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

Provided this limit exists. f(z) is said to be differentiable at z0.

zzfzzf

zfz

)()(lim)( 00

00

Some Exercises

Show that

1) f(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.

z

Solutions

1.0 0

0 0

0 0

0 0

( ) ( )( ) lim

lim lim

h

h h

f z h f zf z

hz h z h

h h

h 0

X-axis

Y-axis

If we go along X-axis A tends to 1.

A

If we go along Y-axis A tends to -1.

That implies the limit does not exist.

Properties of Derivatives

Rule. Chain''

.0if,''

'

'''

. constant anyfor ''

'''

000

020

00000

00000

00

000

zgzgfzgfdzd

zgzg

zgzfzfzgz

gf

zgzfzgzfzfg

czcfzcf

zgzfzgf

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 , then it is analytic.

z

Example 1

1 2 2

1 2 2

1

1( )

( 1)

,2 2

12 2( )

12 2

1 1( )

1 1

x iyf z

x yz z z z

let x yi

z z z zi

if zz z z z

iz

f zzz z z z

Thus f1(z) is analytic at all points except z=1.

Example 2

2 22

2 2

2

2

( ) 3 1 3

,2 2

( ) 3 1 32 2 2 2

( ) 3 1

f z x y x i y

z z z zlet x y

iz z z z z z z z

f z ii i

f z zz z

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 is at best awkward. Observe:

z

5 4 3 5

53 2 5

1( )

z z z z z zf z

z z z z

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

zzfzzf

zfz

)()(lim)( 00

00

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

x

yxivyxuyxxivyxxuzf

x

),(),(),(),(lim)(' 00000000

00

xyxvyxxv

ix

yxuyxxuzf

xx

),(),(lim

),(),(lim)(' 0000

0

0000

00

But the limits of the bracketed expression are just the first partial derivatives of u and v with respect to x, so that:

).,(),()(' 00000 yxxv

iyxxu

zf

Cauchy-Riemann Equations (3)If it approaches along the y-axis, then z = iy, and we obtain

0 0 0 00 0

( , ) ( , )'( ) lim

y

u x y y u x yf z

i y

And, therefore

).,(),()(' 00000 yxyv

yxyu

izf

0 0 0 0

0

( , ) ( , )limy

v x y y v x yi

i 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

xv

yu

yv

xu

and

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

1,1,2,2

)()()( 22

xv

yu

yyv

xxu

xyiyxzf

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 2Prove that f (z) is entire and find its derivative.

yexv

yeyu

yeyv

yexu

yieyezf

xxxx

xx

sin,sin,cos,cos

:Solution

sincos)(

The first partials are continuous and satisfy the Cauchy-Riemann equations at every point.

.sincos)(' yieyexv

ixu

zf xx

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

.02

2

2

2

yx

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.

ExampleConstruct an analytic function whose real part is: .3),( 23 yxyxyxu

Solution: First verify that this function is harmonic.

.066

6 and16

6and33

2

2

2

2

2

2

2

222

xxyu

xu

and

xyu

xyyu

xxu

yxxu

Example, Continued

16)2(

33)1( 22

xyyu

xv

andyxxu

yv

Integrate (1) with respect to y:

)(3),()3(

33

33

32

22

22

xyyxyxv

yyxv

yyxv

Example, Continued

Now take the derivative of v(x,y) with respect to x:

).('6 xxyxv

According to equation (2), this equals 6xy – 1. Thus,

.3),(

.)(and,)(So

.1ly,Equivalent.1)('and

16)('6

32 CyyxyxvAnd

Cxxxx

xx

xyxxy

Example, Continued

The desired analytic function f (z) = u + iv is:

Cxyyxiyxyxzf 3223 33)(

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 is that z1 = z2 + 2ki, where k is an

integer. Thus ez is a periodic function.

21 ee zz

How is the case with multi-valued functions like z1/n, logz etc??