Cauchy Integrals: evaluate Sc [ cos(z) / (z^2 (z - 2)) ] dz

lastlydreaming

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Oct 7, 2006
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Is this correct?

Evaluate

Sc [ cos(z) / (z^2 (z - 2)) ] dz

where C is the rectangle with vertices + and - 1 and + and - i.

So here z= 0 or z = 2. At z= 0 it is in the region, whereas z= 2 is not.
Hence

f(zo) = 1 / 2pi i S (integral) f(z)/(z-zo) dz.

S (integral) cos z / z^2 (z-2) dz
z' = (z-2) (sin 2)- cos (2) / (z-2)^2
z'(0) = -1/4


Hence 1 / 2pi i (-1/4) = -pi i/ 2

Thanks.
 
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