Complex Numbers: conjugate roots, polynomial factorisation & Argand geometry — CJC 2025 H2 Math Prelim P2
What this question tests
Question
One of the roots of the equation \[ 2z^4 - 14z^3 + 33z^2 - 26z + p = 0, \quad \text{where } p \text{ is a constant,} \] is \(3+\mathrm{i}\).
(a) Based on the above information only, a student claims that the equation has a root \(3-\mathrm{i}\). State, with a reason, why the student's claim may not be true.
(b) Show that \(p=10\).
For the rest of this question, do not use a calculator.
(c) Find the roots of the equation \(2z^4-14z^3+33z^2-26z+10=0\) and mark them clearly on a single labelled Argand diagram.
(d) The points of the Argand diagram in part (c) form the vertices of a quadrilateral. Identify the type of quadrilateral and determine its area.
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(a) Why the conjugate need not be a root
The conjugate root theorem only applies when every coefficient of the polynomial is real.
The student's claim may not be true as \(p\) may not be real.
(b) Showing \(p=10\)
Substitute the known root \(z=3+\mathrm{i}\) into the equation and simplify, using \((3+\mathrm{i})^2 = 8+6\mathrm{i}\).
\[ \begin{aligned} 2z^4 - 14z^3 + 33z^2 - 26z + p &= 0 \\ 2(3+\mathrm{i})^4 - 14(3+\mathrm{i})^3 + 33(3+\mathrm{i})^2 - 26(3+\mathrm{i}) + p &= 0 \\ 2(8+6\mathrm{i})(8+6\mathrm{i}) - 14(3+\mathrm{i})(8+6\mathrm{i}) + 33(8+6\mathrm{i}) - 78 - 26\mathrm{i} + p &= 0 \\ 2(28+96\mathrm{i}) - 14(18+26\mathrm{i}) + 264 + 198\mathrm{i} - 78 - 26\mathrm{i} + p &= 0 \\ 56 + 192\mathrm{i} - 252 - 364\mathrm{i} + 264 + 198\mathrm{i} - 78 - 26\mathrm{i} + p &= 0 \\ -10 + p &= 0 \\ p &= 10 \end{aligned} \](c) Finding all four roots
Now the coefficients are real, so the conjugate \(3-\mathrm{i}\) is also a root; their product gives a real quadratic factor to divide out.
Since all coefficients are real and \(3+\mathrm{i}\) is a root, \(3-\mathrm{i}\) is also a root.
\[ [z-(3+\mathrm{i})][z-(3-\mathrm{i})] = (z-3)^2+1 = z^2-6z+10 \]Dividing \(2z^4-14z^3+33z^2-26z+10\) by \(z^2-6z+10\):
\[ 2z^4-14z^3+33z^2-26z+10 = (z^2-6z+10)(2z^2-2z+1) \]Solving \(2z^2-2z+1=0\):
\[ z = \dfrac{2\pm\sqrt{4-8}}{4} = \dfrac{2\pm\sqrt{-4}}{4} = \dfrac{2\pm 2\mathrm{i}}{4} = \dfrac{1}{2}\pm\dfrac{1}{2}\mathrm{i} \]The four roots are: \(3+\mathrm{i}\), \(3-\mathrm{i}\), \(\tfrac{1}{2}+\tfrac{1}{2}\mathrm{i}\), \(\tfrac{1}{2}-\tfrac{1}{2}\mathrm{i}\). On the Argand diagram, \(3\pm\mathrm{i}\) sit at \((3,\pm1)\) and \(\tfrac{1}{2}\pm\tfrac{1}{2}\mathrm{i}\) sit at \((0.5,\pm0.5)\), all symmetric about the real axis.
(d) Type of quadrilateral and area
The two vertical chords are parallel but of different lengths, so the figure is a trapezium; apply the trapezium area formula.
The quadrilateral is a trapezium. The vertical side joining \(3+\mathrm{i}\) and \(3-\mathrm{i}\) has length \(2\); the vertical side joining \(\tfrac{1}{2}+\tfrac{1}{2}\mathrm{i}\) and \(\tfrac{1}{2}-\tfrac{1}{2}\mathrm{i}\) has length \(1\). Both are vertical (parallel), and the horizontal distance between them is \(3-\tfrac{1}{2}=2.5\).
\[ \text{Area} = \tfrac{1}{2}(1+2)(2.5) = 3.75 \text{ units}^2 \quad\left(=\tfrac{15}{4}\text{ units}^2\right) \]