Complex Numbers Revision 2026 | Math Academy Tuition Centre

Complex Numbers — 2025 Prelim Solutions

20 questions drawn from 2025 JC Preliminary Examinations — Math Academy

Q1
RVHS / P1 / Q3

Find the roots of the equation \(z^2 – (1+2\mathrm{i})z + 1 + 7\mathrm{i} = 0\), giving your answers in the cartesian form \(a + b\mathrm{i}\).

Solution:

In the above equation, \(a = 1\), \(b = -1 – 2\mathrm{i}\), \(c = 1 + 7\mathrm{i}\).

\[ z = \frac{1 + 2\mathrm{i} \pm \sqrt{(-1-2\mathrm{i})^2 – 4(1)(1+7\mathrm{i})}}{2} = \frac{1 + 2\mathrm{i} \pm \sqrt{-7 – 24\mathrm{i}}}{2} \]

To find \(\sqrt{-7 – 24\mathrm{i}}\), let \(\sqrt{-7 – 24\mathrm{i}} = x + y\mathrm{i}\), \(x, y \in \mathbb{R}\).

Comparing real and imaginary parts of \((x+y\mathrm{i})^2 = -7-24\mathrm{i}\):

\[ x^2 – y^2 = -7 \quad\text{—}(1) \qquad 2xy = -24 \implies x = \frac{-12}{y} \quad\text{—}(2) \]

Substituting (2) into (1):

\[ \frac{144}{y^2} – y^2 = -7 \implies y^4 – 7y^2 – 144 = 0 \implies (y^2-16)(y^2+9)=0 \implies y = \pm 4 \]

When \(y = 4\), \(x = -3\). When \(y = -4\), \(x = 3\). The square roots of \(-7-24\mathrm{i}\) are \(3-4\mathrm{i}\) and \(-3+4\mathrm{i}\).

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

\[ \boxed{z = 2 – \mathrm{i} \quad\text{or}\quad z = -1 + 3\mathrm{i}} \]

Q2
VJC / P1 / Q7

(a) Find \(z\) satisfying \(\dfrac{4|z|}{15 – z^*} = 5\mathrm{i}\).

(b)(i) Given one value of \(w\) satisfying \((w-\mathrm{i})^3 = -\mathrm{i}\) is \(2\mathrm{i}\), find the two other values of \(w\).

(b)(ii) Show \(W_1\), \(W_2\), \(W_3\) lie on a circle with centre \(A = k\mathrm{i}\), stating \(k\).

Solution:

(a)

Let \(z = a + b\mathrm{i}\). Comparing real and imaginary parts of \(4\sqrt{a^2+b^2} = -5b + (75-5a)\mathrm{i}\):

\[ 75 – 5a = 0 \implies a = 15 \]

\[ 4\sqrt{225+b^2} = -5b \implies 9b^2 = 3600 \implies b = -20 \]

\[ \boxed{z = 15 – 20\mathrm{i}} \]

(b)(i)

Expanding \((w-\mathrm{i})^3 = -\mathrm{i}\) gives \(w^3 – 3\mathrm{i}w^2 – 3w + 2\mathrm{i} = 0\). Since \(2\mathrm{i}\) is a root:

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

\[ w = \frac{\mathrm{i} \pm \sqrt{3}}{2} \]

\[ \boxed{w = \frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \quad\text{or}\quad w = -\frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i}} \]

(b)(ii)

Setting \(|W_1 A| = |W_2 A|\) with \(A = k\mathrm{i}\) and solving gives \(k = 1\). Then \(W_1 A = W_2 A = W_3 A = 1\), so all three points lie on a circle of radius 1 centred at \(\mathrm{i}\), with \(\boxed{k=1}\).

Q3
EJC / P1 / Q8

The complex number \(w\) satisfies \(w^2 = 2\mathrm{i}\) and \(0 \leq \arg(w) \leq \dfrac{\pi}{2}\).

(a) Find \(w\). (b) Given \(w\) is a root of \(z^3 – z^2 + (1-\mathrm{i})z + s = 0\), find the other roots and \(s\). (c) Hence find roots of \(-\mathrm{i}v^3 + v^2 + (1+\mathrm{i})v + s = 0\).

