22M.1.AHL.TZ1.9

pestleMathematicsAAHLPaper 122M· ahl-1-12-complex-numbers-cartesian-form-and-argand-diag, sl-3-5-unit-circle-definitions-of-sin-cos-tan-exact-trig-ratios-ambiguous-case-of-sine-rulesource ↗

Consider the complex numbers $z_{1} = 1 + b i$ and $z_{2} = (1 - b^{2}) - 2 b i$ , where $b \in ℝ, b \neq 0$ .

Find an expression for $z_{1} z_{2}$ in terms of $b$ .

[3]

Hence, given that $arg (z_{1} z_{2}) = \frac{π}{4}$ , find the value of $b$ .

[3]

Markscheme / solution

$z_{1} z_{2} = (1 + b i) ((1 - b^{2}) - (2 b) i)$

$= (1 - b^{2} - 2 i^{2} b^{2}) + i (- 2 b + b - b^{3})$ M1

$= (1 + b^{2}) + i (- b - b^{3})$ A1A1

Note: Award A1 for $1 + b^{2}$ and A1 for $- b i - b^{3} i$ .

[3 marks]

$arg (z_{1} z_{2}) = arctan (\frac{- b - b^{3}}{1 + b^{2}}) = \frac{π}{4}$ (M1)

EITHER
$arctan (- b) = \frac{π}{4}$ (since $1 + b^{2} \neq 0$ , for $b \in ℝ$ ) A1

$- b - b^{3} = 1 + b^{2}$ (or equivalent) A1

THEN

$b = - 1$ A1

[3 marks]

Examiners’ report

Part (a) was generally well done with many completely correct answers seen. Part (b) proved to be challenging with many candidates incorrectly equating the ratio of their imaginary and real parts to $\frac{π}{4}$ instead of $\tan \frac{π}{4}$ . Stronger candidates realized that when $θ = \frac{π}{4}$ , it forms an isosceles right-angled triangle and equated the real and imaginary parts to obtain the value of b .

[N/A]