assig12 - = 0 can be solved as follows First divide by a to...

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MATH 135 Algebra, Assignment 12 Not to be handed in 1: Express each of the following complex numbers in cartesian form. (a) 4 e i 5 π/ 3 (b) (1 + i 3) 10 (c) 5 e i θ , where θ = tan - 1 2 2: Express each of the following complex numbers in the polar form re i θ . (a) - 2 + 2 i (b) (1 - i ) 2 (1 + i 3) (c) - 3 - i 3: Solve each of the following for z C . Express your answers in the polar form re i θ . (a) z 3 + 8 i = 0 (b) z = z - i z + 2 + i (c) z 4 = 8 z 4: For each of the following polynomials f ( x ), first solve f ( z ) = 0 for z C , and then factor f ( x ) over the real numbers. (You may find it useful to read section 9.2 in the text book). (a) f ( x ) = x 6 + 7 x 3 - 8 (b) f ( x ) = x 6 + 1 (c) f ( x ) = x 4 + 4 x 3 + 6 x 2 + 4 x + 5 5: An equation of the form ax 3 + bx 2 + cx + d = 0, where a, b, c, d C with a 6
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Unformatted text preview: = 0 can be solved as follows. First, divide by a to obtain an equation of the form x 3 + Bx 2 + Cx + D = 0. Next, make the substitution x = z-B 3 . This will convert the equation to the form z 3 + pz + q = 0. Thirdly, make the substitution z = w-p 3 w . This will convert the equation to the form w 3 + r w-3 + q = 0. Finally, multiply through by w 3 to obtain w 6 + q w 3 + r = 0, which we can solve for w 3 using the Quadratic Formula. (You may find section 9.6 useful). (a) Solve x 3-3 x + 1 = 0 for x ∈ R . (b) Solve x 3 + 3 x 2-3 x-7 = 0 for x ∈ R ....
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This note was uploaded on 07/25/2011 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.

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