Mathematical Methods for Physics and Engineering : A Comprehensive Guide

COMPLEX VARIABLES

Forfto be differentiable at the pointz, expressions (24.3) and (24.4) must

be identical. It follows from equating real and imaginary parts thatnecessary

conditions for this are

∂u

∂x

=

∂v

∂y

and

∂v

∂x

=−

∂u

∂y

. (24.5)

These two equations are known as theCauchy–Riemann relations.

We can now see why for the earlier examples (i)f(z)=x^2 −y^2 +i 2 xymight

be differentiable and (ii)f(z)=2y+ixcould not be.

(i)u=x^2 −y^2 ,v=2xy:

∂u

∂x

=2x=

∂v

∂y

and

∂v

∂x

=2y=−

∂u

∂y

,

(ii)u=2y,v=x:

∂u

∂x

=0=

∂v

∂y

but

∂v

∂x

=1=−2=−

∂u

∂y

.

It is apparent that forf(z) to be analytic something more than the existence

of the partial derivatives ofuandvwith respect toxandyis required; this

something is that they satisfy the Cauchy–Riemann relations.

We may enquire also as to thesufficientconditions forf(z) to be analytic in

R. It can be shown§that a sufficient condition is that the four partial derivatives

exist,are continuousand satisfy the Cauchy–Riemann relations. It is the addi-

tional requirement of continuity that makes the difference between the necessary

conditions and the sufficient conditions.

In which domain(s) of the complex plane isf(z)=|x|−i|y|an analytic function?

Writingf=u+ivit is clear that both∂u/∂yand∂v/∂xarezeroinallfourquadrants
and hence that the second Cauchy–Riemann relation in (24.5) is satisfied everywhere.
Turning to the first Cauchy–Riemann relation, in the first quadrant (x>0,y>0) we
havef(z)=x−iyso that
∂u
∂x

=1,

∂v

∂y

=− 1 ,

which clearly violates thefirst relation in (24.5). Thusf(z) is not analytic in the first
quadrant.
Following a similiar argument for the other quadrants, we find
∂u
∂x

=−1or +1 forx<0andx>0, respectively,

∂v

∂y

=−1or +1 fory>0andy<0, respectively.

Therefore∂u/∂xand∂v/∂yare equal, and hencef(z) is analytic only in the second and
fourth quadrants.

§See, for example, any of the references given on page 824.