Show that $(\omega -1)({\omega }^2+\omega +1)={\omega }^3-1$.

Hence, deduce that ${\omega }^2+\omega +1=0$.

Show that ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2=3$.

By factorizing $z^4-1$, or otherwise, deduce that ${\omega }^3+{\omega }^2+\omega +1=0$.

Show that ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2\times {\text{P}}_0{\text{P}}_3=4$.

Suggest a value for ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2\times \dots \times {\text{P}}_0{\text{P}}_{n-1}$.

Write down expressions for ${\text{P}}_0{\text{P}}_2$ and ${\text{P}}_0{\text{P}}_3$ in terms of $\omega$.

Hence, write down an expression for ${\text{P}}_0{\text{P}}_{n-1}$ in terms of $n$ and $\omega$.

Express $z^{n-1}+ z^{n-2}+ \dots +z+1$ as a product of linear factors over the set $\mathrm{ℂ}$.

Hence, using the part (g)(i) and part (f) results, or otherwise, prove your suggested result to part (e).

METHOD 1

attempts to expand $(\omega -1)({\omega }^2+\omega +1)$            (M1)

$={\omega }^3+{\omega }^2+\omega -{\omega }^2-\omega -1$           A1

$={\omega }^3-1$           AG

METHOD 2

attempts polynomial division on $\frac{{\omega }^3-1}{\omega -1}$            M1

$={\omega }^2+\omega +1$           A1

so $(\omega -1)({\omega }^2+\omega +1)={\omega }^3-1$           AG

Note: In part (a), award marks as appropriate where $\omega$ has been converted into Cartesian, modulus-argument (polar) or Euler form.

[2 marks]

(since $\omega$ is a root of $z^3=1$)$\Rightarrow {\omega }^3-1=0$           R1

and $\omega \ne 1$           R1

$\Rightarrow {\omega }^2+\omega +1=0$           AG

Note: In part (a), award marks as appropriate where $\omega$ has been converted into Cartesian, modulus-argument (polar) or Euler form.

[2 marks]

METHOD 1

attempts to find either  ${\text{P}}_0{\text{P}}_1$ or ${\text{P}}_0{\text{P}}_2$             (M1)

accept any valid method

e.g.  $2 \sin \frac{\mathrm{\pi }}{3}, 1^2+1^2-2 \cos \frac{2\mathrm{\pi }}{3}, \frac{1}{\sin {\displaystyle \frac{\mathrm{\pi }}{6}}}=\frac{{\text{P}}_0{\text{P}}_1}{\sin {\displaystyle \frac{2\mathrm{\pi }}{3}}}$from either ${\text{ΔOP}}_0{\text{P}}_1$ or ${\text{ΔOP}}_0{\text{P}}_2$

e.g. use of Pythagoras’ theorem

e.g. $\left|1-{\text{e}}^{\text{i}\frac{2\mathrm{\pi }}{3}}\right|, \left|1-\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\text{i}\right)\right|$ by calculating the distance between $2$ points

${\text{P}}_0{\text{P}}_1\mathrm{=}\sqrt{3}$           A1

${\text{P}}_0{\text{P}}_2\mathrm{=}\sqrt{3}$           A1

Note: Award a maximum of M1A1A0 for any decimal approximation seen in the calculation of either ${\text{P}}_0{\text{P}}_1$ or ${\text{P}}_0{\text{P}}_2$ or both.

so ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2=3$            AG

METHOD 2

attempts to find ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2=\left|1-\omega \right|\left|1-{\omega }^2\right|$             (M1)

${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2=\left|{\omega }^3-{\omega }^2-\omega +1\right|$           A1

$=\left|1-\left({\omega }^2+\omega +1\right)+2\right|$  and since ${\omega }^2+\omega +1=0$           R1

so ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2=3$            AG

[3 marks]

METHOD 1

$z^4-1=\left(z-1\right)\left(z^3+z^2+z+1\right)$           A1

($\omega$ is a root hence) ${\omega }^4-1=0$ and $\omega \ne 1$           R1

$\Rightarrow {\omega }^3+{\omega }^2+\omega +1=0$            AG

Note: Condone the use of $\omega$ throughout.

