- #1

jjangub

- 24

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## Homework Statement

1) Find the Laurent series for (z^2)*cos(1/3z) in the region [tex]\left|z\right|[/tex]

2) Find the Laurent series expansion of (z^2 - 1)^(-2) valid in the following region

a) 0 < [tex]\left|z - 1\right|[/tex] < 2

b) [tex]\left|z + 1\right|[/tex] > 2

## Homework Equations

## The Attempt at a Solution

I did all the work and got the answer, so I basically just want to check that I got it right.

It is pretty hard to write all the steps that I did, so I will briefly write it down.

1) Since z takes infinitely many inverse within the range, we use w = 1/z

then (z^2)*cos(1/3z) becomes cos(w/3)/(w^2).

i) we find the taylor series of cos(w/3),

1 - (w^2)/((3^2)*(2!)) + (w^4)/((3^4)*(4!)) - (w^6)/((3^6)*(6!)) + ...

ii) multiply the series in i) by 1/(w^2),

1/(w^2) - 1/((3^2)*(2!)) + (w^2)/((3^4)*(4!)) - (w^4)/((3^6)*(6!)) + ...

iii) since the series starts at third term, we can just simplify first and second term, and find series,

-1/18 + z^2 + [tex]\sum[/tex]( n=2 to infinity) ((-1)^n)*(w^n)) / ((3^2n)*(2n!))

iiii) now sub back w = 1/z, then the Laurent Series is,

f(z) = -1/18 + z^2 + [tex]\sum[/tex]( n=2 to infinity) ((-1)^n) / ((z^n)*(3^2n)*(2n!))

2)a) since z converges everywhere in this range, the Laurent Series is simply

f(z) = 1 / (4*(z+1)) + 1 / (4*(z+1)^2) - 1 / (4*(z - 1)) + 1 / (4*(z-1)^2)

b) Since z takes infinitely many inverse within the range, I did same way as 1).

let w = 1/z and find the taylor series of each term in a) (4 total) and sub back w = 1/z at the last step. After all this, I got,

f(z) = [tex]\sum[/tex](n=1 to infinity) ((-1)^(n+1)) / (4*z^n) + [tex]\sum[/tex](n=2 to infinity) ((n-1)*(-1)^n) / (4*z^n) - [tex]\sum[/tex](n=1 to infinity) (1/(4*z^n)) + [tex]\sum[/tex](n=2 to infinity) (n-1) / (4*z^n)

Sorry about the mass...I tried to use the code and symbol, but I just failed to do like other people...

Please tell me if I did something wrong.

Thank you.