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...

• Notes
• mjg68
• 6

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
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 .

#### You've reached the end of your free preview.

Want to read all 6 pages?

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern