Page 22 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 2: Complex Functions

Choose z = (x, 0) = x. Then, Notes


0
f’z( ) = lim 0 f(z    z) f(z )
0
z
0
z




= lim u(x   x,y ) iv(x   x,y ) u(x ,y ) iv(x ,y )
0
0
0
0
0
0
0
0
 z 0  x


0 
= lim  u(x   x,y )– u(x  y )  i v(x , x,y ) v(x ,y )
0
0
0
0
0
0
0
0
 z    x  x  
=  u (x ,y ) i  v (x ,y )

0
0
0
0
 x  x
Next, choose z = (0, y) = iy. Then,

f’(z ) = lim 0 f(z    z) f(z )
0
0
z
0
z




= lim u(x ,y   y) iv(x ,y   y) u(x ,y ) iv(x ,y )
0
0
0
0
0
0
0
0

 y 0 i y
 v(x ,y   y) v(x ,y ) u(x ,y   y) u(x ,y )


0
0
0
0
0
0
0
0
= lim   i 

 y 0 i y  y
 
=  v (x ,y ) i  u (x ,y )

0
0
0
0
 y dy
We have two different expressions for the derivative f’(z ), and so
0
 v (x ,y ) i  u (x ,y )  v (x ,y ) i  u (x ,y )


 x 0 0 dy 0 0 =  y 0 0  y 0 0
or,
 u (x ,y )  v (x ,y ),
0 =
 x 0  y 0 0
 u (x ,y ) i  v (x ,y )
0 = 
 x 0  x 0 0
These equations are called the Cauchy-Riemann Equations.
We have shown that if f has a derivative at a point z , then its real and imaginary parts satisfy
0
these equations. Even more exciting is the fact that if the real and imaginary parts of f satisfy
these equations and if in addition, they have continuous first partial derivatives, then the function
f has a derivative. Specifically, suppose u(x, y) and v(x, y) have partial derivatives in a
neighborhood of z = (x , y ), suppose these derivatives are continuous at z , and suppose
0
0
0
0
 u (x ,y )  v (x ,y ),
0 =
 x 0  y 0 0
 u (x ,y )  u (x ,y )
0 =
 y 0  x 0 0
LOVELY PROFESSIONAL UNIVERSITY 15