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lina-09-10-09-s1

# lina-09-10-09-s1 - Linear Algebra Geometry solutions to...

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Linear Algebra & Geometry: solutions to sheet 2 1. ( a ) 1 1 = 2 u ( π/ 4) ( b ) 0 - 5 = 5 u (3 π/ 2) ( c ) - 3 4 = 5 u ( θ ) , where θ = π/ 2 + arccos(4 / 5) = 2 . 214 . . . . 2. (a) 1 2 1 3 (b) 0 2 (c) 10 - 3 / 2 - 1 / 2 = - 5 3 1 3. Let v 1 , v 2 , · · · , v n R 2 be n arbitrary vectors in R 2 . We apply the triangle inequality n - 1 times k v 1 + v 2 + · · · + v n k ≤ k v 1 k + k v 2 + v 3 + · · · + v n k ≤ k v 1 k + k v 2 k + k v 3 + v 4 + · · · + v n k · · · ≤ k v 1 k + k v 2 k + · · · + k v n k Equality holds for instance if v 1 = v 2 = v 3 = · · · = v n . 4. (a) (1 + i) + (2 - 3i) = 3 - 2i (b) (1 + i)(1 + 2i) - 3 = 1 - 2 + 3i - 3 = - 4 + 3i (c) 1 1 - i = 1+i (1 - i)(1+i) = 1+i 2 = 1 2 + 1 2 i (c) 2+i 1+i = (2+i)(1 - i) (1+i)(1 - i) = 3 - i 2 = 3 2 - 1 2 i (e) (2 - 3i)(3 - 2i) 1+i = - 13i 1+i = - 13i(1 - i) (1+i)(1 - i) = - 13 - 13i 2 = - 13 2 - 13 2 i (f) 1 (1+i)(2 - i) = 1 3+i = 3 - i 9+1 = 3 10 - 1 10 i 5. (i) (a) We have z 1 + z 2 = x 1 + x 2 +i( y 1 + y 2 ) and so z 1 + z 2 = x 1 + x 2 - i( y 1 + y 2 ) = x 1 - i y 1 + x 2 - i y 2 = ¯ z 1 + ¯ z 2 (b) We have ¯ z 1 ¯ z 2 = ( x 1 - i y 1 )( x 2 - i y 2 ) = x 1 x 2 - y 1 y 2 - i( x 1 y 2 + x 2 y 1 ) and z 1 z 2 = x 1 x 2 - y 1 y 2 + i( x 1 y 2 + x 2 y 1 ) = x 1 x 2 - y 1 y 2 - i( x 1 y 2 + x 2 y 1 ) (c) we have z 1 /z 2 = ( z 1 ¯ z 2 ) / ( z 2 ¯ z 2 ) = ( z 1 ¯ z 2 ) / | z 2 | 2 and using (b) and ( d ) we get z 1 /z 2 = ¯ z 1 z 2 / | z 2 | 2 = ¯ z 1 / ¯ z 2 (d) ¯ ¯ z = x - i y = x + i y = z (ii) (e)

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lina-09-10-09-s1 - Linear Algebra Geometry solutions to...

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