Complex Numbers: solving a quadratic with complex coefficients & finding complex square roots — RVHS 2025 H2 Math Prelim P1

Apply the quadratic formula

Treat the equation as a standard quadratic in \(z\) with complex coefficients \(a=1\), \(b=-(1+2\mathrm{i})\), \(c=1+7\mathrm{i}\).

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

Find \(\sqrt{-7-24\mathrm{i}}\)

Let the square root equal \(x+\mathrm{i}y\), square it, and compare real and imaginary parts.

Let \(\sqrt{-7-24\mathrm{i}} = x + \mathrm{i}y\), with \(x,y\in\mathbb{R}\). 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 \(-3+4\mathrm{i}\).

Combine to get the roots

Substitute the square root back into the quadratic formula and simplify each sign.

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