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Find the principal part of 1/sin(z) at z0=0
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[QUOTE="jackmell, post: 4538666, member: 246161"] Try to learn how to use long division of polynomials. It comes up not too infrequently in Complex Analysis. You know: [tex]\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots[/tex] Ok, now just divide that into 1. I know that sounds simple and unfortunately we can's display what the long-division process looks like here. I'll start it for you: z goes into 1 a total of 1/z right? Then 1/z times that expression is going to be 1-z^2/6+z^4/120+. . . Now subtract that from I dont' remember what they call the various pieces of long division but you'll get x^2/2-x^4/120 and so on. Now do it again. Eventually you'll run out of principal terms and start getting Taylor terms. Well, that's your principal part. Also, Mathematica does give the power series for Csc(z). Edit: our subset of Latex does not support multicolumn which is needed to format nicely polynomial division. However, if you're interested, here's a link to see what it look's like if you already don't know. This one is similar to this problem: [url]http://tex.stackexchange.com/questions/123731/replicate-a-typesetting-of-polynomial-long-division?rq=1[/url] [/QUOTE]
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Find the principal part of 1/sin(z) at z0=0
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