Solution:

(a)

Let \((a+b\mathrm{i})^2 = 2\mathrm{i}\). Then \(a^2-b^2=0\) and \(2ab=2\), so \(a=b=1\). Since \(0\leq\arg(w)\leq\frac{\pi}{2}\):

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

(b)

Writing \(z^3 – z^2 + (1-\mathrm{i})z + s = [z-(1+\mathrm{i})](z^2 + Bz + C)\) and comparing coefficients: \(B = \mathrm{i}\), \(C = 0\), \(s = 0\).

\[ z^2 + \mathrm{i}z = z(z+\mathrm{i}) = 0 \]

\[ \boxed{z = 0 \quad\text{or}\quad z = -\mathrm{i}, \qquad s = 0} \]

(c)

Substituting \(z = \mathrm{i}v\) into the original equation gives \(-\mathrm{i}v^3 + v^2 + (1+\mathrm{i})v + s = 0\). Then \(v = -\mathrm{i}z\):

\[ \boxed{v = 1-\mathrm{i}, \quad v = 0, \quad v = -1} \]

Q4
TJC / P1 / Q8

(a) Solve \(2z + |w| = 2 – 4\mathrm{i}\) and \(\mathrm{i}z – w = 2\mathrm{i}\) for \(z\) and \(w\).

(b) Given \(z_1 = 2-\mathrm{i}\) and \(z_3 = -3+2\mathrm{i}\), find \(z_2\) and \(z_4\) such that \(ABCD\) is an anti-clockwise square.

Solution:

(a)

Let \(w = a+b\mathrm{i}\). Comparing real and imaginary parts gives \(a = 2\), and solving \(-2-2b = \sqrt{4+b^2}\) gives \(b = -\frac{8}{3}\).

\[ |w| = \sqrt{4+\tfrac{64}{9}} = \tfrac{10}{3}, \qquad z = \frac{2-4\mathrm{i}-\frac{10}{3}}{2} \]

\[ \boxed{z = -\tfrac{2}{3} – 2\mathrm{i}, \qquad w = 2 – \tfrac{8}{3}\mathrm{i}} \]

(b)(ii)

Rotating \(\overrightarrow{BC}\) anticlockwise by \(\frac{\pi}{2}\) about \(B\): \(\mathrm{i}(z_3 – z_2) = z_1 – z_2\), so \((1-\mathrm{i})z_2 = 4+2\mathrm{i}\).

\[ z_2 = \frac{4+2\mathrm{i}}{1-\mathrm{i}} \cdot \frac{1+\mathrm{i}}{1+\mathrm{i}} = \frac{2+6\mathrm{i}}{2} = 1+3\mathrm{i} \]

\[ z_4 = z_3 + (z_1 – z_2) = -3+2\mathrm{i} + 1-4\mathrm{i} = -2-2\mathrm{i} \]

\[ \boxed{z_2 = 1+3\mathrm{i}, \qquad z_4 = -2-2\mathrm{i}} \]

Q5
TMJC / P1 / Q9

Given \(-2+2\mathrm{i}\) is a root of \(z^3 + az^2 + bz – 16\sqrt{2} = 0\) (\(a\), \(b\) real).

(a) Find \(a\), \(b\), and the other two roots. (b) Sketch on Argand diagram with moduli and arguments; state geometrical relationship between \(A\) and \(B\). (c) Hence prove \(\tan\frac{3\pi}{8} = 1+\sqrt{2}\).

Solution:

(a)

Since coefficients are real, \(-2-2\mathrm{i}\) is also a root. The quadratic factor is \([z-(-2+2\mathrm{i})][z-(-2-2\mathrm{i})] = z^2+4z+8\).

\[ z^3+az^2+bz-16\sqrt{2} = (z^2+4z+8)(z-2\sqrt{2}) \]

\[ \boxed{a = 4-2\sqrt{2},\quad b = 8-8\sqrt{2},\quad \text{roots: } -2+2\mathrm{i},\ -2-2\mathrm{i},\ 2\sqrt{2}} \]

(b)

All three roots have modulus \(2\sqrt{2}\). \(\arg(z_1) = \frac{3\pi}{4}\), \(\arg(z_2) = -\frac{3\pi}{4}\), \(\arg(z_3) = 0\). \(B\) is the reflection of \(A\) about the real axis.

