Question

Do Task 212

Task 211 (C). Find the Laurent series of exp z exp-, and exp-2 at zo = 0. From the definition of the coefficients for the Laurent series off at zo, we see that a-1 = Res(f, zo). Sometimes it is easier to find the Laurent series than the residue directly Task 212 (C). Using the results of Task 211, find Res (exp 1,0), Res(-exp z,0), and Res(exp "In fact, given a function f(z) that is holomorphic on an annulus 0 s R1 < Iz-zol < R2 oo, a Laurent series for f in this can be obtained from the same formulas as in Definition 9.2.1 by integrating over a circle lz -zol-p, with R1 <p<R2 and the Laurent series will converge to f(z) on this annulus. For example, the series ++3 +.converges only when zl > 1. What function does it converge to? At O, use the geometric series to expand 1/(1-2), At1, write z -1-(1 -2) and use the geometric series to expand 1/z. 46 9.3 Classification of singularities Definition 9.3.1. Let f be holomorphic, and suppose zo is an isolated singularity of f. Let an, t the coefficients of the Laurent series of f at zo, as in Definition 9.2.1. Then zo is called: Z, be . a removable singularity if an -0 for all n <0;

Answer 1

Res (exp 1/2, 0) Because We know that exp alpha = 1 + alpha + alpha^2/angle 2 + alpha^3/angle 3 + alpha^4/angle 4 exp 1/2 = 1 + 1/2 + (1/2)^2 1/angle 2 + (1/2)^3 1/angle 3 + (1/2)^4 1/ angle 4 + ..... exp 1/2 = 1 + 1/2 + 1/angle 2 1/angle 2 + 1/angle 3 1/angle 3 + ...... (exp 1/2: 2 = 0) = coefficient of 1/2 in ex 1/2 Residence exp (1/2), z = 0 = 1 lessthanorequalto Res (Exp 1/2, 0) = 1 Because 2 exp 1/2 = 2 {1 + 1/2 + 1/angle 2 1/angle 3 + 1/angle 3 1/angle 3 b + 1/angle 3 1/angle 4 + .....} 2 exp 1/2 = 2 + 1 + 1/2 1/angle 2 + 1/angle 2 1/angle 3 + 1/angle 3 1/angle 4 + ....... Res (2 exp 1/z 0) = coefficient of 1/2 in 2 exp = 1/angle 2