• This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure. METHOD 1 ${w^3} = 8{\text{j}}$ writing$8{\text{j}} = 8\left( {{\text{cos}}\left( {\frac{\pi }{2} + 2\pi k} \right) + {\text{j}}\,{\text{sin}}\left( {\frac{\pi }{2} + 2\pi k} \right)} \right)$(M1) Note: Award M1 for an attempt to find cube roots of $w$ using modulus-argument form. cube roots$w = 2\left( {{\text{cos}}\left( {\frac{{\frac{\pi }{2} + 2\pi k}}{3}} \right) + {\text{j}}\,{\text{sin}}\left( {\frac{{\frac{\pi }{2} + 2\pi k}}{3}} \right)} \right)$(M1) i.e. $w = \sqrt 3 + {\text{j,}}\,\, - \sqrt 3 + {\text{j,}}\,\, - 2{\text{j}}$A2 Note: Award *A2 *for all 3 correct, A1 for 2 correct. Note: Accept$w = 1.73 + {\text{j}}$ and$w = - 1.73 + {\text{j}}$.

METHOD2 ${w^3} + {\left( {2{\text{j}}} \right)^3} = 0$ $\left( {w + 2{\text{j}}} \right)\left( {{w^2} - 2w{\text{j}} - 4} \right) = 0$M1 $w = \frac{{2{\text{j}} \pm \sqrt {12} }}{2}$M1 $w = \sqrt 3 + {\text{j,}}\,\, - \sqrt 3 + {\text{j,}}\,\, - 2{\text{j}}$A2 Note:AwardA2for all 3 correct,A1for 2 correct. **Note:**Accept$w = 1.73 + {\text{j}}$ and$w = - 1.73 + {\text{j}}$.

[4 marks]

${w_1} = - 2{\text{j}}$ $\frac{z}{{z - {\text{j}}}} = - 2{\text{j}}$*** M1*** $z = - 2{\text{j}}\left( {z - {\text{j}}} \right)$ $z\left( {1 + 2{\text{j}}} \right) = - 2$ $z = \frac{{ - 2}}{{1 + 2{\text{j}}}}$A1 $z = - \frac{2}{5} + \frac{4}{5}{\text{j}}$A1 Note: Accept $a = - \frac{2}{5}{\text{,}}\,\,b = \frac{4}{5}$. [3 marks]