HW3sol - z = x iy e z = e x(cos y i sin y = e x(cos y-i sin...

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P ROBLEM SET 3 Partial solution Sec 1.4 p. 10 Arg z is not continuous at - 1 . Note that (for example) for ε = 1 for any δ > 0 there exists a point (in fact there exists infinitely many points, but it is enough to find one) w in B ( - 1 , δ ) such that | Arg w - Arg( - 1) | ≥ ε = 1 . In particular w = - 1+ δ 2 i is in B ( - 1 , δ ) since | w - ( - 1) | = δ/ 2 < δ , and | Arg w - Arg( - 1) | = | Arg w - ( - π/ 2) | > π/ 2 > 1 since Arg w > 0 . Sec 1.4 p. 14 Note that (for z 6 = - 2 i ) f ( z ) = z 3 - ( - 2 i ) 3 z - ( - 2 i ) = ( z - ( - 2 i ))( z 2 - 2 iz - 4) z - ( - 2 i )) = z 2 - 2 iz - 4 Hence lim z →- 2 i = ( - 2 i ) 2 - 2 i ( - 2 i ) - 4 = - 12 . Sec 1.5 p. 15a Let
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Unformatted text preview: z = x + iy e z = e x (cos y + i sin y ) = e x (cos y-i sin y ) = e x (cos(-y ) + i sin(-y )) = e z 4. Let z ∈ B ( 1 2 , 1 2 ) . Then | z-1 2 | < 1 / 2 . That is ( z-1 2 )( z-1 2 ) < 1 4 . Multiplying out yields: z z-1 2 ( z + z ) < . In other words z z < Re z . Therefore Re ± 1 z ² = Re ± z z z ² = Re z z z > 1 ....
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