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IB HL Math Homework #3 Solutions Assigned: 9/10/07, Monday Due: 9/13/07, Thursday 1) Find the value of (1 3i ) 7 using DeMoivre's Theorem. (1 3i ) 7 (2cis ( 4 7 28 4 )) 2 7 cis ( ) 128(cis 3 3 3 ) 64 64 3i 2) Find the four solutions to the following equation: z 4 128 128 3i . z 4 128 128 3i 2 z 4 256cis ( 2k ) 3 2 z (256cis ( 2k ))1 / 4 3 2 2k z 2561 / 4 cis ( ), k 0,1,2,3 12 4 k z 4cis ( ), k 0,1,2,3 6 2 2 7 5 z 4cis ( ) , or , z 4cis ( ) , or z 4cis ( ) , or z 4cis ( ) 6 3 6 3 z 2 3 2i , or z 2 2 3i , or z 2 3 2i , or z 2 2 3i 3) Let the roots of the quadratic equation x2 – 5x + 3 = 0 be r and s. Find the quadratic equation with a leading coefficient of 1 with the roots 2r and 2s. We know that r + s = 5 and rs = 3. It follows that 2r + 2s = 2(5) = 10 and (2r)(2s) = 4(3) = 12. Thus, the desired quadratic is x2 – 10x + 12 = 0. 4) Let the roots of the quadratic equation x2 – 5x + 3 = 0 be r and s. Find the quadratic equation with a leading coefficient of 1 with the roots r2 and s2. r2+s2 = (r+s)2 – 2rs = 52 – 2(3) = 19. r2s2 = (rs)2 = 32 = 9 Thus, the desired quadratic equation is x2 – 19x + 9 = 0. 5) Let the roots of the quadratic equation x2 – 5x + 3 = 0 be r and s. Find the quadratic equation with a leading coefficient of 1 with the roots r3 and s3. r3+s3 = (r+s)3 – 3r2s – 3rs2 = (r+s)3 – 3rs(r+s) = 53 – 3(3)5 = 125 – 45 = 80. r3s3 = (rs)3 = 33 = 27, so the desired quadratic equation is x2 – 80x + 27 = 0. 6) Let the roots of the quadratic equation x2 – 5x + 3 = 0 be r and s. Find the quadratic equation with a leading coefficient of 1 with the roots 1/r and 1/s. 1 1 sr 5 r s rs 3 7) The cubic equation x3 – 7x2 + 17x – 15 = 0 has the root 2 + i. What are the other two roots? Another root must be 2 – i. These sum to 4. The sum of the roots is 7, thus the last root is 3. The three roots are 2 + i, 2 – i, and 3.