math185-hw10sol - MATH 185 COMPLEX ANALYSIS FALL 2007\/08 PROBLEM SET 10 SOLUTIONS For a C 0 < r < R we write A(a r R:={z C | r < |z a| < R We also write

math185-hw10sol - MATH 185 COMPLEX ANALYSIS FALL 2007/08...

This preview shows page 1 - 3 out of 6 pages.

MATH 185: COMPLEX ANALYSIS FALL 2007/08 PROBLEM SET 10 SOLUTIONS For a C , 0 < r < R , we write A ( a ; r, R ) := { z C | r < | z - a | < R } . We also write C × = A (0; r, ) and D * ( a, r ) = A ( a ; 0 , r ). The number of zeroes, counted with multiplicity, of f in Ω is denoted Z f (Ω). You may use without proof any results that had been proved in the lectures. 1. Let the Laurent expansion of cot( πz ) on A (0; 1 , 2) be cot( πz ) = X n = -∞ a n z n . Compute a n for n < 0. Solution. Let Γ be the circle ∂D (0 , r ) traversed once counter-clockwise and 1 < r < 2. Note that Ind(Γ; z ) = 1 for all z D (0 , r ), ie. the bounded component of Γ. By the integral formula for Laurent coefficients, a - k = 1 2 πi Z Γ cot( πz ) z - k +1 dz = 1 2 πi Z Γ z k - 1 cot( πz ) dz for k N . For k = 1, z k - 1 cot( πz ) = cot( πz ) = cos( πz ) sin( πz ) has three isolated (non-removable) singularities in the bounded component of Γ, namely, - 1 , 0 , 1. So by the residue theorem 1 , a - 1 = 1 2 πi Z Γ cot( πz ) dz = Res(cot( πz ); - 1) + Res(cot( πz ); 0) + Res(cot( πz ); 1) = cos( πz ) π cos( πz ) z = - 1 + cos( πz ) π cos( πz ) z =0 + cos( πz ) π cos( πz ) z =1 = 3 π . For k 2, lim z 0 z [ z k - 1 cot( πz )] = lim z 0 z sin( πz ) [ z k - 1 cos( πz )] = 1 π lim z 0 πz sin( πz ) × h lim z 0 z k - 1 cos( πz ) i = 0 Date : December 14, 2007 (Version 1.1). 1 We use result that Res( ϕ/ψ ; a ) = ϕ ( a ) 0 ( a ) if ϕ ( a ) 6 = 0, ψ ( a ) = 0 and ψ 0 ( a ) 6 = 0. 1
Image of page 1
In other words, 0 is a removable singularity of z k - 1 cot( πz ) for k 2. Note that the residue about any removable singularity is 0. So by the residue theorem 1 , a - k = 1 2 πi Z Γ z k - 1 cot( πz ) dz = Res( z k - 1 cot( πz ); - 1) + Res( z k - 1 cot( πz ); 0) + Res( z k - 1 cot( πz ); 1) = z k - 1 cos( πz ) d dz sin( πz ) z = - 1 + 0 + z k - 1 cos( πz ) d dz cos( πz ) z =1 = ( - 1) k - 1 π + 1 π = 0 if k is odd, 2 π if k is even, for k 2. 2. (a) For n = 0 , 1 , 2 , . . . , compute 1 2 πi Z Γ n dz z 3 sin z where Γ n is the circle ∂D (0 , r n ) traversed once counter-clockwise and r n = ( n + 1 2 ) π .
Image of page 2
Image of page 3

You've reached the end of your free preview.

Want to read all 6 pages?

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes
A+ icon
Ask Expert Tutors