2016 H2 Maths Paper 2 Q4 — Worked Solution

Part (a)(i)

2 marks

The locus $|z - 3 - i| = 1$ is a circle centre $3 + i$ , radius $1$ . The locus $\arg z = \alpha$ , where $\tan\alpha = 0.4$ , is a half-line (ray) from the origin, in the first quadrant since $\alpha$ is acute, at angle $\alpha \approx {21.8}^{\circ}$ to the positive real axis. Since $\left| z_{1} \right| < \left| z_{2} \right|$ , $z_{1}$ is the nearer intersection point to the origin.

Part (a)(ii)

4 marks

Approach

$z_{1}$ and $z_{2}$ both lie on the circle $|z - 3 - i| = 1$ , so any point equidistant from $z_{1}$ and $z_{2}$ lies on the perpendicular bisector of the chord $z_{1}z_{2}$ — and that perpendicular bisector passes through the centre $3 + i$ , since the centre is itself equidistant from every point on the circle. The chord $z_{1}z_{2}$ lies along the ray at angle $\alpha$ , so the perpendicular bisector through the centre has direction $\alpha + \frac{\pi}{2}$ ; the required points are where this line, extended both ways from the centre, meets the circle again — i.e. the centre displaced by $\text{±}1$ (the radius) in the direction $\alpha + \frac{\pi}{2}$ .

Since $\tan\alpha = 0.4 = \frac{2}{5}$ , take a right triangle with opposite $2$ , adjacent $5$ , hypotenuse $\sqrt{4 + 25} = \sqrt{29}$ :

\sin\alpha = \frac{2}{\sqrt{29}},\quad\quad\cos\alpha = \frac{5}{\sqrt{29}}

A unit vector in the direction $\alpha + \frac{\pi}{2}$ is $\left( - \sin\alpha,\ \cos\alpha \right)$ , since rotating by $\frac{\pi}{2}$ sends $\left( \cos\alpha,\sin\alpha \right) \mapsto \left( - \sin\alpha,\cos\alpha \right)$ .

The required points of $|z - 3 - i| = 1$ are the centre $3 + i$ displaced by $\text{±}1$ in this direction:

z = (3 + i) \pm i\left( \cos\alpha + i\sin\alpha \right)

= 3 + i \pm \left( - \sin\alpha + i\cos\alpha \right)

= \left( 3 \mp \frac{2}{\sqrt{29}} \right) + i\left( 1 \pm \frac{5}{\sqrt{29}} \right)

$\therefore$ the two values of $z$ are

z = \left( 3 - \frac{2}{\sqrt{29}} \right) + i\left( 1 + \frac{5}{\sqrt{29}} \right)\quad\text{and}\quad z = \left( 3 + \frac{2}{\sqrt{29}} \right) + i\left( 1 - \frac{5}{\sqrt{29}} \right).

To check your answer: both points satisfy

|z - 3 - i| = \left| \pm \left( - \sin\alpha + i\cos\alpha \right) \right| = \sqrt{\sin^{2}\alpha + \cos^{2}\alpha} = 1

, confirming they lie on the given circle. ✓

Part (b)(i)

3 marks

Approach

To find cube roots we write $w$ in polar form first, then use $w^{1/3} = r^{1/3}e^{i(\theta + 2k\pi)/3}$ for $k = 0, \pm 1$ , choosing $k$ so that each resulting argument lands in $( - \pi,\pi\rbrack$ .

Since $w = 2 - 2i$ :

r = |w| = \sqrt{2^{2} + 2^{2}} = 2\sqrt{2}

$w$ lies in the fourth quadrant, with $\tan\theta = \frac{- 2}{2} = - 1$ , so $\theta = - \frac{\pi}{4}$ .

\therefore w = 2\sqrt{2}\, e^{- i\frac{\pi}{4}}

The cube roots of $w$ have modulus $r^{1/3} = \left( 2\sqrt{2} \right)^{1/3} = \left( 2^{3/2} \right)^{1/3} = 2^{1/2} = \sqrt{2}$ , and arguments $\frac{\theta + 2k\pi}{3}$ for $k = 0, \pm 1$ :

k = 0:\quad\frac{- \pi/4}{3} = - \frac{\pi}{12}

k = 1:\quad\frac{- \pi/4 + 2\pi}{3} = \frac{7\pi}{12}

k = - 1:\quad\frac{- \pi/4 - 2\pi}{3} = - \frac{3\pi}{4}

All three arguments lie in $( - \pi,\pi\rbrack$ .

$\therefore$ the cube roots of $w$ are

\sqrt{2}\, e^{- i\frac{\pi}{12}},\quad\quad\sqrt{2}\, e^{i\frac{7\pi}{12}},\quad\quad\sqrt{2}\, e^{- i\frac{3\pi}{4}}.

Part (b)(ii)

3 marks

Approach

We write $w^{\text{*}}w^{n}$ in polar form so its argument is a linear expression in $n$ , then find the smallest positive integer $n$ for which that argument reduces to $\frac{\pi}{2}$ inside the principal range, remembering to add or subtract multiples of $2\pi$ as needed.

Since $w = 2\sqrt{2}\, e^{- i\pi/4}$ , its conjugate is $w^{\text{*}} = 2\sqrt{2}\, e^{i\pi/4}$ .

w^{\text{*}}w^{n} = 2\sqrt{2}\, e^{i\pi/4} \cdot \left( 2\sqrt{2} \right)^{n}e^{- in\pi/4}

= \left( 2\sqrt{2} \right)^{n + 1}\, e^{i\frac{\pi}{4}(1 - n)}

We require $\arg\left( w^{\text{*}}w^{n} \right) = \frac{\pi}{2}$ , i.e.

\frac{\pi}{4}(1 - n) = \frac{\pi}{2} + 2k\pi,\quad k \in \mathbb{Z}

1 - n = 2 + 8k

n = - 1 - 8k

For $n$ to be a smallest positive whole number, take $k = - 1$ :

n = - 1 - 8( - 1) = 7

To check your answer: at

n = 7

,

\frac{\pi}{4}(1 - 7) = - \frac{3\pi}{2}

, and

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

, the required argument. ✓

$\therefore$ the smallest positive whole number value of $n$ is $7$ .

2016 H2 Maths Paper 2, Question 4

Part (a)(i)

Part (a)(ii)

Part (b)(i)

Part (b)(ii)

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