complex_tut_sol1

complex_tut_sol1 - Complex Numbers Complex Numbers tutorial...

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Complex Numbers Complex Numbers tutorial solutions 1. a) ( 29 ( 29 ( 29 ( 29 4 j7 2 j3 8 j14 j12 21 4 j7 2 j3 29 j2 - + = - + + - + = - b) ( 29 ( 29 ( 29 ( 29 4 j3 2 j 4 j3 2 j 2 j 2 j 4 j3 8 j6 j4 3 2 j 4 1 4 j3 5 j10 1 j2 2 j 5 + + + = - - + + + + - = - + + + = = + - 2. 2 2 2 2 Let z x jy z x jy zz x y and z z j2y zz 3(z z) x y j6y 13 j12 = + ∴ = - = + - = + - = + + = + Equating real and imaginary parts 2 2 2 2 Real gives x y 13 and imaginarygives 6y 12 y 2 and x 4 13 x 9 x 3 z 3 j2 or z 3 j2 + = = ∴ = + = = = ± ∴ = + = - + 3. r 10 and 60 x r cos and y r sin x 10cos60 y 10sin 60 3 x 5 y 10 2 y 5 3 = θ = ° = θ = θ = ° = ° = = = 1
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Complex Numbers 4. If z = 2 – j2 is a root then z 2 j2 = + is a root and (z – (2 – j2))(z – (2 + j2)) is a factor (z – (2 – j2))(z – (2 + j2)) = z 2 – z(2 – j2) – z(2 + j2) + (2 – j2) (2 + j2) = z 2 – z(2 – j2 + 2 + j2) + 4 + 4 = z 2 – 4z + 8 Therefore z 2 – 4z + 8 is a factor 2 3 2 3 2 2 2 2z 1 z 4z 8 2z 9z 20z 8 2z 8z 16z z
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complex_tut_sol1 - Complex Numbers Complex Numbers tutorial...

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