Evaluate the following integrals-Residue theorem

∫γ dz/((z^n)(z^2+1)) where n is a nonnegative integer
a)where γ is the circle of radius 3/2 centered at i
b)where γ is the circle of radius 2 centered at 1

Does anyone could help me with this exercise please?

The integrand has singularities at z = 0 and z = ±i.

Since z = ±i are simple poles, they have residues
lim(z → ±i) (z - ±i) * 1/(z^n (z^2 + 1))
= lim(z → ±i) 1/(z^n (z + ±i))
= 1/((±i)^n * (±2i)).
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Next, note that z = 0 is a pole of order n.

Via Laurent Series at z = 0,
1/(z^n (z^2 + 1))
= (1/z^n) * 1/(1 - (-z^2))
= (1/z^n) * Σ(k = 0 to ∞) (-z^2)^k
= Σ(k = 0 to ∞) (-1)^k z^(2k-n).

This will have nonzero residue when 2k-n = -1
<==> k = (n-1)/2 (is an integer).
<==> n is odd.

If n is odd, we can write n = 2m+1 for some integer m.
==> k = m, and thus the residue equals (-1)^m.
If n is even, then the residue equals 0.
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a) Note that z = 0 and z = i are inside the contour.

So, the integral equals by the Residue Theorem
2πi [1/(2i^(n+1)) + (-1)^m]
= 2πi [1/(2i^(n+1)) + (-1)^((n-1)/2)], if n is odd (and we write n = 2m+1)

If n is even, the integral equals 2πi * 1/(2i^(n+1)).
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b) This time, z = 0, -i, i are all in the contour.

So, the integral equals by the Residue Theorem
2πi [1/(2i^(n+1)) + 1/(2(-i)^(n+1)) + (-1)^m]
= 2πi [1/(2i^(n+1)) + 1/(2(-i)^(n+1)) + (-1)^((n-1)/2)], if n is odd (and we write n = 2m+1)

If n is even, the integral equals 2πi * [1/(2i^(n+1)) + 1/(2(-i)^(n+1))].
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I hope this helps!