Page 79 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Since f (z )  0, we may write it in the form Re and thus,
i
0
lim  1 e i 1 = Re i i.e. lim  1 e i( 1  1 ) = Re i
re i 1 r

 lim  1 = R = |f (z ) |
r 0
and lim (   ) = 
1
1
i.e. lim   lim  = 
1
1
i.e.    =    =  + 
1
1
1
1
Similarly,  =  + .
2
2
Hence, the curves C and C have definite tangents at w making angles  +  and  + 
2
0
1
1
2
respectively with the real axis. The angle between C and C is
1
2
   = ( + )  (  ) =   2
1
1
2
2
1
which is the same as the angle between C and C . Hence the curve C  and C  intersect at the
2
1
1
2
same angle as the curves C and C . Also the angle between the curves has the same sense in the
2
1
two figures. So the mapping is conformal.
Special Case : When f(z ) = 0, we suppose that f(z) has a zero of order n at the point z . Then in
0
0
the neighbourhood of this point (by Taylor’s theorem)
f(z) = f(z ) + a(z  z ) + , where a  0
n+1
0
0
Hence, w  w = a(z  z ) + .
n+1
0
0
1
i.e.  e i 1 = | a | r e i[d + (n +1)q1] +
n+1
1
where,  = arg a
Hence, lim  = [d + (n + 1)  ] =  + (n + 1)  1 |  is constant
1
1
Similarly, lim  = d + (n + 1)  2
2
Thus, the curves C and C still have definite tangent at w , but the angle between the tangents
1
0
2
is
lim(   ) = (n + 1) (   )
1
2
1
2
Thus, the angle is magnified by (n + 1).
Also the linear magnification, R = lim  1 = 0 | lim  1 = R = |f (z )| = 0
r r 0
Therefore, the conformal property does not hold at such points where f (z) = 0
A point z at which f (z ) = 0 is called a critical point of the mapping. The following theorem is
0
0
the converse of the above theorem and is sufficient condition for the mapping to be conformal.
Theorem: If the mapping w = f(z) is conformal then show that f(z) is an analytic function of z.
Proof. Let w = f(z) = u(x, y) + iv(x, y)
Here, u = u(x, y) and v = v(x, y) are continuously differentiable equations defining conformal
transformation from z-plane to w-plane. Let ds and d be the length elements in z-plane and
w-plane respectively so that
ds = dx + dy , d = du + dv 2 (1)
2
2
2
2
2
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