Complex Numbers: real-coefficient cubic roots & Argand square — JPJC 2025 H2 Math Prelim P2

(a) Other roots, and the values of \(p\) and \(q\)

Real coefficients force the conjugate \(3+\mathrm{i}\) to also be a root, giving a known quadratic factor; the remaining linear factor fixes \(p\) and \(q\).

Since the polynomial has real coefficients, if \(3-\mathrm{i}\) is a root then so is \(3+\mathrm{i}\).

\[ \begin{aligned} [w-(3-\mathrm{i})][w-(3+\mathrm{i})] &= [(w-3)+\mathrm{i}][(w-3)-\mathrm{i}] \\ &= (w-3)^{2} - \mathrm{i}^{2} \\ &= w^{2} - 6w + 10. \end{aligned} \]

Write \(5w^{3}+pw^{2}+68w+q = (w^{2}-6w+10)(5w+a)\). Expanding:

\[ (w^{2}-6w+10)(5w+a) = 5w^{3} + aw^{2} - 30w^{2} - 6aw + 50w + 10a. \]

Comparing coefficients:

\[ \begin{aligned} w^{2}: \quad a - 30 &= p \\ w: \quad -6a + 50 &= 68 \\ \text{const}: \quad 10a &= q \end{aligned} \]

From the \(w\) equation: \(a = -3\). Then \(p = -3 - 30 = -33\) and \(q = 10(-3) = -30\).

So \(5w^{3}-33w^{2}+68w-30 = (w^{2}-6w+10)(5w-3) = 0\), and the other roots are \(3+\mathrm{i}\) and \(\dfrac{3}{5}\).

With \(w = 3-\mathrm{i}\), \(w^{2} = 8-6\mathrm{i}\), \(w^{3} = 18-26\mathrm{i}\). Substituting into \(5w^{3}+pw^{2}+68w+q=0\):

\[ 5(18-26\mathrm{i}) + p(8-6\mathrm{i}) + 68(3-\mathrm{i}) + q = 0, \]

i.e. \(90 - 130\mathrm{i} + 8p - 6p\mathrm{i} + 204 - 68\mathrm{i} + q = 0\). Comparing real and imaginary parts:

\[ \begin{aligned} \text{Im}: \quad -130 - 6p - 68 &= 0 \;\Rightarrow\; p = -33, \\ \text{Re}: \quad 90 + 8(-33) + 204 + q &= 0 \;\Rightarrow\; q = -30. \end{aligned} \]

Factorising as in Method 1 gives the other roots \(3+\mathrm{i}\) and \(\tfrac{3}{5}\).

(b) Argand diagram and the relationship between \(Z_1\), \(Z_2\), \(Z_3\)

Adding \(z_1\) and \(z_2\) places \(Z_3\) at the opposite vertex of the parallelogram with \(O\), \(Z_1\), \(Z_2\) (the parallelogram law).

\(z_{3}=z_{1}+z_{2}=(-1+2\mathrm{i})+(2+\mathrm{i})=1+3\mathrm{i}\). Plotting \(O(0,0)\), \(Z_{1}(-1,2)\), \(Z_{2}(2,1)\) and \(Z_{3}(1,3)\), the point \(Z_3\) is the fourth vertex of the parallelogram \(OZ_2Z_3Z_1\): \(\overrightarrow{OZ_3}=\overrightarrow{OZ_1}+\overrightarrow{OZ_2}\).

(b)(i) Evaluate \(\dfrac{z_1}{z_2}\)

Multiply numerator and denominator by the conjugate of \(z_2\) to clear the imaginary part from the denominator.

\[ \begin{aligned} \frac{z_{1}}{z_{2}} &= \frac{-1+2\mathrm{i}}{2+\mathrm{i}} = \frac{(-1+2\mathrm{i})(2-\mathrm{i})}{(2+\mathrm{i})(2-\mathrm{i})} \\ &= \frac{-2+\mathrm{i}+4\mathrm{i}-2\mathrm{i}^{2}}{4-\mathrm{i}^{2}} = \frac{5\mathrm{i}}{5} = \mathrm{i}. \end{aligned} \]

So \(\dfrac{z_1}{z_2}=\mathrm{i}\), i.e. \(k=1\).

(b)(ii) Show \(OZ_2Z_3Z_1\) is a square

From part (i), \(z_1=\mathrm{i}\,z_2\): multiplying by \(\mathrm{i}\) rotates by \(90^\circ\) and preserves length, so the two sides at \(O\) are equal and perpendicular — a parallelogram with these properties is a square.

From (i), \(z_{1}=\mathrm{i}\,z_{2}\). Hence \(\overrightarrow{OZ_1}\) is \(\overrightarrow{OZ_2}\) rotated through \(90^\circ\) anticlockwise, so

\[ |OZ_1| = |z_1| = |\mathrm{i}\,z_2| = |z_2| = |OZ_2|, \qquad \angle Z_1 O Z_2 = 90^\circ. \]

Since \(z_{3}=z_{1}+z_{2}\), the quadrilateral \(OZ_2Z_3Z_1\) is a parallelogram (opposite sides \(\overrightarrow{OZ_2}=\overrightarrow{Z_1Z_3}\) and \(\overrightarrow{OZ_1}=\overrightarrow{Z_2Z_3}\)). A parallelogram with two adjacent sides equal in length and meeting at a right angle is a square. Therefore \(OZ_2Z_3Z_1\) is a square.