# 475 answers 5 - ANSWERS PAMPHLET 201 Solutions are A3.8 1,...

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ANSWERS PAMPHLET 201 Solutions are 2 1 , 1 2 1 i p 3 2 , 1 2 2 i p 3 2 . A3.8 e 1 1 i 5 e 1 ? e i 5 e (cos 1 1 i sin 1) . e p i 6 2 5 cos 2 1 i sin 2 5 i. e 2 2 i 5 e 2 ? e 2 i 5 e 2 (cos( 2 ) 1 i sin( 2 )) 52 e 2 . A3.9 e z 1 ? e z 2 5 ˆ 1 1 z 1 1 z 2 1 2! 1 z 3 1 3! 1??? 1 1 z 2 1 z 2 2 2! 1 z 3 3 3! ! 5 1 1 ( z 1 1 z 2 ) 1 ˆ z 2 1 2! 1 z 1 z 2 1 z 2 2 2! ! 1 1 3! ± z 3 1 1 3! 1! 2! z 1 z 2 2 1 3! 2! 1! z 2 1 z 2 1 z 3 2 1 1 4! ± z 4 1 1 4! 3! 1! z 3 1 z 2 1 4! 2! 2! z 2 1 z 2 2 1 4! 1! 3! z 1 z 3 2 1 z 4 1 5 1 1 ( z 1 1 z 2 ) 1 1 2! ( z 1 z 2 ) 2 1 1 3! ( z 1 1 z 2 ) 3 1 1 4! ( z 1 1 z 2 ) 4 5 e z 1 1 z 2 . The coefﬁcient of z j 1 z k 2 in the big expansion above is 1 j ! ? 1 k ! 5 1 ( j 1 k )! ? ( j 1 k )! j ! k ! . But, ( j 1 k )! 6 ( j ! k !) is precisely the coefﬁcient of z j 1 z k 2 in the expansion of ( z 1 1 z 2 ) j 1 k . A3.10 If x n 5 c 1 3 n 1 c 2 ? 1 n ,then x n 1 1 5 3 c 1 3 n 1 1 ? c 2 ? 1 n and x n 1 2 5 9 c 1 3 n 1 1 ? c 2 ? 1 n . 4 x n 1 1 2 3 x n 5 (12 c 1 3 n 1 4 c 2 ? 1 n ) 2 (3 c 1 3 n 1 3 c 2 ? 1 n ) 5 9 c 1 3 n 1 c 2 ? 1 n 5 x n 1 2 . Let m 5 n 1 2 . Since x n 1 2 5 4 x n 1 1 2 3 x n , x m 5 4 x m 2 1 2 3 x m 2 2 .

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202 MATHEMATICS FOR ECONOMISTS A3.11 a )I f x n 5 k 1 (2 1 3 i ) n 1 k 2 (2 2 3 i ) n , x n 1 1 5 (2 1 3 i ) k 1 (2 1 3 i ) n 1 (2 2 3 i ) k 2 (2 2 3 i ) n x n 1 2 5 (2 1 3 i ) 2 k 1 (2 1 3 i ) n 1 (2 2 3 i ) 2 k 2 (2 2 3 i ) n 5 ( 2 5 1 12 i ) k 1 (2 1 3 i ) n 1 ( 2 5 2 12 i ) k 2 (2 2 3 i ) n . 4 x n 1 1 2 13 x n 5 [4(2 1 3 i ) 2 13] k 1 (2 1 3 i ) n 1 [4(2 2 3 i ) 2 13] k 2 (2 2 3 i ) n 5 ( 2 5 1 12 i ) k 1 (2 1 3 i ) n 1 ( 2 5 2 12 i ) k 2 (2 2 3 i ) n 5 x n 1 2 . b, c ) By DeMoivre’s formula, (2 6 3 i ) n 5 (5 n cos n u 0 6 i 5 n sin n 0 ), where tan 0 5 3 6 2. Now, ( c 1 1 ic 2 )(5 n cos n 0 1 i 5 n sin n 0 ) 1 ( c 1 2 ic 2 )(5 n cos n 0 2 i 5 n sin n 0 ) 5 ( c 1 5 n cos n 0 2 c 2 5 n sin n 0 ) 1 i ( c 2 5 n cos n 0 1 c 1 5 n sin n 0 ) 1 ( c 1 5 n cos n 0 2 c 2 5 n sin n 0 ) 2 i ( c 2 5 n cos n 0 1 c 1 5 n sin n 0 ) 5 2( c 1 5 n cos n 0 2 c 2 5 n sin n 0 ) 5 5 n ( C 1 cos n 0 1 C 2 sin n 0 ) , where C 1 5 2 c 1 and C 2 52 2 c 2 .
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