Complex numbers: RI 2024 Promo Q7

(a)[5]

Since $a,b\in\mathbb R$, then all the coefficients of the polynomial are real, complex roots will occur in conjugate pairs. Given $1+3\mathrm{i}$ is a root, then $1-3\mathrm{i}$ is also a root. $$\begin{aligned} [x-(1+3\mathrm{i})][x-(1-3\mathrm{i})] &=[(x-1)-3\mathrm{i}][(x-1)+3\mathrm{i}]\\ &=(x-1)^2-(3\mathrm{i})^2\\ &=x^2-2x+10. \end{aligned}$$ $$x^3+ax^2+18x+b=(x^2-2x+10)(x+p).$$ Compare coefficients of $$\begin{aligned} x &:18=-2p+10\Rightarrow p=-4,\\ x^2 &:a=p-2\Rightarrow a=-6,\\ x^0 &:b=-40. \end{aligned}$$ The other roots are $4$ and $1-3\mathrm{i}$.

Alternative:

Since $x=1+3\mathrm{i}$ is a root of the given equation, $$\begin{aligned} (1+3\mathrm{i})^3+a(1+3\mathrm{i})^2+18(1+3\mathrm{i})+b&=0,\\ (1+3\mathrm{i})(-8+6\mathrm{i})+a(-8+6\mathrm{i})+18(1+3\mathrm{i})+b&=0,\\ -8-24\mathrm{i}+6\mathrm{i}-18-8a+6a\mathrm{i}+18+54\mathrm{i}+b&=0,\\ (-8-8a+b)+(6a+36)\mathrm{i}&=0. \end{aligned}$$ Equating real and imaginary parts: $$6a+36=0\Rightarrow a=-6, \qquad \therefore b=8a+8=-40.$$ Since the coefficients of the equation $x^3+ax^2+18x+b=0$ are all real numbers, the complex roots occur in conjugate pairs and so $1-3\mathrm{i}$ is also a root. Let the third root be a real number $k$.

Note that

$$\begin{aligned} [x-(1+3\mathrm{i})][x-(1-3\mathrm{i})] &=[(x-1)-3\mathrm{i}][(x-1)+3\mathrm{i}]\\ &=(x-1)^2-(3\mathrm{i})^2\\ &=x^2-2x+10. \end{aligned}$$ Therefore, we have $$x^3-6x^2+18x-40=(x^2-2x+10)(x-k).$$ Comparing constants, we have $k=4$.

So $a=-6$, $b=-40$ and the other roots are $1-3\mathrm{i}$ and $4$.

(b)[5] $$\begin{aligned} w^*+z&=4-6\mathrm{i}\quad------(1),\\ w-2z&=1+10\mathrm{i}\quad------(2). \end{aligned}$$ $2(1)+(2):\quad2w^*+w=9-2\mathrm{i}$.

Sub $w=c+d\mathrm{i}:\quad3c-d\mathrm{i}=9-2\mathrm{i}$.

Comparing real part: $3c=9\Rightarrow c=3$.

Comparing imaginary part: $d=2$.

$\therefore w=3+2\mathrm{i}$.

From (1): $z=4-6\mathrm{i}-(3-2\mathrm{i})=1-4\mathrm{i}$.

$\therefore w=3+2\mathrm{i}$, $z=1-4\mathrm{i}$.

Alternative 1

Substitute $z=4-6\mathrm{i}-w^*$ into (2): $$w-2(4-6\mathrm{i}-w^*)=1+10\mathrm{i} \Rightarrow2w^*+w=9-2\mathrm{i}.$$ Sub $w=c+d\mathrm{i}:\quad3c-d\mathrm{i}=9-2\mathrm{i}$.

Comparing real part: $3c=9\Rightarrow c=3$.

Comparing imaginary part: $d=2$.

$\therefore w=3+2\mathrm{i}$.

From (1): $z=4-6\mathrm{i}-(3-2\mathrm{i})=1-4\mathrm{i}$.

$\therefore w=3+2\mathrm{i}$, $z=1-4\mathrm{i}$.

Alternative 2

Let $w=a+b\mathrm{i}$ and $z=g+h\mathrm{i}$. Then we have

$$\begin{aligned} a-b\mathrm{i}+g+h\mathrm{i}&=4-6\mathrm{i}\\ \Rightarrow a+g&=4\quad----------(1),\\ h-b&=-6\quad----------(2), \end{aligned}$$ and $$\begin{aligned} a+b\mathrm{i}-2(g+h\mathrm{i})&=1+10\mathrm{i}\\ \Rightarrow a-2g&=1\quad----------(3),\\ b-2h&=10\quad----------(4). \end{aligned}$$ From (1) and (3), $3g=3\Rightarrow g=1$ and $a=3$.

From (2) and (4), $-h=4\Rightarrow h=-4$ and $b=2$.

$\therefore w=3+2\mathrm{i}$ and $z=1-4\mathrm{i}$.

Two typographical errors in the alternative expansion have been corrected: 18(1 + 3i) = 18 + 54i. The resulting values a = −6 and b = −40 are unchanged.