Page 23 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes We shall see that f is differentiable at z .
0

0
0
= f(z   z) f(z )
 z



= [u(x   x,y   y) u(x ,y )] i[v(x   x,y   y) v(x ,y )]
0
0
0
0
0
0
0
0

 x i y

Observe that
u(x + x, y + y) – u(x , y ) = [u(x + x, y + y) – u(x , y + y)] + [u(x , y + y) – u(x ,y ].
0 0 0 0 0 0 0 0 0 0 0 0
Thus,
 u

u(x + x, y + y) – u(x , y + y) = x  x ( ,y   y),
0
0
0
0
0
and,
 u ( ,y   y) =  u (x ,y )   ,

 x 0  x 0 0 1
where
lim   0.
 z 0 1
Thus,
u(x + x, y + y) – u(x , y + y) = x     u x  (x ,y )   1    .
0
0
0
0
0
0
Proceeding similarly, we get

0
0
= f(z   z) f(z )
 z
[u(x   x,y   y) u(x ,y )] i[v(x   x,y   y) v(x ,y )]



= 0 0 0 0 0 0 0 0

 x i y

 x   du (x ,y )    i dv (x ,y ) i 2     y   du (x ,y )    i dv (x ,y ) i 4 


dx
0
0
3
dy
0
0
0
dx
0
0
1
0
dy
=     ,.

 x i y

where   0 as z  0. Now, unleash the Cauchy-Riemann equations on this quotient and
i
obtain,

0
= f(z   z) f(z )
0
 z

 x    u  i  v    i y    u  i  v   stuff
=  x   x   x  x   
 x i y  x i y




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