Complex Numbers: conjugate-root pairs, symmetry of polynomial roots and reciprocal-root substitution — HCI 2025 H2 Math Prelim P1

(a) No real roots when \(k\) is purely imaginary

Argue by contradiction: a real root would force \(k\) to be real.

Assume \(\mathrm{f}(x) = 0\) has a real root \(x \in \mathbb{R}\). Then \(x^4 - 6x^2 + k = 0\), i.e. \(k = -x^4 + 6x^2\). The right-hand side is real, but \(k\) is purely imaginary (and non-zero), so \(k \notin \mathbb{R}\) — a contradiction. Hence \(\mathrm{f}(z) = 0\) cannot have real roots when \(k\) is purely imaginary.

(b) Show \(\mathrm{f}(-z) = \mathrm{f}(z)\)

Substitute \(-z\) and use the even powers.

\[ \begin{aligned} \mathrm{f}(-z) &= (-z)^4 - 6(-z)^2 + k\\ &= z^4 - 6z^2 + k\\ &= \mathrm{f}(z). \quad\textbf{(Shown)} \end{aligned} \]

(c) Determine \(k\) and the remaining roots

Substitute \(z = 2+\mathrm{i}\) to solve for \(k\), then use the conjugate and even-symmetry properties.

\[ \begin{aligned} (2+\mathrm{i})^2 &= 2^2 + 2(2)(\mathrm{i}) + \mathrm{i}^2 = 3 + 4\mathrm{i}\\ (2+\mathrm{i})^4 &= \left[(2+\mathrm{i})^2\right]^2\\ &= (3+4\mathrm{i})^2\\ &= 3^2 + 2(3)(4\mathrm{i}) + (4\mathrm{i})^2\\ &= -7 + 24\mathrm{i} \end{aligned} \] \[ \begin{aligned} (2+\mathrm{i})^4 - 6(2+\mathrm{i})^2 + k &= 0\\ -7 + 24\mathrm{i} - 6(3+4\mathrm{i}) + k &= 0\\ k &= 25 \end{aligned} \]

Since all coefficients of \(\mathrm{f}(z) = 0\) are real and \(2+\mathrm{i}\) is a root, then \(2-\mathrm{i}\) is also a root.

Since \(\mathrm{f}(-z) = \mathrm{f}(z)\), then \(-z\) is also a root of \(\mathrm{f}(z) = 0\).

The remaining roots are \(2-\mathrm{i}\), \(-2-\mathrm{i}\) and \(-2+\mathrm{i}\).

(d) Product of all the roots \(D\)

Pair each root with its conjugate so each product is a real modulus-squared.

\[ \begin{aligned} (2+\mathrm{i})(2-\mathrm{i}) &= |2+\mathrm{i}|^2 = 5\\ (-2+\mathrm{i})(-2-\mathrm{i}) &= |-2+\mathrm{i}|^2 = 5\\ \therefore D &= 5(5) = 25 \end{aligned} \] \[ \begin{aligned} (2+\mathrm{i})(2-\mathrm{i}) &= 2^2 - \mathrm{i}^2 = 5\\ (-2+\mathrm{i})(-2-\mathrm{i}) &= (-2)^2 - \mathrm{i}^2 = 5\\ \therefore D &= 5(5) = 25 \end{aligned} \]

(e) Find \(w_1\)

Divide \(\mathrm{f}(z)=0\) by \(z^4\) to see the new equation is the same one with \(w=\tfrac{1}{z}\).

\(z^4 - 6z^2 + k = 0\), so \(\dfrac{k}{z^4} - \dfrac{6}{z^2} + 1 = 0\).

Replace \(\dfrac{1}{z}\) with \(w\), i.e. \(w = \dfrac{1}{z}\):

\[ \begin{aligned} w_1 &= \frac{1}{2+\mathrm{i}}\\ &= \frac{2-\mathrm{i}}{2^2-\mathrm{i}^2}\\ &= \frac{2}{5} - \frac{1}{5}\mathrm{i} \end{aligned} \] \[ \begin{aligned} 25w^4 - 6w^2 + 1 &= 0\\ 25^2 w^4 - 6(25)w^2 + 25 &= 0\\ (5w)^4 - 6(5w)^2 + 25 &= 0\\ 5w &= 2 + \mathrm{i}\\ w &= \frac{2}{5} + \frac{1}{5}\mathrm{i} \end{aligned} \]