Page 123 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Then there exists a neighbourhood of w in the w-plane in which the function w = f(z) has a
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unique inverse z = F(w) in the sense that the function F is single-valued and analytic in that
neighbourhood such that F(w ) = z and
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F(w) = .
f'(z)
Proof. Consider the function f(z) – w . By hypothesis, f(z ) w = 0. Since f (z )  0, f is not a
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constant function and therefore, neither f(z) – w not f (z) is identically zero. Also f(z) – w is
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analytic at z = z and so it is analytic in some neighbourhood of z . Again, since zeros are
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isolated, neither f(z) – w nor f (z) has any zero in some deleted neighbourhood of z . Hence,
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there exists R > 0 such that f(z) – w is analytic for |z – z |  R and f(z) – w  0, f (z)  0 for
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0 < |z – z |  R. Let D denote the open disc
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{z : |z – z | < R}
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and C denotes its boundary
{z : |z – z | = R}.
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Since f(z) – w for |z – z |  R, we conclude that | f(z) – w | has a positive minimum on the circle
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C. Let
min | f(z) – w | = m
z C 0

and choose d such that 0 < d < m.
We now show that the function f(z) assumes exactly once in D every value w in the open disc
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T = { w : |w – w | < d}. We apply Rouche’s theorem to the functions w – w and f(z)-w . The
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0
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condition of the theorem are satisfied, since
|w – w | < d < m = |f(z) – w |  |f(z) – w | on C.
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Thus, we conclude that the functions.
f(z) – w and (f(z) – w ) + (w – w ) = f(z) – w 1
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have the same number of zeros in D. But the function f(z) - w has only one zero in D i.e. a simple
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zeros at z , since (f(z) – w ) = f(z)  0 at z .
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Hence, f(z) – w must also have only one zero, say z in D. This means that the function f(z)
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assumes the value w, exactly once in D. It follows that the function w = f(z) has a unique inverse,
say z = F(w) in D such that F is single-valued and w = f {F(w)}. We now show that the function
F is analytic in D. For fix w in D, we have f(z) = w for a unique z in D. If w is in T and F(w) = z,
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then
F(w) F(w ) z z 1


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w w 1  f(z) f(z ) (2)


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It is noted that T is continuous. Hence, z  z whenever w  w . Since z  D, as shown above
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f (z ) exists and is zero. If we let w  w, then (2) shows that
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F(w ) = f'(z ) .
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Thus F(w) exists in the neighbourhood T of w so that the function F is analytic there.
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