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Math Help - locus of complex no. question no. 2

  1. #1
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    locus of complex no. question no. 2

    p294 q16 b c
    question:
    it is given that the equation
    z^2-2z+k =0 (k is real ) --------- (*)
    has no real roots.
    a) find the range of k
    b) find the quadratic equation whose roots are the cubes of the roots of (*) and show that the discriminant of this equation is 4(1-k)(4-k)^2
    if this equation has real roots , deduce the value of k.
    c) find, in terms of k , the squares of the roots of (*), expressing the answers in the form x+iy where x and y are real.
    as k varies, find the equation of the locus of the points in the Argand plane representing the squares of the roots of (*).

    i have problems in b) and c)
    b) 4(1-k)(4-k)^2 \ge 0
    (k-1)(k-4)^2 \le 0
    k \le 1 or  k = 4
    But the answer says it is k = 4. i don't understand.
    c)i get the two points 2 - k + 2\sqrt{1-k} and 2 - k - 2\sqrt{1-k} ie -2 + 2 \sqrt 3 i and -2 - 2 \sqrt 3 i
    but i don't know how to represent the locus of them as k varies. it seems that i have problems in the idea of locus.
    thanks!
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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    b) 4(1-k)(4-k)^2 \ge 0
    (k-1)(k-4)^2 \le 0
    k \le 1 or  k = 4
    But the answer says it is k = 4. i don't understand.
    Isn't that because (*) is supposed to have no real roots ?
    And you should have found in a) that k has to be superior (never equal) to 1. So that eliminates solution k \le 1.

    But I don't see where your working comes from... Have you dealt with the first part of the question ?
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