Complex integral around unit circle at origin for 1/z is 2𝛑i, For 1/z^n it is zero as per Cauchy s first even though pole there. Why?

The function f(z) = 1/z, is not analytic at the origin; it has a type of singularity there called a pole. Accepting that result and the reason given, it seemed reasonable that if f(z) = 1/z^n, around a unit circle, then ∮(1/z^n)dz would also NOT be zero. But the result of calculation is that ∮(1/z^n)dz = 0,... show more The function f(z) = 1/z, is
not analytic at the origin; it has a type of singularity there called a pole.
Accepting that result and the reason given, it seemed reasonable that
if f(z) = 1/z^n, around a unit circle, then ∮(1/z^n)dz would also NOT be zero.
But the result of calculation is that ∮(1/z^n)dz = 0, (unless n = 1)
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