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    For part d, i can use real = imaginary and solve it equal to -6.
    Attachment 534385534387
    but for part b in this question, basically the arg(z) = 45 degrees, so equal to pie/4. Therefore i used the same technique as in part d, real = imaginary. But i didnt get it right... can someone explain to me, why i cant use real equal imaginary in part b but i can use it in part d?
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    You should be able to. Are you using Z in the form a + bi ?
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    (Original post by AmarPatel98)
    You should be able to. Are you using Z in the form a + bi ?
    yes:
    5p/(9+p^2) = (6-p^2)/(9+p^2)
    p^4+5p^3+3p^2+45p-54=0

    have I done it wrong?
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    (Original post by alesha98)
    yes:
    5p/(9+p^2) = (6-p^2)/(9+p^2)
    p^4+5p^3+3p^2+45p-54=0

    have I done it wrong?
    No, but cross multiplying isn't particularly useful.

    You have two fractions that are equal. They have the same denominator. Therefore they must have the same numertor and:

    5p = 6-p^2

    etc.
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    (Original post by ghostwalker)
    No, but cross multiplying isn't particularly useful.

    You have two fractions that are equal. They have the same denominator. Therefore they must have the same numertor and:

    5p = 6-p^2

    etc.
    I have missed the most basic step ... thankyou so much
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    (Original post by alesha98)
    yes:
    5p/(9+p^2) = (6-p^2)/(9+p^2)
    p^4+5p^3+3p^2+45p-54=0

    have I done it wrong?
    As ghostwalker said, starting with this: 5p/(9+p^2) = (6-p^2)/(9+p^2), you can multiply both sides by (9+p^2) so you get

    5p = 6-p^2, as ghostwalker said.

    Then just solve the quadratic for p.

    Hope this helps
 
 
 
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