4.1.17 Problem (*). Throughout, suppose that zo E C with Re(zo) < 0 and Im(zo) > 0. In particular, zo (-∞0, 0]. (i) First suppose that Re(zo) <0. Show that | Im(zo)| < |zo|, and so the largest ball on which the Taylor series for Log centered at zo converges is not wholly contained in the domain of Log. = (ii) Put o Arg(zo); since Re(zo) < 0 and Im(zo) > 0, we have π/2 < 0o < . Show that 00π<0< < (00-π) + 2π, so [0, π] (00-π, (00-π) + 2π]. (iii) Let U = {z € C | Im(z) > 0}, so U is the upper half-plane. With 00 as in the previous step, explain why Arg(2) = argo-(2) for all z EU and therefore Logu = logo-u (iv) With zo and o as above, conclude that the Taylor series for Log centered at zo converges to S(z) := Log(zo) + (-1)(z-zo) = kzb k=1 Log(2), z = B(zo; |zo) nu Log(z) + 2πi, z € B (20;|20|) \ U. [Hint: argue first that the Taylor series for Log and logo- centered at zo have the same coefficients. Then, by considering the branch cut for logo- explain why logo0o- is analytic on B(zo; zo). Conclude that the Taylor series for logo-converges to S on B(zo; |zo). What does this imply about S(z) for z E B(zo; |zo) nu? To determine the value of S(z) for z = B(zoizol) \U, use the fact that arge-(z) = Arg(z) +27 if Re(z) <0 and Im(z) < 0.] (v) Now suppose that Re(zo) 20. Show that B(zo; |zol) n(-∞, 0] =Ø. [Hint: show that if x > 0, then |-x-zol> |zol.] Conclude that in this case, the largest ball on which the Taylor series for Log centered at zo converges is wholly contained in the domain of Log, and so this situation is less exciting than the above. (Are you indeed less excited now?)