Unformatted text preview: 2+2 n ) πi . = ⇒ log( i 1 / 2 ) = (1 / 4 + n ) πi . And 1 / 2 log i = 1 / 2(1 / 2 + 2 n ) πi = (1 / 4 + n ) πi . Hence log( i 1 / 2 ) = 1 / 2 log i . (b) log( i 2 ) = log(1) = (1 + 2 n ) πi . But 2 log i = 2(1 / 2 + 2 n ) πi = (1 + 4 n ) πi . Hence log( i 2 ) 6 = 2 log i . (4) (32.2) (a) i i = e iLogi = e i ( πi/ 2) = eπ/ 2 . (b) e 2 (1√ 3 i ) = e · e (2 πi/ 3) . = ⇒ [ e 2 (1√ 3 i )] 3 πi = e 3 πiLog [ e 2 (1√ 3 i )] = e 3 πi (12 πi/ 3) = e 3 πi · e 2 π 2 =e 2 π 2 since e 3 πi =1. (c) (1i ) 4 i = e 4 iLog (1i ) = e 4 i (ln √ 2πi/ 4) = e π +2 ln √ 2 = e π (cos(2 ln 2) + i sin(2 ln 2)) since 4 ln √ 2 = 2 ln 2. Date : today. 1...
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 Fall '07
 Lim
 Math, Open set, main result

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