(c)

Since \(OA = OC = 2\sqrt{2}\), let \(D = z_1 + z_3 = (2\sqrt{2}-2)+2\mathrm{i}\). Then \(OD\) bisects \(\angle AOC\), so \(\arg(z_D) = \frac{3\pi}{8}\).

\[ \tan\frac{3\pi}{8} = \frac{2}{2\sqrt{2}-2} = \frac{1}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1} = 1+\sqrt{2} \qquad\square \]

Q6
DHS / P1 / Q8

\(z = 2\cos t + 3\mathrm{i}\sin t\), \(0 \leq t < 2\pi\).

(a) Cartesian equation of locus. (b) Possible values of \(\arg(z_1+z_2)\) when \(|z_1|=|z_2|\). (c) Is \(\alpha\) necessarily real? (d) Given \(\alpha\) not real and \(|z_1|=|z_2|=\frac{\sqrt{26}}{2}\), find \(\alpha\) and \(\beta\).

Solution:

(a)

\[ \cos t = \frac{x}{2},\quad \sin t = \frac{y}{3} \implies \boxed{\frac{x^2}{4} + \frac{y^2}{9} = 1} \quad\text{(ellipse)} \]

(b)

\(\arg(z_1+z_2)\) is either \(0\), \(\dfrac{\pi}{2}\), or undefined (when \(z_2 = -z_1\)).

(c)

It is not necessary for \(\alpha\) to be real.

(d)

Since \(\alpha\) is not real, \(z_1 = (a,b)\) and \(z_2 = (-a,b)\). Using \(a^2+b^2 = \frac{26}{4}\) and the ellipse equation \(\frac{a^2}{4}+\frac{b^2}{9}=1\):

\[ b^2 = \frac{9}{2},\quad a^2 = 2,\quad \alpha = -(z_1+z_2) = -2b\mathrm{i} = -3\sqrt{2}\,\mathrm{i} \]

\[ \boxed{\alpha = -3\sqrt{2}\,\mathrm{i}, \qquad \beta = -\frac{13}{2}} \]

Q7
SAJC / P1 / Q10

(a) Solve \(|z| + 5w = 0\) and \(\mathrm{i}z – 4w = -4+7\mathrm{i}\).

(b)(ii) Given \(C\), \(D\) represent \(v-u\) and \((v-u)^*\) with \(\angle CDA = 90^\circ\), find \(\operatorname{Im}(v)\).

Solution:

(a)

From (1): \(w = -\frac{1}{5}|z|\). Substituting into (2) and comparing parts gives \(x = 7\), then \(y = 24\).

\[ z = 7+24\mathrm{i},\quad |z| = 25,\quad w = -\frac{25}{5} = -5 \]

\[ \boxed{z = 7+24\mathrm{i}, \qquad w = -5} \]

(b)(ii)

Since \(C\) and \(D\) are complex conjugates, \(CD\) is vertical. For \(\angle CDA = 90^\circ\), \(DA\) must be horizontal. Since \(A\) and \(B = -u\) are equidistant from the real axis, \(BC\) is also horizontal, meaning \(v\) is real.

\[ \boxed{\operatorname{Im}(v) = 0} \]

Q8
ACJC / P1 / Q10

(a) Root \(2+3\mathrm{i}\) of \(\omega^4 – 2\omega^3 + 10\omega^2 + p\omega + q = 0\). Find \(p\), \(q\), other roots.

(b) \(u = -\sqrt{2}+\mathrm{i}\sqrt{2}\), \(|v|=3\), \(0<\arg v<\frac{\pi}{4}\). Find \(|u|\), \(\arg u\); show \(|u+v|^2 = a + b\cos(\theta+K)\).

Solution:

(a)

Since coefficients are real, \(2-3\mathrm{i}\) is also a root. Quadratic factor: \(\omega^2-4\omega+13\).

\[ \omega^4-2\omega^3+10\omega^2+p\omega+q = (\omega^2-4\omega+13)(\omega^2+2\omega+5) \]

\[ \boxed{p = 6,\quad q = 65,\quad \text{other roots: } -1\pm 2\mathrm{i}} \]

(b)(i)

\[ |u| = 2, \qquad \arg u = \pi – \frac{\pi}{4} = \frac{3\pi}{4} \]

(b)(iii)

\(\angle AOB = \frac{3\pi}{4} – \theta\), so \(\angle OAC = \theta + \frac{\pi}{4}\). By cosine rule on \(\triangle OAC\):

\[ |u+v|^2 = 4 + 9 – 2(2)(3)\cos\!\left(\theta+\frac{\pi}{4}\right) = 13 – 12\cos\!\left(\theta+\frac{\pi}{4}\right) \]

\[ \boxed{a = 13,\quad b = -12,\quad K = \frac{\pi}{4}} \]

Q9
HCI / P1 / Q12

\(\mathrm{f}(z) = z^4 – 6z^2 + k\), \(k \neq 0\).

