Complex Differentiation
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Transcript Complex Differentiation
Complex Differentiation
Mohammed Nasser
Department of statistics
RU
Derivatives
• Differentiation of complex-valued functions
is completely analogous to the real case:
• Definition. Derivative. Let f(z) be a
complex-valued function defined in a
neighborhood of z0. Then the derivative of
f(z) at z0 is given by
f ( z0 z) f ( z0 )
f ( z0 ) lim
z0
z
Provided this limit exists. f(z) is said to be
differentiable at z0.
Some Exercises
Show that
1) f(z)= z is nowwhere differentiable.
2) g(z)=zn has derivative nzn-1.
3) h(z)=ez has derivative ez.
4) l(z)=|z|2 is nowhere differentiable except z=0
5) Every real-valued function of complex variable is
either non-differentiable or differentiable with
derivative equal to 0.
Solutions
1.
f (z 0 h ) f (z 0 )
f (z 0 ) lim
h 0
h
z0 h z0
h
lim
lim
h 0
h 0
h
h
h
A
0
Yaxis
Xaxis
If we go along Y-axis
A tends to -1.
If we go along X-axis
A tends to 1.
That implies the limit does
not exist.
Properties of Derivatives
f g' z0 f ' z0 g' z0
cf' z0 cf' z0 for any constant c.
fg' z0 f z0 g' z0 f ' z0 gz0
gz0 f ' z0 f z0 g' z0
f
, if gz0 0.
' z0
2
gz0
g
d
f gz0 f ' gz0 g' z0 Chain Rule.
dz
Analytic. Holomorphic.
• Definition. A complex-valued function f (z) is said
to be analytic, or equivalently, holomorphic, on
an open set if it has a derivative at every point
of . (The term “regular” is also used.)
• It is important that a function may be
differentiable at a single point only. Analyticity
implies differentiability within a neighborhood of
the point. This permits expansion of the function
by a Taylor series about the point.
• If f (z) is analytic on the whole complex plane,
then it is said to be an entire function.
Rational Function.
• Definition. If f and g are polynomials in z,
then h (z) = f (z)/g(z), g(z) 0 is called a
rational function.
• Remarks.
– All polynomial functions of z are entire.
– A rational function of z is analytic at every
point for which its denominator is nonzero.
– If a function can be reduced to a polynomial
function which does not involve z , then it is
analytic.
Example 1
f1(z )
let x
f1(z )
f1(z )
x 1 iy
(x 1)2 y 2
z z
z z
,y
2
2i
z z
z z
1i
2
2i
2
2
z z
z z
1
2
2
i
z 1
1
zz z z 1 z 1
Thus f1(z) is
analytic at
all points
except
z=1.
Example 2
f2 (z ) x y 3x 1 i 3y
2
2
z z
z z
let x
,y
2
2i
2
2
z z z z
z z
z z
f2 ( z )
3
1 i3
2 2i
2
2i
f2 (z ) zz 3z 1
Thus f2(z) is nowhere analytic.
Testing for Analyticity
Determining the analyticity of a function by
searching for z in its expression that cannot be
removed is at best awkward. Observe:
f (z )
z 5z z 4 z 3 z z 5 1
z z z
3
2
z
5
5
It would be difficult and time consuming to try to
reduce this expression to a form in which you
could be sure that the z could not be removed.
The method cannot be used when anything but
algebraic functions are used.
Cauchy-Riemann Equations (1)
If the function f (z) = u(x,y) + iv(x,y) is
differentiable at z0 = x0 + iy0, then the limit
f ( z0 ) lim
z0
f ( z0 z) f ( z0 )
z
can be evaluated by allowing z to approach
zero from any direction in the complex plane.
Cauchy-Riemann Equations (2)
If it approaches along the x-axis, then z = x,
and we obtain
f ' ( z0 ) lim
x 0
u( x 0 x , y0 ) iv( x 0 x , y0 ) u( x 0 , y0 ) iv( x 0 , y0 )
x
u( x 0 x , y0 ) u( x 0 , y0 )
v( x x , y0 ) v( x 0 , y0 )
f ' ( z0 ) lim
i lim 0
x 0
x 0
x
x
But the limits of the bracketed expression are
just the first partial derivatives of u and v with
respect to x, so that:
u
v
f ' ( z0 ) ( x 0 , y0 ) i ( x 0 , y0 ).
x
x
Cauchy-Riemann Equations (3)
If it approaches along the y-axis, then z = iy,
and we obtain
u (x 0, y 0 y ) u (x 0, y 0 )
f '(z 0 ) lim
y 0
i y
v(x 0, y 0 y ) v(x 0, y 0 )
i lim
y 0
i
y
And, therefore
u
v
f ' ( z0 ) i ( x 0 , y0 ) ( x 0 , y0 ).
y
y
Cauchy-Riemann Equations (4)
By definition, a limit exists only if it is unique.
