Verify that ${\omega }_1=1+{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}$ is a root of this equation.

Find ${\omega }_2$ and ${\omega }_3$, expressing these in the form $a+{\mathrm{e}}^{\mathrm{i}\theta }$, where $a\in \mathrm{ℝ}$ and $\theta >0$.

Plot the points $\mathrm{A}$, $\mathrm{B}$ and $\mathrm{C}$ on an Argand diagram.

Find $\mathrm{AC}$.

By using de Moivre’s theorem, show that $\alpha =\frac{1}{1-{\mathrm{e}}^{\mathrm{i}{\displaystyle \frac{\mathrm{\pi }}{6}}}}$ is a root of this equation.

Determine the value of $\text{Re}\left(\alpha \right)$.

${\left(1+{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}-1\right)}^3$

$={\left({\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}\right)}^3$                   A1

$={\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{2}}$                  A1

$=\cos \frac{\mathrm{\pi }}{2}+\mathrm{i} \sin \frac{\mathrm{\pi }}{2}$

$=\mathrm{i}$                  AG

Note: Candidates who solve the equation correctly can be awarded the above two marks. The working for part (i) may be seen in part (ii).

[2 marks]

${\left(z-1\right)}^3={\mathrm{e}}^{\mathrm{i}\left(\frac{\mathrm{\pi }}{2}+2\mathrm{\pi k}\right)}$                  (M1)

$z-1={\mathrm{e}}^{\mathrm{i}\left(\frac{\mathrm{\pi }}{6}+\frac{4\mathrm{\pi k}}{6}\right)}$                  (M1)

$\left(k=1\right)\Rightarrow {\omega }_2=1+{\mathrm{e}}^{\mathrm{i}\frac{5\mathrm{\pi }}{6}}$                  A1

$\left(k=2\right)\Rightarrow {\omega }_3=1+{\mathrm{e}}^{\mathrm{i}\frac{9\mathrm{\pi }}{6}}$                  A1

[4 marks]

EITHER

attempt to express ${\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}$, ${\mathrm{e}}^{\mathrm{i}\frac{5\mathrm{\pi }}{6}}$, ${\mathrm{e}}^{\mathrm{i}\frac{9\mathrm{\pi }}{6}}$ in Cartesian form and translate 1 unit in the positive direction of the real axis                  (M1)

OR

attempt to express $w_1$, $w_2$ and $w_3$ in Cartesian form                  (M1)

THEN

Note: To award A marks, it is not necessary to see $\mathrm{A}$, $\mathrm{B}$ or $\mathrm{C}$, the $w_1$, or the solid lines

                  A1A1A1

[4 marks]

valid attempt to find ${\omega }_1-{\omega }_3 \left(\mathrm{or} {\omega }_3-{\omega }_1\right)$                     M1

${\omega }_1-{\omega }_3=\left(1+\frac{\sqrt{3}}{2}+\frac{1}{2}\mathrm{i}\right)-\left(1-\mathrm{i}\right)=\frac{\sqrt{3}}{2}+\frac{3}{2}\mathrm{i}$  OR  $\cos \frac{\mathrm{\pi }}{6}+\mathrm{i} \sin \frac{\mathrm{\pi }}{6}+\mathrm{i} \sin \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 2}}$

valid attempt to find $\left|\frac{\sqrt{3}}{2}+\frac{3}{2}\mathrm{i}\right|$                     M1

$=\sqrt{\frac{3}{4}+\frac{9}{4}}$

$\mathrm{AC}=\sqrt{3}$                     A1

[3 marks]

METHOD 1

${\left(z-1\right)}^3=\mathrm{i}{z}^3\Rightarrow {\left(\frac{z-1}{z}\right)}^3=\mathrm{i}$                     M1

${\left(\frac{z-1}{z}\right)}^3={\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{2}}$                     A1

$\frac{\alpha -1}{\alpha }={\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}$                     A1

Note: This step to change from $z$ to $\alpha$ may occur at any point in MS.

$\alpha -1=\alpha {\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}$

$\alpha -\alpha {\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}=1$

$\alpha \left(1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}\right)=1$

$\alpha =\frac{1}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}$                     AG

METHOD 2

${\left(z-1\right)}^3=\mathrm{i}{z}^3\Rightarrow {\left(\frac{z-1}{z}\right)}^3=\mathrm{i}$                     M1

${\left(1-\frac{1}{z}\right)}^3={\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{2}}$                     A1

$1-\frac{1}{z}={\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}$                     A1

Note: This step to change from $z$ to $\alpha$ may occur at any point in MS.

