Complex Numbers: roots of a complex polynomial, conjugate-pair argument and substitution — NYJC 2025 H2 Math Prelim P2
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Question
(a) It is given that the roots of the equation \(-\mathrm{i}x^3+5\mathrm{i}x^2+ax+b=0\), where \(a\) and \(b\) are purely imaginary, are \(2+\mathrm{i}\), \(2-\mathrm{i}\) and \(1\). Explain why the complex roots occur in conjugate pairs.
(b) By using (a) and an appropriate substitution, find the roots of the equation \[ \mathrm{i}x^3+5\mathrm{i}x^2+a^*x+b=0, \] where the complex conjugate of \(a\) is denoted by \(a^*\).
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(a) Why the complex roots are conjugates
Write the purely imaginary coefficients as \(\mathrm{i}\) times a real number, then divide through by \(\mathrm{i}\) to expose real coefficients.
Replacing \(a=\lambda\mathrm{i}\) and \(b=\mu\mathrm{i}\) where \(\lambda,\mu\in\mathbb{R}\):
\[ -\mathrm{i}x^3+5\mathrm{i}x^2+\lambda\mathrm{i}x+\mu\mathrm{i}=0 \implies x^3-5x^2-\lambda x-\mu=0. \]Since the coefficients of this equation are all real, the complex roots occur in conjugate pairs.
(b) Roots of the related equation
Because \(a\) is purely imaginary, \(a^*=-a\); the substitution \(x\to -x\) turns the new equation back into the equation from part (a).
The equation is \(\mathrm{i}x^3+5\mathrm{i}x^2+a^*x+b=0\).
Since \(a\) is purely imaginary, \(a^*=-a\). Using the substitution \(x\to -x\):
\[ -\mathrm{i}(-x)^3+5\mathrm{i}(-x)^2+a(-x)+b=0. \]This is exactly the equation in (a), which has roots \(x=2+\mathrm{i},\;2-\mathrm{i},\;1\).
So the substitution gives \(-x = 2+\mathrm{i},\;2-\mathrm{i},\;1\), yielding
\[ x=-2+\mathrm{i},\;-2-\mathrm{i}\ \text{ and }\ -1. \]