Therefore, these two expressions must be
equivalent. Equating real and imaginary parts,
we have that
u v
u
v
and
x y
y
x
must hold at z0 = x0 + iy0 . These equations
are called the Cauchy-Riemann Equations.
Their importance is made clear in the
following theorem.
Cauchy-Riemann Equations (5)
• Theorem. Let f (z) = u(x,y) + iv(x,y) be
defined in some open set containing
the point z0. If the first partial derivatives
of u and v exist in , and are continuous
at z0 , and satisfy the Cauchy-Riemann
equations at z0, then f (z) is differentiable
at z0. Consequently, if the first partial
derivatives are continuous and satisfy the
Cauchy-Riemann equations at all points
of , then f (z) is analytic in .
Example 1
f ( z) ( x 2 y) i ( y2 x )
u
v
u
v
2x ,
2 y,
1,
1
x
y
y
x
Hence, the Cauchy-Riemann equations
are satisfied only on the line x = y, and
therefore in no open disk. Thus, by the
theorem, f (z) is nowhere analytic.
Example 2
Prove that f (z) is entire and find its derivative.
f ( z) ex cos y iex sin y
Solution :
u
v x
u
v
x
x
e cos y,
e cos y,
e sin y,
ex sin y
x
y
y
x
The first partials are continuous and satisfy the
Cauchy-Riemann equations at every point.
u
v
f ' ( z)
i
ex cos y iex sin y.
x
x
Harmonic Functions
• Definition. Harmonic. A real-valued function
(x,y) is said to be harmonic in a domain D if all
of its second-order partial derivatives are
continuous in D and if each point of D satisfies
2 2
2 0.
2
x
y
Theorem. If f (z) = u(x,y) + iv(x,y) is analytic in a
domain D, then each of the functions u(x,y)
and v(x,y) is harmonic in D.
Harmonic Conjugate
• Given a function u(x,y) harmonic in, say,
an open disk, then we can find another
harmonic function v(x,y) so that u + iv is an
analytic function of z in the disk. Such a
function v is called a harmonic conjugate
of u.
Example
Construct an analytic function whose real part is:
u( x , y) x 3 3xy2 y.
Solution: First verify that this function is harmonic.
u
u
2
2
3x 3 y and 2 6 x
x
x
2
u
u
6 xy 1 and 2 6 x
y
y
2
2u 2u
and
2 6 x 6 x 0.
2
x
y
Example, Continued
v u
(1)
3x 2 3 y2 and
y x
v u
( 2)
6 xy 1
x
y
Integrate (1) with respect to y:
v 3x 2 3 y2 y
2
2
y
v
3
x
3
y
( 3) v( x , y) 3x 2 y y3 ( x )
Example, Continued
Now take the derivative of v(x,y) with respect to x:
v
6 xy ' ( x ).
x
According to equation (2), this equals 6xy – 1. Thus,
6 xy ' ( x ) 6 xy 1
and ' ( x ) 1. Equivalent ly,
1.
x
So ( x ) x , and ( x ) x C.
And v( x , y) 3x 2 y y3 C.
Example, Continued
The desired analytic function f (z) = u + iv is:
f ( z) x 3 3xy2 y i 3x 2 y y3 x C
Remember Complex Exponential
• We would like the complex exponential to
be a natural extension of the real case,
with f (z) = ez entire. We begin by
examining ez = ex+iy = exeiy.
• eiy = cos y + i sin y by Euler’s and
DeMoivre’s relations.
• Definition. Complex Exponential Function.
If z = x + iy, then ez = ex(cos y + i sin y).
• That is, |ez|= ex and arg ez = y.
More on Exponentials
• Recall that a function f is one-to-one on a set S if
the equation f (z1) = f (z2), where z1, z2 S,
implies that z1 = z2. The complex exponential
function is not one-to-one on the whole plane.
• Theorem. A necessary and sufficient condition
that ez = 1 is that z = 2ki, where k is an integer.
Also, a necessary and sufficient condition that
ez1 ez2 is that z1 = z2 + 2ki, where k is an
integer. Thus ez is a periodic function.
How is the case with multi-valued
functions like z1/n, logz etc??