$1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}=\frac{1}{\alpha }$

$\alpha =\frac{1}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}$                     AG

METHOD 3

LHS$={\left(z-1\right)}^3={\left(\frac{1}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}-1\right)}^3$

$={\left(\frac{{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}\right)}^3$

$=\frac{\mathrm{i}}{{\left(1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}\right)}^3} \left(=\frac{\mathrm{i}}{{\displaystyle \frac{5}{2}}-{\displaystyle \frac{3\sqrt{3}}{2}}+\mathrm{i}\left(\frac{3\sqrt{3}}{2}-{\displaystyle \frac{5}{2}}\right)}\right)$                      M1A1

Note: Award M1 for applying de Moivre’s theorem (may be seen in modulus- argument form.)

RHS$=\mathrm{i}{z}^3=\mathrm{i}{\left(\frac{1}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}\right)}^3$

$=\frac{\mathrm{i}}{{\left(1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}\right)}^3}$                     A1

 ${\left(z-1\right)}^3=\mathrm{i}{z}^3$                     AG

METHOD 4

${\left(z-1\right)}^3=\mathrm{i}{z}^3$

$z^3-3z^2+3z-1=\mathrm{i}{z}^3$

$\left(1-\mathrm{i}\right)z^3-3z^2+3z-1=0$                     (M1)

$\left(1-\mathrm{i}\right){\left(\frac{1}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}\right)}^3-3{\left(\frac{1}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}\right)}^2+3\left(\frac{1}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}\right)-1$

$=\left(1-\mathrm{i}\right)-3\left(1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}\right)+3{\left(1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}\right)}^2-{\left(1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}\right)}^3$                     (A1)

$=\left(1-\mathrm{i}\right)-3\left(1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}\right)+3\left(1-2{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}+{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{3}}\right)-\left(1-3{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}+3{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{3}}-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{2}}\right)$                     A1

$=0$                     AG

Note: If the candidate does not interpret their conclusion, award (M1)(A1)A0 as appropriate.

[3 marks]

METHOD 1

$\frac{1}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}=\frac{1}{1-\left(\cos {\displaystyle \frac{\mathrm{\pi }}{6}}+\mathrm{i} \sin {\displaystyle \frac{\mathrm{\pi }}{6}}\right)}$                    M1

$=\frac{2}{2-\sqrt{3}-\mathrm{i}}$                     A1

attempt to use conjugate to rationalise                    M1

$=\frac{4-2\sqrt{3}+2\mathrm{i}}{{\left(2-\sqrt{3}\right)}^2+1}$                     A1

$=\frac{4-2\sqrt{3}+2\mathrm{i}}{8-4\sqrt{3}}$                     A1

$=\frac{1}{2}+\frac{1}{4-2\sqrt{3}}\mathrm{i}$

$\Rightarrow \text{Re}\left(\alpha \right)\mathrm{=}\frac{1}{2}$                     A1

Note: Their final imaginary part does not have to be correct in order for the final three A marks to be awarded

METHOD 2

$\frac{1}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}=\frac{1}{1-\left(\cos {\displaystyle \frac{\mathrm{\pi }}{6}}+\mathrm{i} \sin {\displaystyle \frac{\mathrm{\pi }}{6}}\right)}$                    M1

attempt to use conjugate to rationalise                    M1

$=\frac{1}{\left(1-\cos {\displaystyle \frac{\mathrm{\pi }}{6}}\right)-\mathrm{i} \sin \frac{\mathrm{\pi }}{6}}\times \frac{\left(1-\cos \frac{\mathrm{\pi }}{6}\right)+\mathrm{i} \sin \frac{\mathrm{\pi }}{6}}{\left(1-\cos \frac{\mathrm{\pi }}{6}\right)+\mathrm{i} \sin \frac{\mathrm{\pi }}{6}}$                     A1

$=\frac{\left(1-\cos \frac{\mathrm{\pi }}{6}\right)+\mathrm{i} \sin \frac{\mathrm{\pi }}{6}}{{\left(1-\cos \frac{\mathrm{\pi }}{6}\right)}^2+{\sin }^2\frac{\mathrm{\pi }}{6}}$                     A1

