proof to Cauchy-Riemann equations (polar coordinates)

proof to Cauchy-Riemann equations (polar coordinates)

If

f

⁢

(

z

)

is differentialble at

z

0

then the following limit

f

′

⁢

(

z

0

)

=

lim

ξ

→

0

⁡

f

⁢

(

z

0

+

ξ

)

-

f

⁢

(

z

0

)

ξ

will remain the same approaching from any direction. First we fix

θ

as

θ

0

then we take the limit along the ray where the

argument

is equal to

θ

0

. Then

f

′

⁢

(

z

0

)

=

lim

h

→

0

⁡

f

⁢

(

r

0

⁢

e

i

⁢

θ

0

+

h

⁢

e

i

⁢

θ

0

)

-

f

⁢

(

r

0

⁢

e

i

⁢

θ

0

)

h

⁢

e

i

⁢

θ

0

=

lim

h

→

0

⁡

f

⁢

(

(

r

0

+

h

)

⁢

e

i

⁢

θ

0

)

-

f

⁢

(

r

0

⁢

e

i

⁢

θ

0

)

h

⁢

e

i

⁢

θ

0

=

lim

h

→

0

⁡

u

⁢

(

r

0

+

h

,

θ

0

)

+

i

⁢

v

⁢

(

r

0

+

h

,

θ

0

)

-

u

⁢

(

r

0

,

θ

0

)

-

i

⁢

v

⁢

(

r

0

,

θ

0

)

h

⁢

e

i

⁢

θ

0

=

1

e

i

⁢

θ

0

⁢

[

lim

h

→

0

⁡

u

⁢

(

r

0

+

h

,

θ

0

)

-

u

⁢

(

r

0

,

θ

0

)

h

+

i

⁢

lim

h

→

0

⁡

v

⁢

(

r

0

+

h

,

θ

0

)

-

v

⁢

(

r

0

,

θ

0

)

h

]

=

1

e

i

⁢

θ

0

⁢

[

∂

⁡

u

∂

⁡

r

⁢

(

r

0

,

θ

0

)

+

i

⁢

∂

⁡

v

∂

⁡

r

⁢

(

r

0

,

θ

0

)

]

Similarly, if we take the limit along the circle with fixed

r

equals

r

0

. Then

f

′

⁢

(

z

0

)

=

lim

h

→

0

⁡

f

⁢

(

r

0

⁢

e

i

⁢

θ

0

+

r

0

⁢

e

i

⁢

(

θ

0

+

h

)

)

-

f

⁢

(

r

0

⁢

e

i

⁢

θ

0

)

r

0

⁢

e

i

⁢

θ

0

⁢

(

e

i

⁢

h

-

1

)

=

lim

h

→

0

⁡

f

⁢

(

r

0

⁢

e

i

⁢

(

θ

0

+

h

)

)

-

f

⁢

(

r

0

⁢

e

i

⁢

θ

0

)

h

⁢

e

i

⁢

θ

0

=

lim

h

→

0

⁡

u

⁢

(

r

0

,

θ

0

+

h

)

+

i

⁢

v

⁢

(

r

0

,

θ

0

+

h

)

-

u

⁢

(

r

0

,

θ

0

)

-

i

⁢

v

⁢

(

r

0

,

θ

0

)

h

⁢

e

i

⁢

θ

0

=

1

r

0

⁢

e

i

⁢

θ

0

⁢

[

lim

h

→

0

⁡

u

⁢

(

r

0

+

h

,

θ

0

)

-

u

⁢

(

r

0

,

θ

0

)

h

⋅

h

e

i

⁢

h

-

1

+

i

⁢

lim

h

→

0

⁡

v

⁢

(

r

0

+

h

,

θ

0

)

-

v

⁢

(

r

0

,

θ

0

)

h

⋅

h

e

i

⁢

h

-

1

]

=

1

r

0

⁢

e

i

⁢

θ

0

⁢

[

lim

h

→

0

⁡

u

⁢

(

r

0

+

h

,

θ

0

)

-

u

⁢

(

r

0

,

θ

0

)

h

⋅

lim

h

→

0

⁡

h

e

i

⁢

h

-

1

+

i

⁢

lim

h

→

0

⁡

v

⁢

(

r

0

+

h

,

θ

0

)

-

v

⁢

(

r

0

,

θ

0

)

h

⋅

lim

h

→

0

⁡

h

e

i

⁢

h

-

1

]

=

1

r

0

⁢

e

i

⁢

θ

0

⁢

[

∂

⁡

u

∂

⁡

θ

⁢

(

r

0

,

θ

0

)

⁢

1

i

+

∂

⁡

v

∂

⁡

θ

⁢

(

r

0

,

θ

0

)

]

=

1

r

0

⁢

e

i

⁢

θ

0

⁢

[

∂

⁡

v

∂

⁡

θ

⁢

(

r

0

,

θ

0

)

-

i

⁢

∂

⁡

u

∂

⁡

θ

⁢

(

r

0

,

θ

0

)

]

Note: We use l’Hôpital’s rule to obtain the following result used above

lim

h

→

0

⁡

h

e

i

⁢

h

-

1

=

1

i

.

Now, since the limit is the same along the circle and the ray then they are equal:

1

e

i

⁢

θ

0

⁢

[

∂

⁡

u

∂

⁡

r

⁢

(

r

0

,

θ

0

)

+

i

⁢

∂

⁡

v

∂

⁡

r

⁢

(

r

0

,

θ

0

)

]

=

1

r

0

⁢

e

i

⁢

θ

0

⁢

[

∂

⁡

v

∂

⁡

θ

⁢

(

r

0

,

θ

0

)

-

i

⁢

∂

⁡

u

∂

⁡

θ

⁢

(

r

0

,

θ

0

)

]

[

∂

⁡

u

∂

⁡

r

⁢

(

r

0

,

θ

0

)

+

i

⁢

∂

⁡

v

∂

⁡

r

⁢

(

r

0

,

θ

0

)

]

=

1

r

0

⁢

[

∂

⁡

v

∂

⁡

θ

⁢

(

r

0

,

θ

0

)

-

i

⁢

∂

⁡

u

∂

⁡

θ

⁢

(

r

0

,

θ

0

)

]

which implies that

∂

⁡

u

∂

⁡

r

=

1

r

⁢

∂

⁡

v

∂

⁡

θ

∂

⁡

v

∂

⁡

r

=

-

1

r

⁢

∂

⁡

u

∂

⁡

θ

QED

Title

proof to

Cauchy-Riemann equations

(

polar coordinates

)

Canonical name

ProofToCauchyRiemannEquationspolarCoordinates

Date of creation

2013-03-22 14:06:13

Last modified on

2013-03-22 14:06:13

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

5

Author

Daume (40)

Entry type

Proof

Classification

msc 30E99

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