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Unformatted text preview: ¥Length of C < 2πœ
(the circle can be assumed to have a radius smaller than 1..) Since œ is arbitrarily small,
the given integral must be zero, and we are done.
(A) Evaluate O
(C) Evaluate O
±π/2, ±3π/2, ...
(D) Evaluate O
C z e
z-2 dz, where z is any circle enclosing 2.
2 z -1 dz where C is any simple contour enclosing 1 but non of the points z
dz where C is the circle |z - 2i| = 4.
2 17 z/(z+3i)
[To evaluate this, rewrite the integrand as z-3i .] Corollary 5.3 (Analytic Functions have Derivatives of All Orders)
Let f be analytic on simply connected D, let z0 é D and let C be any simple closed
path in D around z0. Then f (z0) exists, and
f (z0) =
2πi ı (z - z )n+1
Proof: Let us start with n = 1: Write
f(w) - f(z0)
f'(z0) = lim
wÆz0 w - z0
Applying the Integral Formula theorem to each term gives:
1 Û f(z)
1 Û f(z)
ı z - w dz
2πi ı z - z
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