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Prove: Let f(z) is analytic at z_{0} and
(z_{0})
0. Then the under the mapping w = f(z) the
tangent at z_{0} to any curve C passing through z_{0} is rotated through the angle arg
(z_{0}).

Proof. See Fig. 1. As a point
moves from z_{0} to z_{0} + Δz along C,
the image point traces out C' in
the w plane, going from w_{0} to w_{0}
+Δw. Assume that the curves C
and C', are defined parametrically
with t as the parameter. Then
corresponding to the path z = z(t)
[or x = x(t), y= y(t)] in the z
plane, we have the path w = w(t)
[or u = u(t), v = v(t)] in the w
plane.

The derivatives dz/dt and dw/dt represent tangent vectors to corresponding points on C and C'.

Now

At points z_{0} and w_{0}, we have

Let

Equation 1) then becomes

Thus

which is what we wished to prove.

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