examples of Cauchy-Riemann equations

examples of Cauchy-Riemann equations

To illustrate the

Cauchy-Riemann equations

, we may consider a few
examples. Let

f

be the squaring

function

, i.e. for any

complex
number

z

, we have

f

⁢

(

z

)

=

z

2

.

We now separate real from

imaginary parts

. Letting

x

and

y

be
real variables we have

f

⁢

(

x

+

i

⁢

y

)

=

(

x

+

i

⁢

y

)

2

=

x

2

+

2

⁢

i

⁢

x

⁢

y

-

y

2

=

(

x

2

-

y

2

)

+

i

⁢

(

2

⁢

x

⁢

y

)

.

Defining

real functions

u

and

v

by

f

⁢

(

x

+

i

⁢

y

)

=

u

⁢

(

x

,

y

)

+

i

⁢

v

⁢

(

x

,

y

)

and taking

derivatives

, we have

u

⁢

(

x

,

y

)

=

x

2

-

y

2

v

⁢

(

x

,

y

)

=

2

⁢

x

⁢

y

∂

⁡

u

∂

⁡

x

=

2

⁢

x

∂

⁡

u

∂

⁡

y

=

-

2

⁢

y

∂

⁡

v

∂

⁡

x

=

2

⁢

y

∂

⁡

v

∂

⁡

y

=

2

⁢

x

Since

∂

⁡

u

/

∂

⁡

x

=

∂

⁡

v

/

∂

⁡

y

and

∂

⁡

u

/

∂

⁡

y

=

-

∂

⁡

v

/

∂

⁡

x

. the
Cauchy-Riemann

relations

are seen to be satisfied.

Next, consider the

complex conjugation

function. In this
case,

x

+

i

⁢

y

¯

=

x

-

i

⁢

y

, so we have

u

⁢

(

x

,

y

)

=

x

and

v

⁢

(

x

,

y

)

=

y

. Taking derivatives,

∂

⁡

u

∂

⁡

x

=

1

∂

⁡

u

∂

⁡

y

=

0

∂

⁡

v

∂

⁡

x

=

0

∂

⁡

v

∂

⁡

y

=

-

1

Because

∂

⁡

u

/

∂

⁡

x

≠

∂

⁡

v

/

∂

⁡

y

,
the Cauchy-Riemann equations are

not

satisfied do the
conjugation function is not

holomorphic

. Likewise, one can show
that the functions

ℜ

,

ℑ

,

a

n

d

|

⋅

|

which appear in

complex analysis

are not holomorphic.

For our next example, we try a

polynomial

. Let

f

⁢

(

Z

)

=

z

3

+

2

⁢

z

+

5

. Writing

z

=

x

+

i

⁢

y

and

f

=

u

+

i

⁢

v

,
we find that

u

⁢

(

x

,

y

)

=

x

3

-

3

⁢

x

⁢

y

2

+

2

⁢

x

+

5

and

v

⁢

(

x

,

y

)

=

3

⁢

x

2

⁢

y

-

y

3

+

2

⁢

y

. Taking

partial derivatives

,
one can confirm that the Cauchy-Riemann equations
are satisfied, so we have a

holomorphic function

.

More generally, we can show that

all

complex
polynomials are holomorphic. Since the Cauchy-Riemann
equations are linear, it suffices to check that

integer

powers are holomorphic. We can do this by an induction

argument

. That

f

⁢

(

z

)

=

z

satisfies the equations is
trivial and we have shown that

f

⁢

(

z

)

=

z

2

also
satisfies them. Let us assume that

f

⁢

(

z

)

=

z

n

happens to satisfy the Cauchy-Riemann equations for a
particular value of

n

and write

(

x

+

i

⁢

y

)

n

=

u

⁢

(

x

,

y

)

+

i

⁢

v

⁢

(

x

,

y

)

(

x

+

i

⁢

y

)

n

+

1

=

u

~

⁢

(

x

,

y

)

+

i

⁢

v

~

⁢

(

x

,

y

)

By elementary algebra, we have

u

~

⁢

(

x

,

y

)

=

x

⁢

u

⁢

(

x

,

y

)

-

y

⁢

v

⁢

(

x

,

y

)

v

~

⁢

(

x

,

y

)

=

y

⁢

u

⁢

(

x

,

y

)

+

x

⁢

v

⁢

(

x

,

y

)

.

