Complex numbers: VJC 2023 Promo Q7

Since all the coefficients are real, $1-\sqrt2\mathrm{i}$ is also a root. $$\begin{aligned} \text{Quadratic factor}&=[x-(1-\sqrt2\mathrm{i})][x-(1+\sqrt2\mathrm{i})]\\ &=(x-1)^2-2(-1)\\&=x^2-2x+3. \end{aligned}$$ Let $3x^3+px^2+qx+3=(x^2-2x+3)(ax+b)$.

By inspection or by comparing coefficient of $x^3$ and constant term, $a=3$ and $b=1$. $$3x^3+px^2+qx+3=(x^2-2x+3)(3x+1)$$ By comparing coefficients, $$p=1-6=-5\qquad q=-2+9=7$$ $$3x^3+px^2+qx+3=(x^2-2x+3)(3x+1)=0$$ Other than $1+\sqrt2\mathrm{i}$, the other roots are $1-\sqrt2\mathrm{i}$ and $-\dfrac13$.

Alternative $$\begin{aligned} 3(1+\sqrt2\mathrm{i})^3+p(1+\sqrt2\mathrm{i})^2+q(1+\sqrt2\mathrm{i})+3&=0\\ 3(1+3\sqrt2\mathrm{i}-6-2\sqrt2\mathrm{i})+p(1+2\sqrt2\mathrm{i}-2)+q(1+\sqrt2\mathrm{i})+3&=0\\ (-12-p+q)+(3+2p+q)\sqrt2\mathrm{i}&=0 \end{aligned}$$ Comparing real and imaginary parts, $$\begin{cases}-12-p+q=0\\3+2p+q=0\end{cases} \Rightarrow\begin{cases}-p+q=12\\2p+q=-3\end{cases} \Rightarrow p=-5,\ q=7.$$ Since all the coefficients are real, $1-\sqrt2\mathrm{i}$ is also a root. $$\begin{aligned} \text{Quadratic factor}&=[x-(1-\sqrt2\mathrm{i})][x-(1+\sqrt2\mathrm{i})]\\ &=(x-1)^2-2(-1)=x^2-2x+3\\ 3x^3-5x^2+7x+3&=0\\ (x^2-2x+3)(3x+1)&=0 \end{aligned}$$ $$x=1\pm\sqrt2\mathrm{i}\quad\text{or}\quad x=-\frac13$$ Other than $1+\sqrt2\mathrm{i}$, the other roots are $1-\sqrt2\mathrm{i}$ and $-\dfrac13$.

$$w+z^*=2-\mathrm{i}\quad\Rightarrow\quad z^*=2-\mathrm{i}-w$$

$$\begin{aligned} w(2-\mathrm{i}-w)&=3-\mathrm{i}\\ w^2-(2-\mathrm{i})w+(3-\mathrm{i})&=0\\ w&=\frac{2-\mathrm{i}\pm\sqrt{(2-\mathrm{i})^2-4(3-\mathrm{i})}}2\\ &=\frac{2-\mathrm{i}\pm\sqrt{4-4\mathrm{i}-1-12+4\mathrm{i}}}2\\ &=\frac{2-\mathrm{i}\pm\sqrt{-9}}2\\ &=\frac{2-\mathrm{i}\pm3\mathrm{i}}{2} \end{aligned}$$ $w=1-2\mathrm{i}$ or $w=1+\mathrm{i}$.

When $w=1-2\mathrm{i}$: $$z^*=2-\mathrm{i}-(1-2\mathrm{i})=1+\mathrm{i},\qquad z=1-\mathrm{i}$$ (reject since it’s given that $|w|<|z|$)

When $w=1+\mathrm{i}$: $$z^*=2-\mathrm{i}-(1+\mathrm{i})=1-2\mathrm{i},\qquad z=1+2\mathrm{i}$$ Hence, $w=1+\mathrm{i}$ and $z=1+2\mathrm{i}$.

The alternate solution’s printed root typo has been corrected: the roots of x² − 2x + 3 are 1 ± √2i.