(a) Purely imaginary \(k\): can \(\mathrm{f}(z)=0\) have real roots? (b) Show \(\mathrm{f}(-z)=\mathrm{f}(z)\). (c) Root \(2+\mathrm{i}\): find \(k\) and remaining roots. (d) Find product \(D\) of all roots. (e) Find \(w_1\) satisfying \(kw^4-6w^2+1=0\).

Solution:

(a)

If \(x\) is real, \(x^4-6x^2\) is real but \(-k\) is purely imaginary — contradiction. So \(\mathrm{f}(z)=0\) cannot have real roots.

(b)

\[ \mathrm{f}(-z) = (-z)^4 – 6(-z)^2 + k = z^4 – 6z^2 + k = \mathrm{f}(z) \qquad\square \]

(c)

\[ (2+\mathrm{i})^2 = 3+4\mathrm{i}, \quad (2+\mathrm{i})^4 = -7+24\mathrm{i} \]

\[ -7+24\mathrm{i} – 6(3+4\mathrm{i}) + k = 0 \implies \boxed{k = 25} \]

Since coefficients are real, \(2-\mathrm{i}\) is also a root. Since \(\mathrm{f}(-z)=\mathrm{f}(z)\), \(-2\pm\mathrm{i}\) are also roots.

\[ \boxed{\text{Remaining roots: } 2-\mathrm{i},\ -2+\mathrm{i},\ -2-\mathrm{i}} \]

(d)

\[ D = (2+\mathrm{i})(2-\mathrm{i}) \cdot (-2+\mathrm{i})(-2-\mathrm{i}) = 5 \times 5 = \boxed{25} \]

(e)

Replace \(z\) by \(\frac{1}{w}\): \(w_1 = \dfrac{1}{2+\mathrm{i}} = \dfrac{2-\mathrm{i}}{5} = \boxed{\dfrac{2}{5} – \dfrac{1}{5}\mathrm{i}}\).

Q10
NYJC / P1 / Q6

\(\arg p = \alpha\), \(\arg q = \beta\), \(0 < \alpha < \frac{\beta}{2}\), \(r = p+q\).

(a) \(|p|=|q|\): describe \(OPRQ\) and find \(\arg(r)\). (b) State \(\angle POQ’\). (b)(i) Find \(\arg(q’)\). (b)(ii) Write \(q’\) in \(a+b\mathrm{i}\) form.

Solution:

(a)

\(OPRQ\) is a rhombus.

\[ \arg(r) = \frac{\alpha + \beta}{2} \]

(b)

\[ \angle POQ’ = \beta – \alpha \]

(b)(i)

\[ \arg(q’) = 2\alpha – \beta \]

(b)(ii)

\[ \boxed{q’ = |q|\cos(\beta – 2\alpha) – \mathrm{i}|q|\sin(\beta – 2\alpha)} \]

Q11
RI / P1 / Q8

(a)(i) Show \(\arg(kw) = \arg(w)\) for real \(k > 1\). (b)(i) Show \(\mathrm{f}(\mathrm{f}(z)) – \mathrm{f}(z) = z(z-2)(z-3)(z+1)\). (b)(ii) Show \((z-2)(z+1)=m\mathrm{i}\). (b)(iii) For \(z = x+2\mathrm{i}\) with \(x > 0\), find \(x\), \(m\), and \(B\).

