 2  i 2 )  e
  i 2   2
2
r
r
r
(re )
Z
 i
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23. Homework
pp. 71-72
Ex. 1, Ex. 2, Ex. 6, Ex. 7, Ex. 8
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24. Analytic Function
 Analytic at a point z0
A function f of the complex variable z is analytic at a
point z0 if it has a derivative at each point in some
neighborhood of z0.
Note that if f is analytic at a point z0, it must be analytic at each
point in some neighborhood of z0
 Analytic function
A function f is analytic in an open set if it has a
derivative everywhere in that set.
Note that if f is analytic in a set S which is not open, it is to be
understood that f is analytic in an open set containing S.
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24. Analytic Function
 Analytic vs. Derivative
 For a point
Analytic  Derivative
Derivative Analytic
 For all points in an open set
Analytic  Derivative
Derivative Analytic
f is analytic in an open set D iff f is derivative in D
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24. Analytic Function
 Singular point (singularity)
If function f fails to be analytic at a point z0 but is analytic at
some point in every neighborhood of z0, then z0 is called a
singular point.
For instance, the function f(z)=1/z is analytic at every point in the
finite plane except for the point of (0,0). Thus (0,0) is the singular
point of function 1/z.
 Entire Function
An entire function is a function that is analytic at each point
in the entire finite plane.
For instance, the polynomial is entire function.
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24. Analytic Function
 Property 1
If two functions are analytic in a domain D, then
 their sum and product are both analytic in D
 their quotient is analytic in D provided the function in the
denominator does not vanish at any point in D
 Property 2
From the chain rule for the derivative of a composite function, a
composition of two analytic functions is analytic.
d
g ( f ( z ))  g '[ f ( z )] f '( z )
dz
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24. Analytic Function
 Theorem
If f ’(z) = 0 everywhere in a domain D, then f (z) must be
constant throughout D.
f '( z)  ux  ivx  vy  iuy  0
ux  uy  0& vx  vy  0
du
 ( gradu ) U gradu=u xi  uy j
ds
U is the unit vector along L
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25. Examples
 Example 1
The quotient
z3  4
f ( z)  2
( z  3)( z 2  1)
is analytic throughout the z plane except for the singular
points
z   3 & z  i
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25. Examples
 Example 3
Suppose that a function f ( z)  u( x, y)  iv( x, y) and its
conjugate f ( z)  u( x, y)  iv( x, y) are both analytic in a given
domain D. Show that f(z) must be constant throughout D.
Proof:
f ( z)  u( x, y)  iv( x, y) is analytic, then
ux  vy , u y  vx
f ( z)  u( x, y)  iv( x, y) is analytic, then
ux  vy , u y  vx
f '( z)  ux  ivx  0
ux  0, vx  0
Based on the Theorem in pp. 74, we have that f is constant throughout D
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25. Examples
 Example 4
Suppose that f is analytic throughout a given region D, and
the modulus |f(z)| is constant throughout D, then the function
f(z) must be constant there too.
Proof:
|f(z)| = c, for all z in D
where c is real constant.
If c=0, then f(z)=0 everywhere in D.
If c ≠ 0, we have
c2
f ( z) 
, f ( z )  0inD
f ( z)
f ( z) f ( z)  c2
Both f and it conjugate are analytic, thus f must be constant in D. (Refer to Ex. 3)
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25. Homework
pp. 77~78
Ex. 2, Ex. 3, Ex. 4, Ex. 6, Ex. 7
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26. Harmonic Functions
 A Harmonic Function
A real-valued function H of two real variables x and y is
said to be harmonic in a given domain of the xy plane if,
throughout that domain, it has continuous partial derivatives of
the first and second order and satisfies the partial differential
equation
H xx ( x, y)  H yy ( x, y)  0
Known as Laplace’s equation.
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26. Harmonic Functions
 Theorem 1
If a function f (z) = u(x, y) + iv(x, y) is analytic in a domain
D, then its component functions u and v are harmonic in D.
Proof: f ( z)  u( x, y)  iv( x, y) is analytic in D
ux  vy & uy  vx
Differentiating both sizes of these equations with respect to x and y respectively, we have
uxx  vyx & uyx  vxx
continuity
uxy  vyy & uyy  vxy
uxx  vxy & uxy  vxx
uxy  vyy & uyy  vxy
uxx  uyy  0& vxx  vyy  0
Theorem in Sec.52:
a function is analytic at a point, then its real and imaginary components
have continuous partial derivatives of all order at that point.
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26. Harmonic Functions
 Example 3
The function f(z)=i/z2 is analytic whenever z≠0 and since
i
i  ( z )2
2 xy  i( x 2  y 2 )
 2

2
2
z
( x 2  y 2 )2
z  ( z)
The two functions
x2  y 2
v( x, y )  2
( x  y 2 )2
2 xy
u ( x, y )  2
( x  y 2 )2
are harmonic throughout any domain in the xy plane that does not contain the origin.
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26. Harmonic Functions
 Harmonic conjugate
If two given function u and v are harmonic in a domain D
and their first-order partial derivatives satisfy the CauchyRiemann equation throughout D, then v is said to be a
harmonic conjugate of u.
Is the definition symmetry for u and v?
Cauchy-Riemann equation
ux  vy & uy  vx
If u is a harmonic conjugate of v, then
ux  vy & uy  vx
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26. Harmonic Functions
 Theorem 2
A function f (z) = u(x, y) + iv(x, y) is analytic in a domain D
if and only if v is a harmonic conjugate of u.
 Example 4
The function f ( z)  z 2 is entire function, and its real and
imaginary components are u( x, y)  x2  y 2 & v( x, y)  2xy
Based on the Theorem 2, v is a harmonic conjugate of u
throughout the plane. However, u is not the harmonic
2
2
g
(
z
)

2
xy

i
(
x

y
) is not an analytic
conjugate of v, since
function.
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26. Harmonic Functions
 Example 5
Obtain a harmonic conjugate of a given function.
u( x, y)  y3  3x2 y
Suppose that v is the harmonic conjugate of the given function
Then ux  vy & uy  vx
ux  6xy  vy
uy  3 y2  3x2  vx
v  3xy 2   ( x)
3 y 2  3x2  (3 y 2   '( x))
 '( x)  3x2   ( x)=x3  C
v  3xy 2  x3  C
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26. Homework
pp. 81-82
Ex. 1, Ex. 2, Ex. 3, Ex. 5
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27. Uniquely Determined Analytic Function
 Lemma
Suppose that
a) A function f is analytic throughout a domain D;
b) f(z)=0 at each point z of a domain or line segment
contained in D.
Then f (z) ≡ 0 in D; that is, f (z) is identically equal to zero
throughout D.
Refer to Chap. 6 for the proof.
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27. Uniquely Determined Analytic Function
 Theorem
A function that is analytic in a domain D is uniquely
determined over D by its values in a domain, or along a
line segment, contained in D.
f ( z)  g ( z)
D
D
f(z)
g(z)
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28. Reflection Principle
 Theorem
Suppose that a function f is analytic in some domain D
which contains a segment of the x axis and whose lower half is
the reflection of the upper half with respect to that axis. Then
f ( z)  f ( z)
for each point z in the domain if and only if f (x) is real for
each point x on the segment.
y
D
x
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