Page 20 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 2: Complex Functions

means that given an  > 0, there is a  so that |f(z) – L| <  whenever 0 < |z – z | < . As you could Notes
0
guess, we say that f is continuous at z if it is true that lim f(z)  f(z ). If f is continuous at each
0
0
0 z
z
point of its domain, we say simply that f is continuous.
Suppose both lim f(z)  limg(z) exist. Then the following properties are easy to establish:
z 0 z z 0 z
lim[f(z) g(z)]  lim f(z) lim g(z)


z 0 z z 0 z z 0 z
lim[f(z)g(z)] lim f(z)lim g(z)

z 0 z z 0 z z 0 z
and
f(z) lim f(z)
lim  z 0 z
z 0 z g(z) limg(z)
z 0 z
provided, of course, that lim g(z)  0.
z 0 z
It now follows at once from these properties that the sum, difference, product, and quotient of
two functions continuous at z are also continuous at z . (We must, as usual, except the dreaded
0
0
0 in the denominator.)
It should not be too difficult to convince yourself that if z = (x, y), z = (x , y ), and f(z) =
0
0
0
u(x, y) + iv(x, y), then

lim f(z)  lim u(x,y) i lim v(x,y)
z 0 z (x,y) ( 0 x ,y ) (x,y) ( 0 x y 0
0
Thus, f is continuous at z = (x , y ) precisely when u and v are.
0
0
0
Our next step is the definition of the derivative of a complex function f. It is the obvious thing.
Suppose f is a function and z is an interior point of the domain of f . The derivative f’(z ) of f is
0
0
f(z) f(z )

f'(z )  lim 0
0

z 0 z z z 0
Example 2:
Suppose f(z) = z . Then, letting z = z – z , we have
2
0
f(z) f(z ) z) f(z )

lim 0 = lim f(z    0
0

z 0 z z z 0  z 0  z
2
= lim f(z   z)  z 2 0
0
 z 0  z
2z z ( z) 2



= lim 0
 z 0  z
= lim(2z   z)
0
 z 0
= 2z 0
No surprise here–the function f(z) = z has a derivative at every z, and it’s simply 2z.
2
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