Solution:

(a)(i)

Since \(u > 0\), \(v > 0\): \(\arg(kw) = \tan^{-1}\dfrac{kv}{ku} = \tan^{-1}\dfrac{v}{u} = \arg(w)\). \(\square\)

(b)(i)

\begin{align*}
\mathrm{f}(\mathrm{f}(z)) – \mathrm{f}(z) &= (z^2-2z)[(z^2-2z) – 3] \\
&= (z^2-2z)(z^2-2z-3) \\
&= z(z-2)(z-3)(z+1) \qquad\square
\end{align*}

(b)(ii)

Since \(\triangle ABC\) is right-angled at \(B\) anticlockwise with \(BC = m \cdot BA\): \(\mathrm{i}[\mathrm{f}(\mathrm{f}(z))-\mathrm{f}(z)] = m[z-\mathrm{f}(z)]\), giving \(\dfrac{z(z-2)(z-3)(z+1)}{z(z-3)} = m\mathrm{i}\), so \((z-2)(z+1) = m\mathrm{i}\). \(\square\)

(b)(iii)

With \(z = x+2\mathrm{i}\): \(x^2-x-6 = 0 \implies x = 3\) (positive), \(m = 4(3)-2 = 10\).

\[ \mathrm{f}(3+2\mathrm{i}) = (3+2\mathrm{i})^2 – 2(3+2\mathrm{i}) = 9+12\mathrm{i}-4-6-4\mathrm{i} \]

\[ \boxed{B \text{ represents } {-1+8\mathrm{i}}} \]

Q12
ASRJC / P2 / Q6

\(p = \sqrt{2}+\sqrt{2}\,\mathrm{i}\), \(\arg(q)=\frac{2\pi}{3}\), \(|q|=2\).

(a) Find \(|p|\), \(\arg(p)\). (c) Deduce \(\operatorname{Re}(q)\), \(\operatorname{Im}(q)\). (d) Describe \(OPRQ\). (e) Show \(\tan\frac{11\pi}{24} = \sqrt{6}+\sqrt{3}+\sqrt{2}+2\).

Solution:

(a)

\[ |p| = 2, \qquad \arg(p) = \frac{\pi}{4} \]

(c)

\[ q = 2\!\left(\cos\frac{2\pi}{3}+\mathrm{i}\sin\frac{2\pi}{3}\right) = -1+\sqrt{3}\,\mathrm{i}, \qquad \operatorname{Re}(q) = -1,\quad \operatorname{Im}(q) = \sqrt{3} \]

(d)

\(OPRQ\) is a rhombus (since \(|p| = |q| = 2\)).

(e)

\[ \arg(p+q) = \frac{\pi/4 + 2\pi/3}{2} = \frac{11\pi}{24}, \qquad p+q = (\sqrt{2}-1)+(\sqrt{2}+\sqrt{3})\,\mathrm{i} \]

\[ \tan\frac{11\pi}{24} = \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-1} = \frac{(\sqrt{2}+\sqrt{3})(\sqrt{2}+1)}{1} = \sqrt{6}+\sqrt{3}+\sqrt{2}+2 \qquad\square \]

Q13
VJC / P2 / Q4

\(|z|=1\), \(\frac{\pi}{2}<\theta<\pi\), \(w = \mathrm{i}\sqrt{3}\,z\).

(a) Geometrical relationship between \(P\), \(Q=w\), \(R=z-w\). (b) Area of \(ORPQ\). (c) \(\theta=\frac{3\pi}{4}\): find \(z\). (d) Show \(z-w = k[(\sqrt{3}-1)+(\sqrt{3}+1)\mathrm{i}]\). (e) Show \(\tan\frac{5\pi}{12} = \frac{\sqrt{3}+1}{\sqrt{3}-1}\).

Solution:

(a)

\(OQ \perp OP\) and \(OQ = \sqrt{3}\,OP\). \(ORPQ\) is a parallelogram (rectangle).

(b)

\[ \text{Area} = |OP| \times |OQ| = 1 \times \sqrt{3} = \sqrt{3} \]

(c)

\[ \boxed{z = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\,\mathrm{i}} \]

(d)

\begin{align*}
z – w &= z(1-\sqrt{3}\,\mathrm{i}) = \frac{1}{\sqrt{2}}(-1+\mathrm{i})(1-\sqrt{3}\,\mathrm{i}) \\
&= \frac{1}{\sqrt{2}}\bigl[(\sqrt{3}-1)+(\sqrt{3}+1)\mathrm{i}\bigr], \qquad k = \frac{1}{\sqrt{2}}
\end{align*}

(e)

\(\arg(z-w) = \frac{3\pi}{4} – \frac{\pi}{3} = \frac{5\pi}{12}\), therefore:

\[ \tan\frac{5\pi}{12} = \frac{\frac{1}{\sqrt{2}}(\sqrt{3}+1)}{\frac{1}{\sqrt{2}}(\sqrt{3}-1)} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \qquad\square \]

Q14
JPJC / P2 / Q4

(a) Root \(3-\mathrm{i}\) of \(5w^3+pw^2+68w+q=0\). Find other roots, \(p\), \(q\).