By elementary

calculus

, we have

∂

⁡

u

~

∂

⁡

x

=

u

+

x

⁢

∂

⁡

u

∂

⁡

x

-

y

⁢

∂

⁡

v

∂

⁡

x

∂

⁡

u

~

∂

⁡

y

=

-

v

+

x

⁢

∂

⁡

u

∂

⁡

y

-

y

⁢

∂

⁡

v

∂

⁡

y

∂

⁡

v

~

∂

⁡

x

=

v

+

y

⁢

∂

⁡

u

∂

⁡

x

+

x

⁢

∂

⁡

v

∂

⁡

x

∂

⁡

v

~

∂

⁡

y

=

u

+

y

⁢

∂

⁡

u

∂

⁡

y

+

x

⁢

∂

⁡

v

∂

⁡

y

so

∂

⁡

u

~

∂

⁡

x

-

∂

⁡

v

~

∂

⁡

y

=

x

⁢

(

∂

⁡

u

∂

⁡

x

-

∂

⁡

v

∂

⁡

y

)

+

y

⁢

(

∂

⁡

u

∂

⁡

y

+

∂

⁡

v

∂

⁡

x

)

∂

⁡

u

~

∂

⁡

y

+

∂

⁡

v

~

∂

⁡

x

=

x

⁢

(

∂

⁡

u

∂

⁡

y

+

∂

⁡

v

∂

⁡

x

)

+

y

⁢

(

∂

⁡

u

∂

⁡

x

-

∂

⁡

v

∂

⁡

y

)

Since the terms in parentheses are zero on account of

u

and

v

satisfying the Cauchy-Riemann equations,
it follows that

u

~

and

v

~

also
satisfy the Cauchy-Riemann equations. By induction,

f

⁢

(

z

)

=

z

n

is holomorphic for all

positive integers

n

.

As our next example, we consider the complex

square root

. As shown in the entry taking square
root algebraically, we have the following equality:

x

+

i

⁢

y

=

x

2

+

y

2

+

x

2

+

(

sign

⁢

y

)

⁢

i

⁢

x

2

+

y

2

-

x

2

Differentiating and simplifying,

u

⁢

(

x

,

y

)

=

x

2

+

y

2

+

x

2

v

⁢

(

x

,

y

)

=

(

sign

⁢

y

)

⁢

x

2

+

y

2

-

x

2

∂

⁡

u

∂

⁡

x

=

2

4

⁢

x

x

2

+

y

2

+

1

x

2

+

y

2

+

x

=

2

4

⁢

x

2

+

y

2

+

x

x

2

+

y

2

∂

⁡

u

∂

⁡

y

=

(

sign

⁢

y

)

⁢

2

4

⁢

y

x

2

+

y

2

x

2

+

y

2

+

x

=

(

sign

⁢

y

)

⁢

2

4

⁢

y

x

2

+

y

2

⁢

x

2

+

y

2

+

x

∂

⁡

v

∂

⁡

x

=

(

sign

⁢

y

)

⁢

2

4

⁢

x

x

2

+

y

2

-

1

x

2

+

y

2

-

x

=

-

(

sign

⁢

y

)

⁢

2

4

⁢

x

2

+

y

2

-

x

x

2

+

y

2

∂

⁡

v

∂

⁡

y

=

2

4

⁢

y

x

2

+

y

2

x

2

+

y

2

-

x

=

2

4

⁢

y

x

2

+

y

2

⁢

x

2

+

y

2

-

x

.

