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evaluate the complex integral 1/(z^3 - 8) where c is the counterclockwise, circular contour |z-1| = 2

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Answer: We can evaluate this integral using the residue theorem. First, we need to find the poles of the integrand within the contour |z-1| = 2.

We have:

z^3 - 8 = (z - 2)(z^2 + 2z + 4)

The roots of the quadratic factor are:

z = (-2 ± sqrt(-4*4))/2 = -1 ± i sqrt(3)

None of these roots are inside the contour, so the only pole is z = 2.

The residue of 1/(z^3 - 8) at z = 2 is:

Res(1/(z^3 - 8), z=2) = 1/(3*2^2) = 1/12

By the residue theorem, the integral is:

∫(|z-1|=2) 1/(z^3 - 8) dz = 2πi Res(1/(z^3 - 8), z=2) = 2πi/12 = πi/6

Therefore, the value of the complex integral is πi/6.

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