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Laurent Series 
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#1
Oct1510, 02:25 PM

P: 24

1. The problem statement, all variables and given/known data
1) Find the Laurent series for (z^2)*cos(1/3z) in the region [tex]\leftz\right[/tex] 2) Find the Laurent series expansion of (z^2  1)^(2) valid in the following region a) 0 < [tex]\leftz  1\right[/tex] < 2 b) [tex]\leftz + 1\right[/tex] > 2 2. Relevant equations 3. 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*(z1)^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) ((n1)*(1)^n) / (4*z^n)  [tex]\sum[/tex](n=1 to infinity) (1/(4*z^n)) + [tex]\sum[/tex](n=2 to infinity) (n1) / (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. 


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