CVR10

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Take the variable z^3 into (z-2) terms.

(z^3)cos(1/(z-2)) = [(z-2)^3 + 6(z-2)^2 + 12(z-2) + 8]cos(1/(z-2))

Let u = (z-2)

So that the equation will be:

= (u^3 + 6u^3 + 12u + 8)cos(1/u)

expans cos(1/u) with the formula Sigma [(-1)^n (1/u)^2n /(2n)!] for n>=0

the series will be

= 1 - [(1/u)^2]/2! + [(1/u)^4]/4! - [(1/u)^6]/6! + ....

Residu is the coefficient of u^-1. Take a look that the multiplication of u^3 with [(1/u)^4]/4! and 12u with - [(1/u)^2]/2! will result the coeff of u^-1.

so that

Residu(f,2) = 1/4! + 12*(-1/2!) = 1/24 - 12(1/2) = 1/24 - 6 = -5 23/24