s1_95-100aSOLUTION

s1_95-100aSOLUTION - ACM95a/100a Last updated October 7,...

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Unformatted text preview: ACM95a/100a Last updated October 7, 2009 Problem Set I Solutions Prepared by Faisal Amlani 1. Show that, for a real number n , each of the two numbers z = 1 ni satisfies the equation z 2 2 z + 1 + n 2 = 0. SOLUTION : For z = 1 + i , z 2 2 z + 1 + n 2 = (1 + ni ) 2 2(1 + ni ) + 1 + n 2 = (1 + 2 ni n 2 ) 2 2 ni + 1 + n 2 = (1 + 2 ni n 2 ) (1 + 2 ni n 2 ) = 0 . For z = 1 i , z 2 2 z + 1 + n 2 = (1 ni ) 2 2(1 ni ) + 1 + n 2 = (1 2 ni n 2 ) 2 + 2 ni + 1 + n 2 = (1 2 ni n 2 ) (1 2 ni n 2 ) = 0 . 2. Write the following complex numbers in the form a + ib (a) bracketleftbigg 2 + i 6 i (1 2 i ) bracketrightbigg 2 (b) (1 i ) 7 SOLUTION : (a) parenleftbigg 2 + i 6 i (1 i 2) parenrightbigg 2 = parenleftbigg 2 + i 1 + 8 i parenrightbigg 2 = 3 + 4 i 63 16 i = 3 + 4 i 63 16 i parenleftbigg 63 + 16 i 63 + 16 i parenrightbigg = 253 4225 i 204 4225 . (b) (1 i ) 7 = ((1 i ) 2 ) 2 (1 i ) 2 (1 i ) = ( 2 i ) 2 ( 2 i )(1 i ) = ( 4)( 2 2 i ) = 8 + 8 i....
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s1_95-100aSOLUTION - ACM95a/100a Last updated October 7,...

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