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]