Download Complex_differentiation

ENGG2420B
Complex functions
and complex differentiation
Kenneth Shum
kshum
ENGG2420B
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What is a function?
Domain
Range
a
b
c
 = f(a)

 = f(b)
 = f(c)

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Real-valued function f(x)=x2
Domain
Range
–2
2
f
4
f
-1
0
f
0
nothing is mapped to -1
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Complex function
Domain is the set of
all complex numbers
Range is the set
of all complex numbers
z = x+i y
w=u+iv
w = f(z)
z-plane
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w-plane
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Notations in this lecture
• z-plane
– Domain of a complex function
– z=x+iy
• x is the real part of z.
• y is the imaginary part of z.
• w-plane
– Image of a complex function
– w = u + i v,
– w is a function of z
• u is the real part of w.
• v is the imaginary part of w.
• u and v are functions of x and y.
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Extension from real to complex
z-plane
f(z) = z2
w-plane
-1-i
1+i
2i
–2
4
2
i
-i
-1
0
0
u = x2 – y2
v=2xy
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Now, every complex number,
including -1, has at least one pre-image.
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f(z) = i z
u = –y
v=x
z-plane
w-plane
3
3
w=iz
2.5
2
2
1.5
1.5
v
y
2.5
1
1
0.5
0.5
0
0
0
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0.5
1
1.5
x
2
2.5
3
-3
-2.5
-2
-1.5
u
-1
-0.5
0
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f(z) = (1+i) z+ 2
u = x–y+2
v = x+y
z-plane
6
2.5
5
y
2
1.5
w = f(z)
4
v
3
3
1
2
0.5
1
0
0
0
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0.5
1
1.5
x
2
2.5
3
-1
w-plane
0
1
2
u
3
4
5
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f(z) = conj(z)
u = –x
v=y
w = conj(z)
3
0
2.5
-0.5
2
-1
1.5
-1.5
v
y
z-plane
1
-2
0.5
-2.5
0
-3
0
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0.5
1
1.5
x
2
2.5
3
0
w-plane
0.5
1
1.5
u
2
2.5
3
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f(z) = z2 (lower left corner is 0)
u = x2 – y2
v=2xy
z-plane
w-plane
20
3
2.5
15
w=
z2
2
v
y
10
1.5
1
5
0.5
0
0
0
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0.5
1
1.5
x
2
2.5
3
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ENGG2012B
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-6
-4
-2
0
u
2
4
6
8
10
f(z) = z2 (lower left corner is 1)
u = x2 – y2
v=2xy
z-plane
w-plane
25
3
2.5
20
w = z2
2
1.5
v
y
15
10
1
5
0.5
0
0
1
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1.5
2
2.5
x
3
3.5
4
-5
0
5
10
15
u
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ENGG2420B
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f(z) = 1/z
u = x/(x2+y2)
v = -y/(x2+y2)
z-plane
w-plane
3
0
-0.5
2.5
w = 1/z
-1
2
1.5
v
y
-1.5
-2
-2.5
1
-3
0.5
-3.5
-4
0
0
0.5
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1
1.5
x
2
2.5
3
0
0.5
1
1.5
2
u
2.5
3
3.5
4
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f(z) = z + (1+i) conj(z)
u = 2x + y
v=x
z-plane
w = z + (1+i) conj(z)
3
w-plane
6
5
2.5
4
2
3
1.5
v
y
2
1
1
0
-1
0.5
-2
0
-3
0
0.5
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1
1.5
x
2
2.5
3
0
1
2
3
4
5
6
7
8
9
u
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Question
• On the left-hand side in each of the previous
slides, the thick red line and the thick blue line
are perpendicular. The angle between the
black line and the red line is arctan(3) = 71.57
degrees.
• On the right-hand sides, these angles at the
intersection point are sometime preserved,
and sometime not. Is there any explanation?
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Review: Differentiation of real-valued
function
y = x2
2
1.5
y
y = 2x-1
1
0.5
0
-1.5
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-1
-0.5
0
x
0.5
1
1.5
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Behaviour of f(x)=x2 near x=1
2
An infinitesimal arrow
in the domain starting
at x=1 is transformed
to an infinitesimal arrow
in the image starting
at y=1, with
scaling factor = 2
y
1.5
1
0.5
0
-1.5
-1
-0.5
0
x
0.5
1
Domain
1.5
Image
f(x) = x2
1
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x
1
y
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Behaviour of f(x)=x2 near x=0
2
An infinitesimal arrow
in the domain starting
at x=1 is transformed
to an arrow of length 0
in the image
y
1.5
1
0.5
0
-1.5
-1
-0.5
0
x
0.5
1
Domain
1.5
Image
f(x) = x2
0
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x
0
y