$=\frac{\left(1-\cos \frac{\mathrm{\pi }}{6}\right)+\mathrm{i} \sin \frac{\mathrm{\pi }}{6}}{1-2 \cos \frac{\mathrm{\pi }}{6}+{\cos }^2\frac{\mathrm{\pi }}{6}+{\sin }^2\frac{\mathrm{\pi }}{6}}$

$=\frac{\left(1-\cos \frac{\mathrm{\pi }}{6}\right)+\mathrm{i} \sin \frac{\mathrm{\pi }}{6}}{2-2 \cos \frac{\mathrm{\pi }}{6}}$                     A1

$=\frac{1}{2}+\frac{\mathrm{i} \sin \frac{\mathrm{\pi }}{6}}{2-2 \cos \frac{\mathrm{\pi }}{6}}$

$\Rightarrow \text{Re}\left(\alpha \right)\mathrm{=}\frac{1}{2}$                     A1

Note: Their final imaginary part does not have to be correct in order for the final three A marks to be awarded

METHOD 3

attempt to multiply through by $-\frac{{\mathrm{e}}^{-\mathrm{i}\frac{\mathrm{\pi }}{12}}}{{\mathrm{e}}^{-\mathrm{i}\frac{\mathrm{\pi }}{12}}}$                    M1

$\frac{1}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}=-\frac{{\mathrm{e}}^{-\mathrm{i}\frac{\mathrm{\pi }}{12}}}{{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{12}}-{\mathrm{e}}^{-\mathrm{i}\frac{\mathrm{\pi }}{12}}}$                     A1

attempting to re-write in r-cis form                    M1

$=-\frac{\cos \left(-{\displaystyle \frac{\mathrm{\pi }}{12}}\right)+\mathrm{i} \sin \left(-{\displaystyle \frac{\mathrm{\pi }}{12}}\right)}{\cos {\displaystyle \frac{\mathrm{\pi }}{12}}+\mathrm{i} \sin {\displaystyle \frac{\mathrm{\pi }}{12}}-\left(\cos \left(-\frac{\mathrm{\pi }}{12}\right)+\mathrm{i} \sin \left(-\frac{\mathrm{\pi }}{12}\right)\right)}$                     A1

$=-\frac{\cos \frac{\mathrm{\pi }}{12}-\mathrm{i} \sin \frac{\mathrm{\pi }}{12}}{2\mathrm{i} \sin {\displaystyle \frac{\mathrm{\pi }}{12}}}$                     A1

$=\frac{1}{2}-\frac{1}{2\mathrm{i}}\cot \frac{\mathrm{\pi }}{12} \left(=\frac{1}{2}+\frac{1}{2}\mathrm{i} \cot \frac{\mathrm{\pi }}{12}\right)$

$\Rightarrow \text{Re}\left(\alpha \right)\mathrm{=}\frac{1}{2}$                     A1

METHOD 4

attempt to multiply through by $\frac{1-{\mathrm{e}}^{-\mathrm{i}\frac{\mathrm{\pi }}{6}}}{1-{\mathrm{e}}^{-\mathrm{i}\frac{\mathrm{\pi }}{6}}}$                    M1

$\frac{1}{1-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}}=\frac{1-{\mathrm{e}}^{-\mathrm{i}\frac{\mathrm{\pi }}{6}}}{1-{\mathrm{e}}^{-\mathrm{i}\frac{\mathrm{\pi }}{6}}-{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{6}}+1}$                     A1

attempting to re-write in r-cis form                    M1

$=\frac{1-\cos \frac{\mathrm{\pi }}{6}-\mathrm{i} \sin \frac{\mathrm{\pi }}{6}}{2-2 \cos {\displaystyle \frac{\mathrm{\pi }}{6}}}$                    A1

attempt to re-write in Cartesian form                    M1

$=\frac{1-{\displaystyle \frac{\sqrt{3}}{2}}-{\displaystyle \frac{1}{2}}\mathrm{i}}{2-\sqrt{3}} \left(=\frac{{\displaystyle \frac{2-\sqrt{3}}{2}}}{2-\sqrt{3}}+\mathrm{i}\frac{{\displaystyle \frac{1}{2}}}{2-\sqrt{3}}\right)$

$\Rightarrow \text{Re}\left(\alpha \right)\mathrm{=}\frac{1}{2}$                     A1

Note: Their final imaginary part does not have to be correct in order for the final A mark to be awarded

[6 marks]