Complex Numbers: Polynomials (complex coeff.) — SAJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
(a) It is given that \(\mathrm{f}(x) = px^6 + qx^4 + r\), where \(p\), \(q\), and \(r\) are real constants.
- Show that if \(x = \alpha\) is a root of \(\mathrm{f}(x)=0\), then \(x=-\alpha\) is also a root.
- Given now that \(x=5\) and \(x=\beta\) are roots of \(\mathrm{f}(x)=0\), where \(\mathrm{Re}(\beta)\neq 0\) and \(\mathrm{Im}(\beta)\neq 0\), write down all the remaining roots.
(b) The complex number \(z\) satisfies the equation \[ z^3 + (3-a)z^2 - (2+6\mathrm{i})z - 6 = 0, \] where \(a\) is a complex number. It is given that one of the roots is \(1+\mathrm{i}\).
Find \(a\) and the other roots of the equation.
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(a)(i) Since \(x = \alpha\) is a root, \(\mathrm{f}(\alpha) = 0\): \[ p\alpha^6 + q\alpha^4 + r = 0 \] \[ \mathrm{f}(-\alpha) = p(-\alpha)^6 + q(-\alpha)^4 + r = p\alpha^6 + q\alpha^4 + r = 0 \]
(a)(ii) \(x = 5\) and \(x = \beta\) are roots. By part (i), \(x = -5\) and \(x = -\beta\) are also roots. Since the coefficients of \(\mathrm{f}\) are all real, \(x = \beta^*\) and \(x = -\beta^*\) are also roots.
(b)
Method 1
Since \(1+\mathrm{i}\) is a root: \[ z^3 + (3-a)z^2 - (2+6\mathrm{i})z - 6 = (z - 1 - \mathrm{i})(z^2 + Bz + C) \]
Comparing constant terms: \[ -6 = (-1-\mathrm{i})C \implies C = \frac{-6}{-1-\mathrm{i}} = \frac{6}{1+\mathrm{i}} = \frac{6(1-\mathrm{i})}{2} = 3 - 3\mathrm{i} \]
Comparing \(z\) coefficients: \[ -(2+6\mathrm{i}) = B(-1-\mathrm{i}) + C = B(-1-\mathrm{i}) + (3-3\mathrm{i}) \] \[ B(-1-\mathrm{i}) = -(2+6\mathrm{i}) - (3-3\mathrm{i}) = -5 - 3\mathrm{i} \] \[ B = \frac{-5-3\mathrm{i}}{-1-\mathrm{i}} = \frac{(5+3\mathrm{i})(1-\mathrm{i})}{(1+\mathrm{i})(1-\mathrm{i})} = \frac{5-5\mathrm{i}+3\mathrm{i}+3}{2} = \frac{8-2\mathrm{i}}{2} = 4 - \mathrm{i} \]
Comparing \(z^2\) coefficients: \[ 3 - a = -1 - \mathrm{i} + B = -1-\mathrm{i} + 4-\mathrm{i} = 3 - 2\mathrm{i} \implies a = 2\mathrm{i} \]
So: \[ z^3 + (3-2\mathrm{i})z^2 - (2+6\mathrm{i})z - 6 = (z-1-\mathrm{i})(z^2 + (4-\mathrm{i})z + 3-3\mathrm{i}) = 0 \]
Solving \(z^2 + (4-\mathrm{i})z + 3-3\mathrm{i} = 0\): \[ (z+3)(z+1-\mathrm{i}) = 0 \implies z = -3 \quad\text{or}\quad z = -1+\mathrm{i} \]
OR using the quadratic formula: \[ z = \frac{-(4-\mathrm{i})\pm\sqrt{(4-\mathrm{i})^2 - 4(3-3\mathrm{i})}}{2} = \frac{-(4-\mathrm{i})\pm\sqrt{15-8\mathrm{i}-12+12\mathrm{i}}}{2} = \frac{-(4-\mathrm{i})\pm\sqrt{3+4\mathrm{i}}}{2} \] \[ \sqrt{3+4\mathrm{i}} = 2+\mathrm{i} \implies z = \frac{-4+\mathrm{i}\pm(2+\mathrm{i})}{2} \implies z = -3 \quad\text{or}\quad z = -1+\mathrm{i} \]