alg-hw-ans-1-4 (2)

alg-hw-ans-1-4 (2) - 01.08 (1 + i)3 = 1 + 3i + 3i2 + i3 = 1...

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01.08 (1 + i ) 3 = 1 + 3 i + 3 i 2 + i 3 = 1 + 3 i - 3 - i = - 2 + 2 i 01.17 z 4 = - 1 = e i π . So z = e i ( π/ 4+2 kπ/ 4) for k = 0 , 1 , 2 , 3. Note that e i θ = cos θ + i sin θ 01.38 e i a e i b = e i ( a + b ) . By Euler’s formula, (cos a + i sin a )(cos b + i sin b ) = cos( a + b ) + i sin( a + b ) That is, (cos a cos b - sin a sin b ) + i (sin a cos b + cos a sin b ) So we get cos( a + b ) = cos a cos b - sin a sin b, sin( a + b ) = sin a cos b +cos a sin b. 02.03 ( b * d ) * c = e * c = a and b * ( d * c ) = b * b = c . So * is not associative. 02.04 The operator * is commutative since the table is symmetric. 02.08 First, a * b = ab + 1 = b * a , So * is commutative. Second, ( a * b ) * c = ( ab + 1) * c = ( ab + 1) c + 1 = abc + c + 1 and a * ( b * c ) = a * ( bc + 1) = a ( bc + 1) + 1 = abc + a + 1. So * is not associative. 02.26 If * is associative and commutative, then ( a * b ) * ( c * d ) = ( c * d ) * ( a * b ) = ( d * c ) * ( a * b ) = [( d * c ) * a ] * b 02.36 Let * be a associative binary operation on S and H = { a S | a * x = x * a for all a S } . We show that H is closed under * . If a, b H , then a * x = x * a and b * x = x * b for all x S . Now ( a * b ) * x = a * ( b * x ) = a * ( x * b ) = ( a * x ) * b = ( x * a ) * b = x * ( a * b ) Hence a * b H . This proves that H is closed under * . 03.03 The map φ ( n ) = 2 n is not an isomorphism from h Z , + i to h Z , + i , since it is not onto. For example, there is no integer n such that φ ( n ) = 1. 03.04 The map φ ( n ) = n + 1 is not an isomorphism from h Z , + i to h Z , + i , since it does not satisfy the homomorphism property: φ ( n + m ) = n + m + 1 but φ ( n ) + φ ( m
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This note was uploaded on 10/12/2010 for the course MATH 5310 taught by Professor Staff during the Spring '08 term at Auburn University.

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alg-hw-ans-1-4 (2) - 01.08 (1 + i)3 = 1 + 3i + 3i2 + i3 = 1...

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