1550251515-Classical_Complex_Analysis__Gonzalez_

Differentiation 325

valid at least in Nr(zo). Taking derivatives on both sides of (6.6-8) with

respect to z, we have

f'(z) =u1 [~(z+zo), ;i(z-zo)] ~

+u2 [~(z+zo), ;i(z-zo)] ;i

+iv1 [~(z+zo), ;i(z-zo)] ~

• iv 2 [ ~ (z + zo), ;i (z - zo)] ;i (6.6-9)

where for convenience we have used subindices to indicate partial deriva-
tives. By the Cauchy-Riemann equations we have v 2 = u 1 and v 1 = -u 2 •
Hence (6.6-9) becomes

f'(z) = u1 [ ~ (z + z 0 ), ;i (z - zo)] -iu2 [ ~ (z + zo), ;i (z - zo)]

and integrating with respect to z we get

f(z)=2u [~(z+zo), ~i(z-zo)] +c (6.6-10)

where C is a complex constant. If the region of analyticity of f contains

the origin, we can take zo = 0, and formula (6.6-10) reduces to

f(z)=2u(~, ;J +c (6.6-11)

Examples

1. Consider the function u = x^3 - 3xy^2 + x^2 - y^2 , which is harmonic in
   ~.2. Using (6.6-11) we get

f(z) =2 [ (i)3 -3 (i) (;J2 + (i)2 -(;J2] +c

= z

3

+ z^2 + C

2. The function u = ex cosy is also harmonic in llt^2 • Again, using ( 6.6-11)
   we have

f(z) = 2ez/^2 cos ;i + C = 2ez/^2 ~ (ez/^2 + e-z/^2 ) + C

= ez + 1 + C = ez + C'