Laurent Expansion of sin(1-1/z)

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SUMMARY

The discussion focuses on finding the Laurent expansion of the function f(z) = sin(1 - 1/z) about z = 0, specifically addressing the annulus of convergence. A suggested approach involves using the sine sum formula: sin(1 - 1/z) = sin(1)cos(1/z) - cos(1)sin(1/z), followed by applying Taylor series expansions for sine and cosine. The correct expression for sin(1/z) is identified as 1/z - (1/z^3) * 3! + (1/z^5) * 5! - ..., which is essential for constructing the Laurent series.

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Homework Statement


Find the Laurent expansion of f(z) = \sin(1-\frac{1}{z}) about z = 0, and state the annulus of convergence.

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The Attempt at a Solution


I tried doing the regular expansion of sin(z), then applying the binomial expansion on the (1-1/z)^n terms, but I can't help but feel that there's a better way to approach the problem. Any thoughts?
 
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I would suggest that you use the sine sum formula to write it as
sin(1-\frac{1}{z})= sin(1)cos(\frac{1}{z})- cos(1)sin(\frac{1}{z})
and then use the Taylor's series for sine and cosin.
 
Boy don't I feel dumb. Thanks!
 
is this sentence correct?
sin(1/z)=1/z-1/z^3*3!+1/z^5*5!+-...
for laurent series?
 

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