Calculate \(\frac{{{z_1}}}{{{z_2}}}\) giving your answer both in modulus-argument form and Cartesian form.

Using your results, find the exact value of tan 75° , giving your answer in the form \(a + \sqrt b \) , a , \(b \in {\mathbb{Z}^ + }\) .

in Cartesian form

\({z_1} = 2 \times - \frac{{\sqrt 3 }}{2} + 2 \times \frac{1}{2}{\text{i}}\)     M1

\( = - \sqrt 3 + {\text{i}}\)     A1

\(\frac{{{z_1}}}{{{z_2}}} = \frac{{ - \sqrt 3 + {\text{i}}}}{{ - 1 + {\text{i}}}}\)

\( = \frac{{\left( { - \sqrt 3 + {\text{i}}} \right)}}{{( - 1 + {\text{i}})}} \times \frac{{( - 1 - {\text{i}})}}{{( - 1 - {\text{i}})}}\)     M1

\( = \frac{{1 + \sqrt 3 }}{2} + \frac{{\left( {\sqrt 3 - 1} \right)}}{2}{\text{i}}\)     A1

in modulus-argument form

\({{\text{z}}_2} = \sqrt 2 {\text{cis}}135^\circ \)     A1

\(\frac{{{z_1}}}{{{z_2}}} = \frac{{2{\text{cis}}150^\circ }}{{\sqrt 2 {\text{cis}}135^\circ }}\)

\( = \sqrt 2 {\text{cis}}15^\circ \)     A1A1

[7 marks]

equating the two expressions for \(\frac{{{z_1}}}{{{z_2}}}\)

\(\cos 15^\circ = \frac{{1 + \sqrt 3 }}{{2\sqrt 2 }}\)     A1

\(\sin 15^\circ = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\)     A1

\(\tan 75^\circ = \frac{{\cos 15^\circ }}{{\sin 15^\circ }} = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\)     M1

\( = \frac{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}\)     A1

\( = 2 + \sqrt 3 \)     A1

[5 marks]