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
z0
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  gz0 
gz0  f ' z0   f z0  g' z0 
f
, if gz0   0.
 ' z0  
2
 gz0 
 g
d
f  gz0   f '  gz0  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
1i
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 2  y 2  3x  1  i 3y
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 in its expression
that cannot be removed
z is at best awkward. Observe:
f (z ) 
z 5z  z 4z 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 could not be removed. The
method cannot be used when anything but algebraic functions are used.
z
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
z0
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 = iy, 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 CauchyRiemann 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 = 2ki, where k is an integer.
Also, a necessary and sufficient condition that
ez1  ez2 is that z1 = z2 + 2ki, where k is an
integer. Thus ez is a periodic function.
How is the case with multi-valued
functions like z1/n, logz etc??