soln11 - MATH 135 Algebra, Solutions to Assignment 11 1:...

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Unformatted text preview: MATH 135 Algebra, Solutions to Assignment 11 1: Express each of the following complex numbers in cartesian form. (a) (2 + i )(3 + 2 i )- (5 + 3 i ) Solution: We have (2 + i )(3 + 2 i )- (5 + 3 i ) = (6 + 4 i + 3 i- 2)- (5 + 3 i ) = (4 + 7 i )- (5 + 3 i ) =- 1 + 4 i . (b) 1 ( 2 + i )(1 + i 2) Solution: We have 1 ( 2 + i )(1 + i 2) = 1 2 + 2 i + i- 2 = 1 3 i =- 1 3 i . (c) (1 + 2 i ) 3 (3 + i ) 2 Solution: We have (1 + 2 i ) 3 (3 + i ) 2 = 1 + 6 i + 12 i 2 + 8 i 3 9 + 6 i + i 2 = 1 + 6 i- 12- 8 i 9 + 6 i- 1 =- 11- 2 i 8 + 6 i =- 11- 2 i 2(4 + 3 i ) 4- 3 i 4- 3 i =- 44 + 33 i- 8 i- 6 2(16 + 9) =- 50 + 25 i 50 =- 1 + 1 2 i. 2: Solve each of the following equations for z C . Express your answers in cartesian form. (a) z + 1 z + i = 3 + i Solution: We have z + 1 z + i = 3 + i ( z + 1) = (3 + i )( z + i ) z + 1 = 3 z + 3 i + iz- 1 2- 3 i = (2 + i ) z z = 2- 3 i 2 + i = 2- 3 i 2 + i 2- i 2- i = 1- 8 i 5 z = 1 5- 8 5 i. (b) z 2 = z Solution: Write z = x + iy with x,y R . Then we have z 2 = z ( x + iy ) 2 = x- iy ( x 2- y 2 ) + i (2 xy ) = x- iy ( x 2- y 2 = x and 2 xy =- y ) . To get 2 xy =- y we need y = 0 or 2 x =- 1 2 . When y = 0, the equation x 2- y 2 = x gives x 2 = x so x = 0 or 1. When 2 x =- 1 so x =- 1 2 , the equation x 2- y 2 = x gives 1 4- y 2 =- 1 2 so y 2 = 3 4 , that is y = 3 2 . Thus z = x + iy = 0, 1,- 1 2 + 3 2 i or- 1 2- 3 2 i ....
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soln11 - MATH 135 Algebra, Solutions to Assignment 11 1:...

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