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if z^3 + (3+2i)z + (-1+ia) has one real root , then the value of a lies in the interval?

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  • if z^3 + (3+2i)z + (-1+ia) has one real root , then the value of a lies in the interval?


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Answer #1 | 23/12 2013 13:05
Let the real root be b. Then b^3 + (3 + 2i)*b + (-1 + ia) = 0 -----> (b^3 + 3b - 1) + (2b + a)*i = 0 -----> b^3 + 3b - 1 = 0 and 2b + a = 0 -----> a = -2b. Now with f(x) = x^3 + 3x - 1 we have f'(x) = 3*x^2 + 3 = 3*(x^2 + 1) > 0 for all x, so f(x) is always increasing. So since f(x) -> -infinity as x -> -infinity and f(x) -> infinity as x -> infinity, we know there is exactly one real root b of f(x). Next, note that f(0) = -1 and f(1/2) = 5/8, so 0 < b < 1/2, and thus since a = -2b we have -1 < a < 0, i.e., a lies on the interval (0,1).
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