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 \(w = \frac{1}{{z + {z^{ - 1}}}}\).