Complex Analysis Help

a) Find the residue of (z-1)^3/ [z(z+2)^3] at z=∞

b) Use TWO methods of evaluating
∫γ (z-1)^3/ [z(z+2)^3] dz
where γ is the circle of radius 3 centered at 0.

Thanks.

a) By definition, this equals the residue of (-1/z^2) * f(1/z) at z = 0.

Since f(z) = (z-1)^3/ [z(z+2)^3], we have
f(1/z) = (1/z - 1)^3/ [(1/z) (1/z + 2)^3]
........= (1/z^2)(1 - z)^3 / [(1/z) * (1/z)^3 (1 + 2z)^3]
........= z^2(1 - z)^3 / (1 + 2z)^3.

So, the residue at infinity equals the residue of (-1/z^2) * z^2(1 - z)^3 / (1 + 2z)^3
= -(1 - z)^3 / (1 + 2z)^3 at z = 0, which equals -1.
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b) By using part a, the integral equals (-2πi) * -1 = 2πi.
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On the other hand, we can compute the residues at the finite poles z = 0, -2.

Since z = 0 is of order 1, its residue equals
lim(z→0) z * (z-1)^3/ [z(z+2)^3] = lim(z→0) (z-1)^3/(z+2)^3 = -1/8.

Since z = -2 is of order 3, its residue equals
lim(z→ -2) (1/2!) * (d^2/dz^2) (z - (-2))^3 * (z-1)^3/ [z(z+2)^3]
= lim(z→ -2) (1/2!) * (d^2/dz^2) (z - 1)^3 / z
= lim(z→ -2) (1/2!) * (d^2/dz^2) (z^2 - 3z + 3 - 1/z)
= lim(z→ -2) (1/2!) * (d/dz) (2z - 3 + 1/z^2)
= lim(z→ -2) (1/2!) * (2 - 2/z^3)
= 9/8.

Hence, the integral equals 2πi * (-1/8 + 9/8) = 2πi by the Residue Theorem.
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I hope this helps!