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G( z ) = b0 + b1( z − a) + ..... , there. Then H( z ) = F( z )G( z ) is analytic in the same disc. ) = = a0 b0 + (a1 b0 + a 0 b 1 )( z − a) + . . + (ak b0 + . . + a0 bk )( z − a) k + . . Is this the Taylor series of H? We know that in r z − a s < S we have ∞ H( z ) = ∑ ( z − a) kH (k)(a) / k ! k =0 and, by Leibnitz’ rule, k k j =0 j =0 H (k)(a) = ∑ F ( j)(a) G (k − j)(a) k ! (k − j)! = k ! ∑ a j bk − j . So the answer to our question is yes. 6. Find the Taylor series of (cos z ) / (1 − z 2) in t z ✉ < 1.

12. Same again, for z − 8sin( z 3). 13. Find the integral once counter-clockwise around ❾ z ❿ = 4 of z − 6(1 − z ) − 1. 14. Find the Laurent series of 1 / ( z 2 − 4) in (a) ➀ z ➁ < 1 (b) 0 < ➂ z −2 ➃ < 4 (c) 4 < ➄ z + 2 ➅ < ∞ (d) 10 < ➆ z ➇ < ∞. 15. Calculate ∫ ➈ z➉ = 3 e 1 / z z 4 dz , the integral being once counter-clockwise. 1 Singularities We say that the complex-valued function f has an isolated singularity at a if f is not defined at a but there is some s > 0 such that f is analytic in the punctured disc { z ∈ : 0 < ➊ z − a ➋ < s } .

K − j)! = k ! ∑ a j bk − j . So the answer to our question is yes. 6. Find the Taylor series of (cos z ) / (1 − z 2) in t z ✉ < 1. 7. Evaluate ∫ exp(z 2) z − 17dz with the integral once counter-clockwise around ✈ z ✇ = 1. 8. Function of a function. Suppose that f (u) is analytic in ① u − b ② < r and that g( z ) is analytic in ③ z − a ④ < s , with g(a) = b . Then if z is close enough to a we have ⑤ g(z ) − b ⑥ < r, and so f (g( z )) = h( z ) is analytic in ⑦ z − a ⑧ < t , for some t > 0. , ❶ z − a ❷ < s.