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The problem is in the Cauchy Integral Formula section in Gamelin's "Complex Analysis". $$ \oint_{|z-1|=3} \frac{dz}{z(z^2-4)e^z} $$ I have trouble with it because -2 is actually on the boundary.

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1 Answer 1

up vote 4 down vote accepted

It should be treated as principal value, i.e

$$ \begin{eqnarray} \operatorname{P.V.} \oint_{|z-1|=3} \frac{\mathrm{d}z}{z (z^2-4) \mathrm{e}^z} &=& \frac{1}{2} \left( \oint_{|z-1|=3^+} \frac{\mathrm{d}z}{z (z^2-4) \mathrm{e}^z} + \oint_{|z-1|=3^-} \frac{\mathrm{d}z}{z (z^2-4) \mathrm{e}^z} \right) \\ &=& 2 \pi i \left( \left. \frac{\mathrm{e}^{-z}}{z^2-4} \right|_{z=0} + \left. \frac{\mathrm{e}^{-z}}{z(z+2)} \right|_{z=2} + \frac{1}{2} \left. \frac{\mathrm{e}^{-z}}{z(z-2)} \right|_{z=-2} \right) \\ &=& 2 \pi i \left( -\frac{1}{4} + \frac{\mathrm{e}^{-2}}{8} + \frac{\mathrm{e}^{2}}{16} \right) \end{eqnarray} $$

Added: Also can be done using Mathematica:

In[217]:= TrigToExp[
 Integrate @@ {1/(z (z^2 - 4) Exp[z]) Dt[z]/Dt[phi] /. 
    z -> 1 + (3) Exp[I phi], {phi, 0, 2 Pi}, PrincipalValue -> True}]

Out[217]= -I Pi/2 + I Pi/(4 E^2) + 1/8 I E^2 Pi
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(Principal value, to be precise.) –  Hans Lundmark Oct 25 '11 at 6:29
@HansLundmark Thank you, I have just corrected the spelling. –  Sasha Oct 25 '11 at 10:26
I dont understand the solution, whats going on here, please teach me –  El Angel Exterminador Apr 30 '13 at 14:09

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