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MA232 Mid_Term Review III

# MA232 Mid_Term Review III - =& 3(13 2(13 =& 39 26...

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1. Find the roots to: z 3 = 8 + 8 i Solution: j 8 + 8 i j = p 8 2 + 8 2 = 8 p 2 = r Both real and imaginary parts are positive so: rcos° = Re ( z 0 ) where z 0 = 8 + 8 i 8 p 2 cos° = 8 cos ° = 1 = p 2 ° = ±= 4 So: z 0 = 8 e ° 4 i = 8 + 8 i ; z 3 = 8 e ° 4 i Roots: z k = 8 1 3 e i ( ° 4 + k 2 ± ) 1 3 for k = 0 ; 1 ; 2 w 0 = 2 e i 3 ( ° 4 +(0)2 ± ) = 2 e i 12 w 1 = 2 e i 3 ( ° 4 +(1)2 ± ) = 2 e 3 4 i w 2 = 2 e i 3 ( ° 4 +(2)2 ± ) = 2 e 17 12 i 2. Put in complex form: 3 e ° 6 i +4 Solution: 3 e ° 6 i +4 = 3 e 4 e ° 6 i = 3 e 4 ( cos ± 6 + isin ± 6 ) = 3 e 4 ( p 3 = 2 + i= 2) = 3 e 4 ( p 3+ i 2 ) 3. Find: det 0 B B @ 1 2 3 4 3 3 2 1 2 3 1 2 3 4 2 1 1 C C A Solution 1

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det 0 B B B @ 1 2 3 4 3 3 2 1 2 3 1 2 3 4 2 1 1 C C C A = det 0 B B B @ 1 2 3 4 0 ° 3 ° 7 ° 11 0 ° 1 ° 5 ° 6 0 ° 2 ° 7 ° 11 1 C C C A = det 0 @ ° 3 ° 7 ° 11 ° 1 ° 5 ° 6 ° 2 ° 7 ° 11 1 A = ° 3(55 ° 42) + 1(77 ° 77) ° 2(42
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Unformatted text preview: = & 3(13) + 2(13) = & 39 + 26 = & 13 4. Find: @ 2 1 3 4 2 1 3 1 2 1 A @ 1 2 3 2 3 1 3 2 1 1 A Answer: @ 13 13 10 11 16 15 11 13 12 1 A 5. Solve for x,y,z: 1 x + & 3 y + 2 z = 6 1 x + 4 y + & 1 z = 4 5 x + 6 y + 1 z = 20 Answer: This is inconsistent and therefore has no solu-tion. 2...
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MA232 Mid_Term Review III - =& 3(13 2(13 =& 39 26...

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