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Unformatted text preview: 2+2 n ) i . = log( i 1 / 2 ) = (1 / 4 + n ) i . And 1 / 2 log i = 1 / 2(1 / 2 + 2 n ) i = (1 / 4 + n ) i . Hence log( i 1 / 2 ) = 1 / 2 log i . (b) log( i 2 ) = log(1) = (1 + 2 n ) i . But 2 log i = 2(1 / 2 + 2 n ) i = (1 + 4 n ) i . Hence log( i 2 ) 6 = 2 log i . (4) (32.2) (a) i i = e iLogi = e i ( i/ 2) = e/ 2 . (b) e 2 (1 3 i ) = e e (2 i/ 3) . = [ e 2 (1 3 i )] 3 i = e 3 iLog [ e 2 (1 3 i )] = e 3 i (12 i/ 3) = e 3 i e 2 2 =e 2 2 since e 3 i =1. (c) (1i ) 4 i = e 4 iLog (1i ) = e 4 i (ln 2i/ 4) = e +2 ln 2 = e (cos(2 ln 2) + i sin(2 ln 2)) since 4 ln 2 = 2 ln 2. Date : today. 1...
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This note was uploaded on 09/01/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at University of California, Berkeley.
 Fall '07
 Lim
 Math

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