A complex number z is given by $z = \frac{{a + {\text{i}}}}{{a - {\text{i}}}},{\text{ }}a \in \mathbb{R}$.

(a)     Determine the set of values of a such that

          (i)     z is real;

          (ii)     z is purely imaginary.

(b)     Show that $\left| z \right|$ is constant for all values of a.

(a)     $\frac{{a + {\text{i}}}}{{a - {\text{i}}}} \times \frac{{a + {\text{i}}}}{{a + {\text{i}}}}$     M1

$= \frac{{{a^2} - 1 + 2a{\text{i}}}}{{{a^2} + 1}}{\text{ }}\left( { = \frac{{{a^2} - 1}}{{{a^2} + 1}} + \frac{{2a}}{{{a^2} + 1}}{\text{i}}} \right)$     A1

          (i)     z is real when $a = 0$     A1

          (ii)     z is purely imaginary when $a =  \pm 1$     A1

Note:     Award M1A0A1A0 for $\frac{{{a^2} - 1 + 2a{\text{i}}}}{{{a^2} - 1}}{\text{ }}\left( { = 1 + \frac{{2a}}{{{a^2} - 1}}{\text{i}}} \right)$ leading to $a = 0$ in (i).

[4 marks]

(b)     METHOD 1

attempting to find either $\left| z \right|$ or ${\left| z \right|^2}$ by expanding and simplifying

eg ${\left| z \right|^2} = \frac{{{{\left( {{a^2} - 1} \right)}^2} + 4{a^2}}}{{{{\left( {{a^2} + 1} \right)}^2}}} = \frac{{{a^4} + 2{a^2} + 1}}{{{{\left( {{a^2} + 1} \right)}^2}}}$     M1

$= \frac{{{{\left( {{a^2} + 1} \right)}^2}}}{{{{\left( {{a^2} + 1} \right)}^2}}}$

${\left| z \right|^2} = 1 \Rightarrow \left| z \right| = 1$     A1

METHOD 2

$\left| z \right| = \frac{{\left| {a + {\text{i}}} \right|}}{{\left| {a - {\text{i}}} \right|}}$     M1

$\left| z \right| = \frac{{\sqrt {{a^2} + 1} }}{{\sqrt {{a^2} + 1} }} \Rightarrow \left| z \right| = 1$     A1

[2 marks]

Total [6 marks]

Part (a) was reasonably well done. When multiplying and dividing by the conjugate of $a - {\text{i}}$, some candidates incorrectly determined their denominator as ${a^2} - 1$.

In part (b), a significant number of candidates were able to correctly expand and simplify $\left| z \right|$ although many candidates appeared to not understand the definition of $\left| z \right|$.