(b)(i) Find \(z_1/z_2\) in form \(k\mathrm{i}\). (b)(ii) Show \(OZ_2Z_3Z_1\) is a square.

Solution:

(a)

Since coefficients are real, \(3+\mathrm{i}\) is also a root. Quadratic factor: \(w^2-6w+10\).

\[ 5w^3+pw^2+68w+q = (w^2-6w+10)(5w-3) \]

\[ \boxed{p = -33,\quad q = -30,\quad \text{other roots: } 3+\mathrm{i} \text{ and } \tfrac{3}{5}} \]

(b)(i)

\[ \frac{z_1}{z_2} = \frac{-1+2\mathrm{i}}{2+\mathrm{i}} \times \frac{2-\mathrm{i}}{2-\mathrm{i}} = \frac{5\mathrm{i}}{5} = \boxed{\mathrm{i}} \]

(b)(ii)

\(z_1 = \mathrm{i}z_2\) means \(Z_1\) is \(Z_2\) rotated \(90^\circ\) anticlockwise: \(\angle Z_1OZ_2 = 90^\circ\) and \(OZ_1 = OZ_2 = \sqrt{5}\). Since \(\overrightarrow{OZ_3} = \overrightarrow{OZ_1}+\overrightarrow{OZ_2}\), \(OZ_1Z_3Z_2\) is a parallelogram. Hence \(OZ_2Z_3Z_1\) is a square. \(\square\)

Q15
NJC / P2 / Q1

\(w = -\mathrm{i}\sqrt{3}\). Find \(|w|\), \(\arg(w)\). Represent \(w\), \(-w\), \(2-w\) on Argand diagram. Find \(\arg(2-w)\).

Solution:

(i)

\[ |w| = \sqrt{3}, \qquad \arg(w) = -\frac{\pi}{2} \]

(ii)

\(-w = \mathrm{i}\sqrt{3}\), \(2-w = 2+\mathrm{i}\sqrt{3}\). Triangle \(OA(-w)B(2-w)\) is isosceles with \(AB\) parallel to the real axis.

(iii)

\[ \arg(2-w) = \tan^{-1}\!\frac{\sqrt{3}}{2} \]

Q16
CJC / P2 / Q4

Root \(3+\mathrm{i}\) of \(2z^4-14z^3+33z^2-26z+p=0\).

(a) Why may \(3-\mathrm{i}\) not be a root? (b) Show \(p=10\). (c) Find all roots. (d) Identify quadrilateral; find area.

Solution:

(a)

The student’s claim may not be true as \(p\) may not be real (the conjugate root theorem requires all real coefficients).

(b)

\[ (3+\mathrm{i})^2 = 8+6\mathrm{i},\quad (3+\mathrm{i})^3 = 18+26\mathrm{i},\quad (3+\mathrm{i})^4 = 28+96\mathrm{i} \]

\[ 2(28+96\mathrm{i}) – 14(18+26\mathrm{i}) + 33(8+6\mathrm{i}) – 26(3+\mathrm{i}) + p = -10+p = 0 \implies p = 10 \qquad\square \]

(c)

Coefficients are real so \(3-\mathrm{i}\) is also a root. Factorising: \((z^2-6z+10)(2z^2-2z+1)=0\).

\[ 2z^2-2z+1=0 \implies z = \tfrac{1}{2}\pm\tfrac{1}{2}\mathrm{i} \]

\[ \boxed{\text{Roots: } 3\pm\mathrm{i},\quad \tfrac{1}{2}\pm\tfrac{1}{2}\mathrm{i}} \]

(d)

Trapezium.

\[ \text{Area} = \tfrac{1}{2}(2+1) \times 2.5 = \boxed{\tfrac{15}{4}} \]

Q17
RVHS / P2 / Q2

\(z_A = 5+6\mathrm{i}\), \(z_B = 9+3\mathrm{i}\). \(ABC\) is isosceles right-angled (clockwise), \(\angle CAB = 90^\circ\). Find \(z_C\); find \(z_D\) such that \(ABDC\) is a parallelogram.

