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Complex Analysis and Differential Geometry
Notes
stand for the sequence g which is such that g(n) = z . For example, i is the sequence g for
n
n
i
which g(n) = .
n
The number L is a limit of the sequence (z ) if given an > 0, there is an integer N such that
n
|zn L| < for all n N . If L is a limit of (z ), we sometimes say that (z ) converges to L. We
n
n
frequently write lim (z ) = L. It is relatively easy to see that if the complex sequence (z ) =
n
n
(u + iv ) converges to L, then the two real sequences (u ) and (v ) each have a limit: (u ) converges
n
n
n
n
n
to ReL and (v ) converges to ImL. Conversely, if the two real sequences (u ) and (v ) each have a
n
n
n
limit, then so also does the complex sequence (u + iv ). All the usual nice properties of limits of
n
n
sequences are thus true:
lim(z ± w ) = lim(z ) ± lim(w );
n
n
n
n
lim(z w ) = lim(z ) lim(wn); and
n
n
n
z lim(z )
lim n = n .
w n lim(w )
n
provided that lim(z ) and lim(w ) exist. (And in the last equation, we must, of course, insist that
n
n
lim(w ) 0.)
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A necessary and sufficient condition for the convergence of a sequence (a ) is the celebrated
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Cauchy criterion: given > 0, there is an integer N so that |a a | < whenever n,m > N .
n m
A sequence (f ) of functions on a domain D is the obvious thing: a function from the positive
n
integers into the set of complex functions on D. Thus, for each z D, we have an ordinary
sequence (fn(z)). If each of the sequences (fn(z)) converges, then we say the sequence of functions
(f ) converges to the function f defined by f(z) = lim(f (z)). This pretty obvious stuff. The sequence
n
n
(f ) is said to converge to f uniformly on a set S if given an > 0, there is an integer N so that
n
|f (z) f(z)| < for all n N and all z S.
n
Notes It is possible for a sequence of continuous functions to have a limit function
that is not continuous. This cannot happen if the convergence is uniform.
To see this, suppose the sequence (f ) of continuous functions converges uniformly to f on a
n
domain D, let z D, and let > 0. We need to show there is a so that |f(z ) f(z)| < whenever
0
0
|z z| < . Lets do it. First, choose N so that |f (z) f(z)| < . We can do this because of the
3
0
N
uniform convergence of the sequence (f ). Next, choose so that |f (z ) f (z)| < 3 whenever
N
0
n
N
|z z| < . This is possible because f is continuous.
0
N
Now then, when |z z| < , we have
0
|f(z ) f( )| = |f(z ) f (z ) + f (z ) f (z) + f (z) f(z)|
z
0
N
N
0
N
0
0
N
|f(z ) f (z )| + |f (z ) f (z)| + |f (z) f(z)|
N
0
N
N
N
0
0
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