MA 513 HW 6

MA 513 HW 6 - z. For | z | > 1 , 1 z 2 (1 − z ) = −...

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MA 513 Homework 6 Solutions p.188: #4; p.197: #8, 13; p. 205: #2, 3, 4; p. 220: #4. p. 188: #4: For 0 < r < 1 , s k =0 ( re ) k = s k =0 r k (cos + i sin ) = 1 1 re = 1 1 r cos θ ir sin θ = 1 r cos θ ir sin θ (1 r cos θ ) 2 + r 2 sin 2 θ . Now equate real and imaginary parts, use cos 2 θ + sin 2 θ = 1 , and subtract 1 from both sides (of the real part) to get the result. p. 197: #8: cos z = sin( z π 2 ) = ( ( z π 2 ) ( z π 2 ) 3 / 3! + ( z π 2 ) 5 / 5! ..... ) #13: 1 4 z z 2 = 1 4 z (1 z/ 4) = 1 4 z (1 + z/ 4 + ( z/ 4) 2 + ( z/ 4) 3 + ... + ( z/ 4) n + ... ) = 1 4 z + s n =0 z n 4 n +2 . p. 205: #2: e z ( z + 1) 2 = 1 e e ( z +1) ( z + 1) 2 = 1 e 1 ( z + 1) 2 p s n =0 ( z + 1) n n ! P . #3: Obvious. #4: Interval 0 < | z | < 1 : use geometric series in
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Unformatted text preview: z. For | z | > 1 , 1 z 2 (1 − z ) = − 1 z 3 (1 − 1 /z ) = − 1 z 3 ∞ s n =0 1 z n . p. 220: #4: sin z = z − z 3 / 6 + ... is convergent for all z ∈ C . Thus, f ( z ) = sin z/z = 1 − z 2 / 6 + ... is convergent for all z n = 0 . At z = 0 , it converges to 1 = f (0) . Therefore, the series is the Taylor series for f ( z ) , which is consequently entire, since the series converges for all z ....
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This note was uploaded on 06/11/2010 for the course MA 513 taught by Professor Staff during the Spring '08 term at N.C. State.

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