Mathematical Methods for Physics and Engineering : A Comprehensive Guide

24.2 The Cauchy–Riemann relations

Show that the functionf(z)=1/(1−z)is analytic everywhere except atz=1.

Sincef(z) is given explicitly as a function ofz, evaluation of the limit (24.1) is somewhat
easier. We find

f′(z) = lim

∆z→ 0

[

f(z+∆z)−f(z)

∆z

]

= lim

∆z→ 0

[

1

∆z

(

1

1 −z−∆z

−

1

1 −z

)]

= lim

∆z→ 0

[

1

(1−z−∆z)(1−z)

]

=

1

(1−z)^2

,

independently of the way in which ∆z→0, providedz= 1. Hencef(z)isanalytic
everywhere except at the singularityz=1.

24.2 The Cauchy–Riemann relations

From examining the previous examples, it is apparent that for a functionf(z)

to be differentiable and hence analytic there must be some particular connection

between its real and imaginary partsuandv.

By considering a general function we next establish what this connection must

be. If the limit

L= lim

∆z→ 0

[

f(z+∆z)−f(z)

∆z

]

(24.2)

is to exist and be unique, in the way required for differentiability, then any two

specific ways of letting ∆z→0 must produce the same limit. In particular, moving

parallel to the real axis and moving parallel to the imaginary axis must do so.

This is certainly a necessary condition, although it may not be sufficient.

If we letf(z)=u(x, y)+iv(x, y)and∆z=∆x+i∆ythen we have

f(z+∆z)=u(x+∆x, y+∆y)+iv(x+∆x, y+∆y),

and the limit (24.2) is given by

L= lim

∆x,∆y→ 0

[

u(x+∆x, y+∆y)+iv(x+∆x, y+∆y)−u(x, y)−iv(x, y)

∆x+i∆y

]

.

If we first suppose that ∆zis purely real, so that ∆y= 0, we obtain

L= lim

∆x→ 0

[

u(x+∆x, y)−u(x, y)

∆x

+i

v(x+∆x, y)−v(x, y)

∆x

]

=

∂u

∂x

+i

∂v

∂x

,

(24.3)

provided each limit exists at the pointz. Similarly, if ∆zis taken as purely

imaginary, so that ∆x= 0, we find

L= lim

∆y→ 0

[

u(x, y+∆y)−u(x, y)

i∆y

+i

v(x, y+∆y)−v(x, y)

i∆y

]

=

1

i

∂u

∂y

+

∂v

∂y

.

(24.4)