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If f(z) has a pole of order mat z = z then 1 dm-3 Res(f(z), zo) = lim m-5 (2 - 2.)m + 1f(z) z - Zodz (m - 1)?! Z 1 dm-1 Res (f(z), zo) = lim (2 - z. )"+(2) f (m-1)! Z- 20 dz m-1 dm R Res(f(z), zo) = = 1 lim (m + 1)?! y (2 - 2.)m+ 24 (2) 1fz - Z - zodz m Res(f(z), zo) = on Lim 1 2-2012 - zobfcz) m( m! Z Save Submit You have used 0 of 1 attempt Multiple Choice 1 point possible (graded) 1 For the function f(z) = (2-1)?(2-3) determine Res(f(z), 1). 2 Res(f(z), 1) 2 3 1 Res(f(z), 1) = 1 4 1 Res(f(z), 1) - 4 5 Res(f(z), 1) = 2 Save
Multiple Choice 1 point possible (graded) Let, f(z) be single-valued and analytic inside and on a simple closed curve C, which have residues given by 0 - 10 6 -10C-1.... Then $ cf(z) dz = 2mila -2 + b = 2nila - 1 + b -1 +0-1+ .....) d b - True False O Maybe / Don t know Save Submit You have used 0 of 1 attempt
Multiple Choice 1 point possible (graded) 1 TT 00 x4 +1 X It's true that so dx = 7. By V2 inspection of the integrand, we see that the corresponding complex valued 1 function flz) = has two poles in 24 +1 IT i the upper half-plane at z = e 4 and Зп і Z2 = e 4 . Now check which of the -2 following is false. 1 1 Res(f(z), 22) = +12zsa Z = 472 4 472 4 O -- Res(f(z), z,) = 1 1 42 42 the poles 21 and z, are of order m = 4. = the integral converges