Linear Algebra Solutions 52

Linear Algebra Solutions 52 - i 2-4(4 3 i =-8-6 i Hence by...

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3. 1 + (1 + i ) + ··· + (1 + i ) 99 = (1 + i ) 100 - 1 (1 + i ) - 1 = (1 + i ) 100 - 1 i = - i ' (1 + i ) 100 - 1 . Now (1 + i ) 2 = 2 i . Hence (1 + i ) 100 = (2 i ) 50 = 2 50 i 50 = 2 50 ( - 1) 25 = - 2 50 . Hence - i ' (1 + i ) 100 - 1 = - i ( - 2 50 - 1) = (2 50 + 1) i . 4. (i) Let z 2 = - 8 - 6 i and write z=x+iy, where x and y are real. Then z 2 = x 2 - y 2 + 2 xyi = - 8 - 6 i, so x 2 - y 2 = - 8 and 2 xy = - 6. Hence y = - 3 /x, x 2 - µ - 3 x 2 = - 8 , so x 4 + 8 x 2 - 9 = 0. This is a quadratic in x 2 . Hence x 2 = 1 or - 9 and consequently x 2 = 1. Hence x = 1 , y = - 3 or x = - 1 and y = 3. Hence z = 1 - 3 i or z = - 1 + 3 i . (ii) z 2 - (3+ i ) z +4+3 i = 0 has the solutions z = (3+ i ± d ) / 2, where d is any complex number satisfying d 2 = (3 +
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Unformatted text preview: i ) 2-4(4 + 3 i ) =-8-6 i. Hence by part (i) we can take d = 1-3 i . Consequently z = 3 + i ± (1-3 i ) 2 = 2-i or 1 + 2 i. (i) The number lies in the Frst quadrant of the complex plane. | 4 + i | = p 4 2 + 1 2 = √ 17 . Also Arg(4 + i ) = α , where tan α = 1 / 4 and 0 < α < π/ 2. Hence α = tan-1 (1 / 4).-¾ 6 ? x y 4 + i α » » » » : 55...
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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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