Complex Numbers: square roots, cubic roots and substitution — EJC 2025 H2 Math Prelim P1

(a) Find \(w\)

Write \(w = a + b\mathrm{i}\), square it, and compare real and imaginary parts with \(2\mathrm{i}\).

\[ \begin{aligned} (a+b\mathrm{i})^2 &= 2\mathrm{i} \\ a^2 - b^2 + 2ab\mathrm{i} &= 2\mathrm{i} \\ a^2 - b^2 &= 0 \quad\text{--- (1)} \\ 2ab &= 2 \quad\text{--- (2)} \end{aligned} \]

From (2), \(a = \dfrac{1}{b}\). Substituting into (1):

\[ \begin{aligned} \left(\frac{1}{b}\right)^2 - b^2 &= 0 \\ 1 - b^4 &= 0 \\ (1+b^2)(1-b^2) &= 0 \\ b^2 &= 1 \quad (\because b \text{ is real}) \end{aligned} \]

So \(b = 1\), \(a = 1\) or \(b = -1\), \(a = -1\) (reject since \(0 \leq \arg(w) \leq \dfrac{\pi}{2}\)).

\[ w = 1 + \mathrm{i} \]

(b) Find the other roots and \(s\)

Factor out the known root \(z-(1+\mathrm{i})\) and compare coefficients to pin down the quadratic factor and \(s\).

\[ z^3 - z^2 + (1-\mathrm{i})z + s = \bigl[z-(1+\mathrm{i})\bigr](z^2+Bz+C) \]

Expanding and comparing coefficients:

\[ \begin{aligned} z^2 &: \quad -1 = -(1+\mathrm{i}) + B &&\implies B = \mathrm{i}\\ z &: \quad 1-\mathrm{i} = -(1+\mathrm{i})\mathrm{i} + C = 1-\mathrm{i}+C &&\implies C = 0\\ \text{const} &: \quad s = -(1+\mathrm{i})C = 0 &&\implies s = 0 \end{aligned} \]

So \(z^2 + \mathrm{i}z = 0\), giving \(z(z+\mathrm{i}) = 0\).

Other roots: \(z = 0\) and \(z = -\mathrm{i}\), with \(s = 0\).

(c) Find the roots in \(v\)

Spot that the new equation is the old one with \(z\) replaced by \(\mathrm{i}v\), so transform each known root.

Replace \(z\) with \(\mathrm{i}v\): \((\mathrm{i}v)^3 - (\mathrm{i}v)^2 + (1-\mathrm{i})(\mathrm{i}v) + s = 0\), which gives

\[ -\mathrm{i}v^3 + v^2 + (1+\mathrm{i})v + s = 0 \]

Hence \(z = \mathrm{i}v \implies v = -\mathrm{i}z\), so:

\[ \begin{aligned} v &= -\mathrm{i}(1+\mathrm{i}) = 1-\mathrm{i}, \\ v &= -\mathrm{i}(0) = 0, \\ v &= -\mathrm{i}(-\mathrm{i}) = -1 \end{aligned} \]