principal parts of some laurent expansions?

Question

Can anyone help with 3 Laurent Expansions for me (10 points):

1) Find the principal part of the Laurent expansion of the function 1/(cos(z)-1)^2 in the annulus 0<|z|<pi.

2) Find the principal part of the Laurent expansion of the function 1/(sinz)^3 in the same annulus

3) Find the principal part of the Laurent expansion of the function 1/(z^2+4) about the point 2i.

Can you show your working please, I know you have use the series for cos and sin for 1 and 2 and some sort of geometric progression for 3 but would appreciate the details.

Thanks for the help.

kb · Accepted Answer

1) 1/(cos z - 1)^2
= 1/[(1 - z^2/2! + z^4/4! - ...) - 1]^2
= 1/(z^2/2! - z^4/4! - ...)^2
= 1/[(1/4)z^4 * (1 - z^2/12 - ...)^2]
= (4/z^4) * [1/(1 - z^2/6 + ...)]
= (4/z^4) * (1 + z^2/6 + (z^4 terms and higher))

So, the principal part is (4/z^4) * (1 + z^2/6) = 4/z^4 + 2/(3z^2).

2) 1/sin^3(z)
= 1/(z - z^3/3! + ...)^3
= (1/z^3) * 1/(1 - z^2/3! + ...)^3
= (1/z^3) * 1/(1 - z^2/2 + ...)
= (1/z^3) * (1 + z^2/2 + (z^3 terms and higher))

So, the principal part is (1/z^3)(1 + z^2/2) = 1/z^3 + 1/(2z).

3) 1/(z^2 + 4)
= 1/[(z - 2i)(z + 2i)]
= (1/(4i)) [1/(z - 2i) - 1/(z + 2i)], by partial fractions
= (1/(4i)) [1/(z - 2i) - 1/((z - 2i) - 4i)]
= (1/(4i)) [1/(z - 2i) - 1/(-4i * (1 - (z - 2i)/(4i))]
= (1/(4i)) {1/(z - 2i) - (1/(-4i)) * Σ(n = 0 to ∞) [(z - 2i)/(4i)]^n}, via geometric series.

So, the principal part is (1/(4i)) * 1/(z - 2i) = (-i/4) * 1/(z - 2i).
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I hope this helps!

principal parts of some laurent expansions?

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