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Unformatted text preview: (1) n ´ 1 z µ n After multiplying both side by 1 /z , we have the desired expression. 1 2 MATH 185 SOLUTIONS OF HW XI (7) (56. 5)(20pts) (a) If  z  < 1, then 1 1z = ∞ X n =0 z n Therefore the Maclaurin series of f ( z ) is z + 1 z1 =z ∞ X n =0 z n∞ X n =0 z n =12 ∞ X n =1 z n (b) If  z  > 1, then  1 /z  < 1 , so we have 1 + 1 /z 11 /z = ∞ X n =0 zn + 1 z ∞ X n =0 zn = 1 + 2 ∞ X n =1 zn (8) (56. 6)(10pts) Since 0 <  z1  < 2, 0 <  1 / ( z1)  < 1, so we have z z3 =z/ 2 1( z1) / 2 =z1 2 ∞ X n =0 ± z1 2 ² n1 2 ∞ X n =0 ± z1 2 ² n Therefore, z ( z1)( z3) =∞ X n =0 ( z1) n 2 n +11 2 ∞ X n =0 ( z1) n1 2 n +1 =3 ∞ X n =0 ( z1) n 2 n +21 2( z1) ....
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 Fall '07
 Lim
 Maclaurin Series, Taylor Series, 10pts, Laurent, 20pts

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