Bak and Newan, Complex Analysis.pdf

Can be seen to be the left half of the annulus

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can be seen to be the left half of the annulus centered at 1 w i th i nner rad i us 1 / e and outer rad i us e . ] Thus, Res 1 ( 1 e z ) n ; 0 = Res 1 (w n ( 1 w) ; 0 = 1 because 1 (w n ( 1 w) = 1 w n ( 1 + w + w 2 + · · · ) has Res = 1 at w = 0 for all n . 6 . f ( z + h ) f ( z ) h = γ ϕ(w) 1 w ( z + h ) 1 w z h d w . Because z γ , we can take l i m h 0 i ns i de the i ntegral and l i m h 0 f ( z + h ) f ( z ) h = γ ϕ(w) (w z ) 2 d w . In part i cular, be- cause f ( z ) = 1 2 π i C f (w) w z d w where C i s a regular curve surround i ng z , i t follows that f ( z ) = 1 2 π i C f (w) (w z ) 2 d w and cont i nu i ng i nduct i vely we can prove Theorem 10 . 11 .
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304 Answers R f f (– R ) f ( R ) R 7 . Cons i der the i mage of the c i rcle | z | = R under the mapp i ng w = f ( z ) . Because f ( z ) i s real i f, and only i f, z i s real, f maps the ent i re upper sem i c i rcle | z | = R ; y > 0 i nto e i ther the upper half-plane or the lower half-plane, and, l i kew i se, f maps the ent i re lower sem i c i rcle | z | = R , y < 0 i nto e i ther the upper or the lower half-plane . Because Arg w i s at most π i n e i ther the upper or lower half-plane, i t follows that Arg f ( z ) , as z traverses | z | = R , i s, at most, 2 π and Z ( f ) i n | z | ≤ R , = 1 2 π Arg f ( z ) 1 . 9 . a . 0 because | 3 e z | ≥ 3 e > | z | on | z | = 1 b . 1 because | z | > 1 3 e z c . On | z | = 2 , | z 4 | > | 5 z 1 | . On | z | = 1 , | 5 z | > | z 4 + 1 | . Hence, there are 3 zeroes i n the annulus . d . Note that, on | z | = 1 , | 5 z 4 | = 5 ≥ | z 6 + 3 z 2 1 | w i th equal i ty poss i ble only at z = ± i . Because z 6 5 z 4 + 3 z 2 1 = 0 at z = ± i , i t follows that there are 4 zeroes i n | z | ≤ 1 . 11 . Res z m f ( z ) f ( z ) ; z k = p · z m k where p i s the order of the zero at z k . 12 . Note that 1 + z + z 2 2 ! + · · · z n n ! e z wh i ch has no zeroes anywhere . Because the convergence i s un i form i n | z | ≤ R , the result follows . 14 . Use the fact that | a n z n | > | a n 1 z n 1 + · · · + a 0 | on the c i rcle | z | = R for suffic i ently large R . 15 . To show that J (λ) i s defined and cont i nuous, note that | f | > | g | throughout γ i mpl i es that f + λ g i s nonzero throughout γ for all λ : 0 λ 1 . 16 . Let f ( z ) = z 2 1 = exp 1 2 z 2 2 ζ d ζ ζ 2 1 . W i thout loss of general i ty, we can assume that the path of i ntegrat i on i s i n the upper half-plane, i f Im z > 0, and i n the lower half-plane, i f Im z < 0 . To show that l i m z x f ( z ) ex i sts for −∞ < x < 1, we must establ i sh that the same l i m i t ex i sts as we approach x through the upper half-plane or the lower half-plane . The d i fference between the l i m i ts equals C 2 ζ ζ 2 1 d ζ where C i s a regular closed curve surround i ng ζ = ± 1 . By the argument pr i nc i ple, C 2 ζ ζ 2 1 d ζ = 2 π i Res 2 ζ ζ 2 1 ; ± 1 = 4 π i Hence, l i m z x z 2 1 ex i sts because e w/ 2 = e (w + 4 π i )/ 2 . By Theorem 7 . 7, then f i s analyt i c i n the plane m i nus [ 1 , 1] .
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Answers 305 17 . Define f ( z ) = 3 ( z 1 )( z 2 )( z 3 ) as exp 1 3 log [ ( z 1 )( z 2 )( z 3 ) ] where log [ ( z 1 )( z 2 )( z 3 ) ] = z 4 [ ( z 1 )( z 2 )( z 3 ) ] ( z 1 )( z 2 )( z 3 ) dz + log 6 for z i n the plane m i nus the i nterval ( −∞ , 3] . Show then that f ( z ) defines a funct i
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  • Spring '17
  • JANE SMITH
  • Math

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