Complex Numbers: Real polynomial, complex root — RVHS 2025 H2 Math Prelim Paper 1

Find the roots of \(z^2 - (1+2\mathrm{i})z + 1 + 7\mathrm{i} = 0\).

Here \(a = 1\), \(b = -(1+2\mathrm{i})\), \(c = 1+7\mathrm{i}\). \[\begin{aligned} z &= \frac{(1+2\mathrm{i}) \pm \sqrt{(-1-2\mathrm{i})^2 - 4(1)(1+7\mathrm{i})}}{2(1)}\\[4pt] &= \frac{1+2\mathrm{i} \pm \sqrt{4\mathrm{i}-3-4-28\mathrm{i}}}{2}\\[4pt] &= \frac{1+2\mathrm{i} \pm \sqrt{-7-24\mathrm{i}}}{2} \end{aligned}\]

Find \(\sqrt{-7-24\mathrm{i}}\). Let \(\sqrt{-7-24\mathrm{i}} = x + \mathrm{i}y\), \(x,y\in\mathbb{R}\).

Then \(-7-24\mathrm{i} = (x+\mathrm{i}y)^2 = (x^2-y^2) + (2xy)\mathrm{i}\).

Comparing real and imaginary parts: \[ x^2 - y^2 = -7 \quad\cdots(1) \qquad 2xy = -24 \implies x = \frac{-12}{y} \quad\cdots(2) \]

Substitute \((2)\) into \((1)\): \[ \left(\frac{-12}{y}\right)^2 - y^2 = -7 \implies \frac{144}{y^2} - y^2 = -7 \implies 144 - y^4 = -7y^2 \] \[ y^4 - 7y^2 - 144 = 0 \implies (y^2-16)(y^2+9) = 0 \] \[ y^2 = 16 \quad \text{or} \quad y^2 = -9 \text{ (reject, } y \in \mathbb{R}\text{)} \implies y = \pm 4 \]

When \(y = 4\): \(x = -3\). When \(y = -4\): \(x = 3\).

So the square roots of \(-7-24\mathrm{i}\) are \(3-4\mathrm{i}\) and \(-3+4\mathrm{i}\).

Hence, \[ z = \frac{1+2\mathrm{i} \pm (3-4\mathrm{i})}{2} \] \[ z = \frac{1+2\mathrm{i}+3-4\mathrm{i}}{2} \quad \text{or} \quad z = \frac{1+2\mathrm{i}-3+4\mathrm{i}}{2} \]