Solution:

(a)

\(AC\) is \(AB\) rotated \(90^\circ\) clockwise about \(A\): \(z_C – z_A = -\mathrm{i}(z_B – z_A)\).

\[ z_C = (5+6\mathrm{i}) – \mathrm{i}(4-3\mathrm{i}) = 5+6\mathrm{i} – 4\mathrm{i} – 3 = \boxed{2+2\mathrm{i}} \]

(b)

\[ z_D = z_B + z_C – z_A = (9+3\mathrm{i})+(2+2\mathrm{i})-(5+6\mathrm{i}) = \boxed{6-\mathrm{i}} \]

Q18
SAJC / P2 / Q2

(a)(i) \(\mathrm{f}(x)=px^6+qx^4+r\): show if \(\alpha\) is a root, so is \(-\alpha\). (a)(ii) Roots \(x=5\) and \(x=\beta\); write all remaining roots.

(b) Root \(1+\mathrm{i}\) of \(z^3+(3-a)z^2-(2+6\mathrm{i})z-6=0\). Find \(a\) and other roots.

Solution:

(a)(i)

\[ \mathrm{f}(-\alpha) = p(-\alpha)^6+q(-\alpha)^4+r = p\alpha^6+q\alpha^4+r = \mathrm{f}(\alpha) = 0 \qquad\square \]

(a)(ii)

Remaining roots: \(-5\), \(\beta^*\), \(-\beta\), \(-\beta^*\).

(b)

Write \(z^3+(3-a)z^2-(2+6\mathrm{i})z-6 = (z-1-\mathrm{i})(z^2+Bz+C)\). Comparing coefficients: \(C = 3-3\mathrm{i}\), \(B = 4-\mathrm{i}\), \(a = 2\mathrm{i}\).

\[ (z-1-\mathrm{i})(z+3)(z+1-\mathrm{i}) = 0 \]

\[ \boxed{a = 2\mathrm{i},\quad \text{other roots: } -3 \text{ and } -1+\mathrm{i}} \]

Q19
NYJC / P2 / Q2

(a) Explain why complex roots of \(-\mathrm{i}x^3+5\mathrm{i}x^2+ax+b=0\) (\(a\), \(b\) purely imaginary) occur in conjugate pairs.

(b) Find roots of \(\mathrm{i}x^3+5\mathrm{i}x^2+a^*x+b=0\).

Solution:

(a)

Since \(a = \lambda\mathrm{i}\) and \(b = \mu\mathrm{i}\), dividing by \(\mathrm{i}\) gives \(x^3-5x^2-\lambda x-\mu=0\) with all real coefficients. Hence complex roots occur in conjugate pairs.

(b)

Since \(a^* = -a\), substituting \(x \to -x\) transforms the equation to the original. So \(-x = 2\pm\mathrm{i},\ 1\):

\[ \boxed{x = -2-\mathrm{i},\quad -2+\mathrm{i},\quad -1} \]

Q20
MI / P2 / Q3

(a) Solve \(z+2z^*+w=6+\mathrm{i}\) and \(\mathrm{i}z-w=3\).

(b) \(\mathrm{f}(x)=x^3+mx^2+nx+5\) has root \(2+\mathrm{i}\). Find where the curve crosses the \(x\)-axis.

Solution:

(a)

From (2): \(w = \mathrm{i}z-3\). Substituting into (1): \((1+\mathrm{i})z+2z^* = 9+\mathrm{i}\). Let \(z = a+b\mathrm{i}\):

\[ 3a-b = 9 \qquad a-b = 1 \implies a=4,\quad b=3 \]

\[ \boxed{z = 4+3\mathrm{i},\qquad w = \mathrm{i}(4+3\mathrm{i})-3 = -6+4\mathrm{i}} \]

(b)

Since coefficients are real, \(2-\mathrm{i}\) is also a root. Quadratic factor: \(x^2-4x+5\).

\[ \mathrm{f}(x) = (x+1)(x^2-4x+5), \quad \text{real root: } x = -1 \]

\[ \boxed{P = (-1,\ 0)} \]