Pulling out a common factor and placing over a common

denominator

,

∂

⁡

u

∂

⁡

x

-

∂

⁡

v

∂

⁡

y

=

2

4

⁢

x

2

+

y

2

⁢

(

x

2

+

y

2

+

x

-

y

x

2

+

y

2

-

x

)

=

2

4

⁢

x

2

+

y

2

⁢

x

2

+

y

2

+

x

⋅

x

2

+

y

2

-

x

-

y

x

2

+

y

2

-

x

=

2

4

⁢

x

2

+

y

2

⁢

(

x

2

+

y

2

)

-

y

2

-

y

x

2

+

y

2

-

x

=

0

∂

⁡

u

∂

⁡

y

+

∂

⁡

v

∂

⁡

x

=

(

sign

⁢

y

)

⁢

2

4

⁢

x

2

+

y

2

⁢

(

y

x

2

+

y

2

+

x

-

x

2

+

y

2

-

x

)

=

(

sign

⁢

y

)

⁢

2

4

⁢

x

2

+

y

2

⁢

y

-

x

2

+

y

2

+

x

⋅

x

2

+

y

2

-

x

x

2

+

y

2

+

x

=

(

sign

⁢

y

)

⁢

2

4

⁢

x

2

+

y

2

⁢

(

x

2

+

y

2

)

-

x

2

-

y

x

2

+

y

2

+

x

=

0

,

so the Cauchy-Riemann equations are satisfied. More
generally, it can be shown that all complex

algebraic
functions

and

fractional powers

satisfy the
Cauchy-Riemann equations. However, as suggested by
the above derivation, a direct verification could
be tedious, so it is better to use an indirect approach.

Finally, we finish up with two examples of

transcendental
functions

, the complex

exponential

and the

complex logarithm

,
The complex exponential is defined as

exp

⁡

(

x

+

i

⁢

y

)

=

exp

⁡

(

x

)

⁢

(

cos

⁡

y

+

i

⁢

sin

⁡

y

)

. Hence
we have

u

⁢

(

x

,

y

)

=

exp

⁡

(

x

)

⁢

cos

⁡

(

y

)

v

⁢

(

x

,

y

)

=

exp

⁡

(

x

)

⁢

sin

⁡

(

y

)

∂

⁡

u

∂

⁡

x

=

exp

⁡

(

x

)

⁢

cos

⁡

(

y

)

∂

⁡

u

∂

⁡

y

=

-

exp

⁡

(

x

)

⁢

sin

⁡

(

y

)

∂

⁡

v

∂

⁡

x

=

exp

⁡

(

x

)

⁢

sin

⁡

(

y

)

∂

⁡

v

∂

⁡

y

=

exp

⁡

(

x

)

⁢

cos

⁡

(

y

)

Thus we see that the

complex exponential function

is holomorphic.

The complex logarithm may be defined as

log

⁡

(

x

+

i

⁢

y

)

=

1

/

2

⁢

log

⁡

(

x

2

+

y

2

)

+

i

⁢

arctan

⁡

(

y

/

x

)

. Hence we have

u

⁢

(

x

,

y

)

=

1

/

2

⁢

log

⁡

(

x

2

+

y

2

)

v

⁢

(

x

,

y

)

=

arctan

⁡

(

y

/

x

)

∂

⁡

u

∂

⁡

x

=

x

x

2

+

y

2

∂

⁡

u

∂

⁡

y

=

y

x

2

+

y

2

∂

⁡

v

∂

⁡

x

=

-

y

/

x

2

1

+

(

y

/

x

)

2

=

-

y

x

2

+

y

2

∂

⁡

v

∂

⁡

y

=

1

/

x

1

+

(

y

/

x

)

2

=

x

x

2

+

y

2

Hence the complex logarithm is holomorphic.

Title

examples of Cauchy-Riemann equations

Canonical name

ExamplesOfCauchyRiemannEquations

Date of creation

2013-03-22 17:38:05

Last modified on

2013-03-22 17:38:05

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

19

Author

rspuzio (6075)

Entry type

Example

Classification

msc 30A99

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