2020 H2 Maths Paper 1 Q4 — Worked Solution

Part (i)

4 marks

Approach

Since we may not use a calculator, we convert each of $z_{1}$ , $z_{2}$ , $z_{3}$ to modulus-argument form and combine using $\left| \frac{z_{1}}{z_{2}z_{3}} \right| = \frac{\left| z_{1} \right|}{\left| z_{2} \right|\left| z_{3} \right|}$ and $\arg\left( \frac{z_{1}}{z_{2}z_{3}} \right) = \arg z_{1} - \arg z_{2} - \arg z_{3}$ , rather than multiplying the complex numbers out in cartesian form.

For $z_{1} = 1 + \sqrt{3}i$ , the point $\left( 1,\sqrt{3} \right)$ lies in the first quadrant, so

\left| z_{1} \right| = \sqrt{1^{2} + \left( \sqrt{3} \right)^{2}} = 2,\ \ \arg\left( z_{1} \right) = \arctan\left( \frac{\sqrt{3}}{1} \right) = \frac{\pi}{3}

For $z_{2} = 1 - i$ , the point $(1, - 1)$ lies in the fourth quadrant, so

\left| z_{2} \right| = \sqrt{1^{2} + ( - 1)^{2}} = \sqrt{2},\ \ \arg\left( z_{2} \right) = - \frac{\pi}{4}

For $z_{3} = 2\left( \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} \right)$ , reading directly from the given polar form,

\left| z_{3} \right| = 2,\ \ \arg\left( z_{3} \right) = \frac{\pi}{6}

For the quotient, moduli divide and arguments subtract.

\left| \frac{z_{1}}{z_{2}z_{3}} \right| = \frac{\left| z_{1} \right|}{\left| z_{2} \right|\left| z_{3} \right|}

= \frac{2}{\sqrt{2} \times 2}

= \frac{1}{\sqrt{2}}

= \frac{\sqrt{2}}{2}

\arg\left( \frac{z_{1}}{z_{2}z_{3}} \right) = \arg\left( z_{1} \right) - \arg\left( z_{2} \right) - \arg\left( z_{3} \right)

= \frac{\pi}{3} - \left( - \frac{\pi}{4} \right) - \frac{\pi}{6}

= \frac{4\pi}{12} + \frac{3\pi}{12} - \frac{2\pi}{12}

= \frac{5\pi}{12}

Since $\frac{5\pi}{12}$ already lies in $( - \pi,\pi\rbrack$ , no adjustment is needed.

To check your answer:

\frac{5\pi}{12} = 75^{\circ}

lies between

\frac{\pi}{3} = 60^{\circ}

and

\frac{\pi}{3} + \frac{\pi}{4} = 105^{\circ}

, the two extremes before subtracting

\frac{\pi}{6}

. ✓

$\therefore$ $\frac{z_{1}}{z_{2}z_{3}} = \frac{\sqrt{2}}{2}\left( \cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12} \right)$ .

Part (ii)

3 marks

Approach

A complex number with modulus $1$ that is purely imaginary must be $i$ or $- i$ . We set $\frac{z_{1}z_{4}}{z_{2}z_{3}}$ equal to each in turn and divide by the result of part (i) to isolate $z_{4}$ .

Since $\left| \frac{z_{1}z_{4}}{z_{2}z_{3}} \right| = 1$ and $\frac{z_{1}z_{4}}{z_{2}z_{3}}$ is purely imaginary, it equals either $i = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2}$ or $- i = \cos\left( - \frac{\pi}{2} \right) + i\sin\left( - \frac{\pi}{2} \right)$ .

Writing $\frac{z_{1}z_{4}}{z_{2}z_{3}} = \frac{z_{1}}{z_{2}z_{3}} \times z_{4}$ and using part (i),

z_{4} = \frac{z_{1}z_{4}/\left( z_{2}z_{3} \right)}{z_{1}/\left( z_{2}z_{3} \right)}

Case 1: $\frac{z_{1}z_{4}}{z_{2}z_{3}} = i = 1\left( \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} \right)$ .

\left| z_{4} \right| = \frac{1}{\sqrt{2}/2} = \sqrt{2}

\arg\left( z_{4} \right) = \frac{\pi}{2} - \frac{5\pi}{12} = \frac{6\pi}{12} - \frac{5\pi}{12} = \frac{\pi}{12}

Case 2: $\frac{z_{1}z_{4}}{z_{2}z_{3}} = - i = 1\left( \cos\left( - \frac{\pi}{2} \right) + i\sin\left( - \frac{\pi}{2} \right) \right)$ .

\left| z_{4} \right| = \sqrt{2}

\arg\left( z_{4} \right) = - \frac{\pi}{2} - \frac{5\pi}{12} = - \frac{6\pi}{12} - \frac{5\pi}{12} = - \frac{11\pi}{12}

Since $- \frac{11\pi}{12}$ already lies in $( - \pi,\pi\rbrack$ , no adjustment is needed.

To check your answer: for Case 1,

\frac{5\pi}{12} + \frac{\pi}{12} = \frac{\pi}{2}

, the required argument of

\frac{z_{1}z_{4}}{z_{2}z_{3}}

; for Case 2,

\frac{5\pi}{12} + \left( - \frac{11\pi}{12} \right) = - \frac{\pi}{2}

. ✓

$\therefore$ $z_{4} = \sqrt{2}\left( \cos\frac{\pi}{12} + i\sin\frac{\pi}{12} \right)$ or $z_{4} = \sqrt{2}\left( \cos\left( - \frac{11\pi}{12} \right) + i\sin\left( - \frac{11\pi}{12} \right) \right)$ .

2020 H2 Maths Paper 1, Question 4

Part (i)

Part (ii)

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