Consider $w = \frac{z}{{{z^2} + 1}}{\text{ where }}z = x + {\text{i}}y{\text{ , }}y \ne 0{\text{ and }}{z^2} + 1 \ne 0$ .

Given that $\operatorname{Im} w = 0$, show that $\left| z \right| = 1$.

METHOD 1

Substituting $z = x + {\text{i}}y$ to obtain $w = \frac{{x + y{\text{i}}}}{{{{(x + y{\text{i}})}^2} + 1}}$     (A1)

$w = \frac{{x + y{\text{i}}}}{{{x^2} - {y^2} + 1 + 2xy{\text{i}}}}$     A1

Use of $({x^2} - {y^2} + 1 + 2xy{\text{i)}}$ to make the denominator real.     M1

${\text{ = }}\frac{{(x + y{\text{i)}}({x^2} - {y^2} + 1 - 2xy{\text{i)}}}}{{{{({x^2} - {y^2} + 1)}^2} + 4{x^2}{y^2}}}$     A1

$\operatorname{Im} w = \frac{{y({x^2} - {y^2} + 1) - 2{x^2}y}}{{{{({x^2} - {y^2} + 1)}^2} + 4{x^2}{y^2}}}$     (A1)

$= \frac{{y(1 - {x^2} - {y^2})}}{{{{({x^2} - {y^2} + 1)}^2} + 4{x^2}{y^2}}}$     A1

$\operatorname{Im} w = 0 \Rightarrow 1 - {x^2} - {y^2} = 0$ i.e. $\left| z \right| = 1{\text{ as }}y \ne 0$     R1AG     N0

[7 marks]

METHOD 2

$w({z^2} + 1) = z$     (A1)

$w({x^2} - {y^2} + 1 + 2{\text{i}}xy) = x + yi$     A1

Equating real and imaginary parts

$w({x^2} - {y^2} + 1) = x{\text{ and }}2wx = 1,{\text{ }}y \ne 0$     M1A1

Substituting $w = \frac{1}{{2x}}$ to give $\frac{x}{2} - \frac{{{y^2}}}{{2x}} + \frac{1}{{2x}} = x$     A1

$- \frac{1}{{2x}}({y^2} - 1) = \frac{x}{2}$ or equivalent     (A1)

${x^2} + {y^2} = 1$, i.e. $\left| z \right| = 1{\text{ as }}y \ne 0$     R1AG

[7 marks]

This was a difficult question that troubled most candidates. Most candidates were able to substitute z = x + yi into w but were then unable to make any further meaningful progress. Common errors included not expanding ${(x + {\text{i}}y)^2}$ correctly or not using a correct complex conjugate to make the denominator real. A small number of candidates produced correct solutions by using $w = \frac{1}{{z + {z^{ - 1}}}}$.