Linear Algebra Solutions 54

Linear Algebra Solutions 54 - Hence Arg z = 5 4 3 6 12 5 12...

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π 4 + π 3 - π 6 5 12 . Hence Arg z = 5 12 and the polar decomposition of z is z = 4 2 µ cos 5 π 12 + i sin 5 π 12 . (ii) Let z = (1+ i ) 5 (1 - i 3) 5 ( 3+ i ) 4 . Then | z | = | (1 + i ) | 5 | (1 - i 3) | 5 | ( 3 + i ) | 4 = ( 2 ) 5 2 5 2 4 = 2 7 / 2 . Arg z Arg(1 + i ) 5 + Arg(1 - 3 i ) 5 - Arg( 3 + i ) 4 (mod 2 π ) 5Arg(1 + i ) + 5Arg(1 - 3 i ) - 4Arg( 3 + i ) 5 π 4 + 5 µ - π 3 - 4 π 6 - 13 π 12 11 π 12 . Hence Arg z = 11 π 12 and the polar decomposition of z is z = 2 7 / 2 µ cos 11 π 12 + i sin 11 π 12 . 7. (i) Let z = 2(cos π 4 + i sin π 4 ) and w = 3(cos π 6 + i sin π 6 ). (Both of these numbers are already in polar form.) (a) zw = 6(cos( π 4 + π 6 ) + i sin( π 4 + π 6 )) = 6(cos 5 π 12 + i sin 5 π 12 ). (b) z w = 2 3 (cos( π 4 - π 6
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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