Answer Happy

Consider the following complex-variable function: F(z) = (z + 2y) ezt z²+2yz+w²¹ where , wo are real and positive. For convenience, use w²: = =w₁² - 2². a) Find its singularities and compute the residues. Hint: there are three cases: w² > 0, w² = 0 and w² < 0. [8] b) Consider the case w²> 0. Using the residues calculated above, evaluate I = 1 (z + 2y) ezt 2πi 2² + 2yz + wo $₂ zdz, where y is composed of a straight line from -iR to iR, a semicircular arc on the left half-plane, and R is large enough so all singularities are inside 7. [7] c) By invoking Jordan's lemma, justify why, in the limit R→ ∞, the integral along the semicircular arc vanishes, so that r+ix 1 (z + 2y) ezt f(t) = 27 i 1100 e-t dz = 2² + 2yz + w₁2dz = [cos(wt) + [cos(wt) + ² sin(wt)] e−7. [5] -i∞ d) Now consider the case w=0. Show that r+ixo 1 (z + 2y) ezt f(t) = [5] 2πi 2² + 2yz +w²dz = (1 + yt)e¯vt. = Sie