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Behaviour of f(x)=x2 near x=-0.5
2
y
1.5
An infinitesimal arrow
in the domain starting
at x=1 is transformed
to an infinitesimal arrow
in the image starting
at y=1, with
scaling factor = -1
1
0.5
0
-1.5
-1
-0.5
0.5
0
x
1
Domain
1.5
Image
f(x) = x2
-0.5
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x
-0.5
y
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Derivative for real function
• Let f(x) be a real-valued function.
– Domain and image are the real numbers
• The function f is said to be differentiable at x0 if we can find a
real number s, such that for any small real number h, we have
f(x0+ h)  f(x0)+ s h.
• The scaling factor s may vary with the reference point x0.
– If we change the reference point x0, the value of s may change.
• The scaling factor s is called the derivative of f(z) at x0, and is
usually denoted by f’(x0).
• Formally, f’(x0) is expressed as the limit
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Complex differentiation
(We draw a complex number as a vector here)
Domain
z-plane
z0 is mapped to w0.
f(z0) = w0
Image
w-plane
w = f(z)
z0
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A complex function f is said to be
differentiable at z0, if whenever we draw
a short vector h from z0, (the angle is
arbitrary), the corresponding displacement
in the image is well approximated by a
complex multiple of h,
i.e., f(z0+ h) – f(z0)  c h.
The complex scaling factor c is called
the derivative of f at z0
w0
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An example of a non-differentiable
function
Domain
z-plane
Image
w-plane
z0 is mapped to w0.
f(z0) = w0
w = f(z)
z0
w0
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Complex derivative
• Let f(z) be a complex function.
– Domain and image are complex planes
• The function f is said to be differentiable at a point z0 if we
can find a complex number c, such that for any small complex
number h, we have the approximation
f(z0+ h)  f(z0)+ c  h.
• The complex scaling factor c may vary with the point z0.
– If we change the reference point z0, the value of c may change.
• The complex scaling factor c is called the derivative of f(z) at
z0, and is usually denoted by f’(z0).
• Formally, the derivative of f(z) at z0 is defined as a limit
Here, h is a complex number.
Addition, subtraction
and division are all
complex arithmetic.
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Analytic function
• A complex function is said to be analytic if it is
differentiable at all points in the domain of
the function.
• An analytic function is sometime called a
holomorphic function.
• If f(z) is differentiable at all points in the
complex plane, than f(z) is called an entire
function.
• For example, f(z) = z2 is an entire function, and
the derivative is given by f’(z)=2z.
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Answer to the question
• Suppose that there are two curves
intersecting at z0 in the domain z-plane,
making an angle  at the intersection point.
• If f(z) is differentiable at z0 with non-zero
derivative, then image of the two curves in the
w-plane have angle  at the intersection point
w0 as well.
• In complex analysis, an angle-preserving
mapping is called a conformal mapping.
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Summary of the examples given in
this lecture
Function
Differentiable ?
Derivative = ?
f(z) = j z
Differentiable everywhere
f’(z) = j
f(z) = (1+j) z+ 2
Differentiable everywhere
f’(z) = 1+j
f(z) = conj(z)
Not differentiable
N/A
f(z) = z2
Differentiable everywhere
f’(z) = 2 z
f(z) = 1/z
Differentiable everywhere
except the origin
f’(z) = - 1/z2
f(z) = cos(z)
Differentiable everywhere
f’(z) = - sin(z)
f(z) = z + (1+j) conj(z) Not differentiable
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ENGG2420B
N/A
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A necessary condition for being
differentiable
Cauchy-Riemann condition (cartesian form)
Consider a complex function
f(z) = u(x,y) + i v(x,y)
Suppose f(z) is defined at a point z0 = x0+i y0 and
its neighborhood.
Then f(z) is differentiable at z0 if
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Application
• We can show that the complex conjugate
function is not differentiable anywhere by
verifying that the Cauchy-Riemann condition
fails.
• conj(z) = x – i y
– u(x,y) = x
– v(x,y) = –y
• Check that ux(x,y)=1, vy(x,y) = –1.
– They are not equal for any x and any y.
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