METHOD 2

considers the sum of roots of $z^4-1=0$             (M1)

the sum of roots is zero (there is no $z^3$ term)         A1

$\Rightarrow {\omega }^3+{\omega }^2+\omega +1=0$            AG

METHOD 3

substitutes for $\omega$             (M1)

e.g. $\text{LHS}={\text{e}}^{\text{i}\frac{3\text{π}}{2}}+{\text{e}}^{\mathrm{\pi i}}+{\text{e}}^{\text{i}\frac{\text{π}}{2}}+1$

$=-\text{i}-1+\text{i}+1$         A1

Note: This can be demonstrated geometrically or by using vectors. Accept Cartesian or modulus-argument (polar) form.

$\Rightarrow {\omega }^3+{\omega }^2+\omega +1=0$            AG

METHOD 4

${\omega }^3+{\omega }^2+\omega +1=\frac{{\omega }^4-1}{\omega -1}$         A1

$=\frac{0}{\omega -1}=0$ as $\omega \ne 1$           R1

$\Rightarrow {\omega }^3+{\omega }^2+\omega +1=0$            AG

[2 marks]

METHOD 1

${\text{P}}_0{\text{P}}_{\mathrm{2}}=2$           A1

attempts to find either ${\text{P}}_0{\text{P}}_1$ or ${\text{P}}_0{\text{P}}_3$             (M1)

Note: For example ${\text{P}}_0{\text{P}}_1\mathrm{=}\left|1-\mathrm{i}\right|$ and ${\text{P}}_0{\text{P}}_3\mathrm{=}\left|1+\mathrm{i}\right|$.
         Various geometric and trigonometric approaches can be used by candidates.

${\text{P}}_0{\text{P}}_1\mathrm{=}\sqrt{2}, {\text{P}}_0{\text{P}}_3\mathrm{=}\sqrt{2}$           A1A1

Note: Award a maximum of A1M1A1A0 if labels such as ${\text{P}}_0{\text{P}}_1$ are not clearly shown.
         Award full marks if the lengths are shown on a clearly labelled diagram.
         Award a maximum of A1M1A1A0 for any decimal approximation seen in the calculation of either ${\text{P}}_0{\text{P}}_1$ or ${\text{P}}_0{\text{P}}_3$ or both.

${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2\times {\text{P}}_0{\text{P}}_3=4$            AG

METHOD 2

attempts to find ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2\times {\text{P}}_0{\text{P}}_3=\left|1-\omega \right|\left|1-{\omega }^2\right|\left|1-{\omega }^3\right|$            M1

${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2\times {\text{P}}_0{\text{P}}_3=\left|-{\omega }^6+{\omega }^5+{\omega }^4-{\omega }^2-\omega +1\right|$         A1

$=\left|-\left(-1\right)+{\omega }^5+1-\left(-1\right)-\omega +1\right|$  since ${\omega }^6={\omega }^2=-1$ and ${\omega }^4=1$        A1

$=\left|{\omega }^5-\omega +4\right|$ and since ${\omega }^5=\omega$           R1

so ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2\times {\text{P}}_0{\text{P}}_3=4$            AG

METHOD 3

${\text{P}}_0{\text{P}}_{\mathrm{2}}=2$           A1

attempts to find ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_3=\left|1-\omega \right|\left|1-{\omega }^3\right|$            M1

${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_3=\left|{\omega }^4-{\omega }^3-\omega +1\right|$         A1

$=\left|2-\left(-\omega \right)-\omega \right|$  since ${\omega }^4=1$ and ${\omega }^3=-\omega$               R1     

so ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2\times {\text{P}}_0{\text{P}}_3=4$            AG

[4 marks]

$\left({\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2\times \dots \times {\text{P}}_0{\text{P}}_{n-1}\right)=n$         A1

[1 mark]

${\text{P}}_0{\text{P}}_2=\left|1-{\omega }^2\right|, {\text{P}}_0{\text{P}}_3=\left|1-{\omega }^3\right|$         A1A1

[2 marks]

${\text{P}}_0{\text{P}}_{n-1}=\left|1-{\omega }^{n-1}\right|$         A1A1

 
Note: Accept $\left|1-\omega \right|$ from symmetry.

[1 mark]

$z^n-1=\left(z-1\right)\left(z^{n-1}+ z^{n-2}+ \dots +z+1\right)$

considers the equation $z^{n-1}+ z^{n-2}+ \dots +z+1=0$         (M1)

the roots are $\omega , {\omega }^2, \dots , {\omega }^{n-1}$         (A1)

so $\left(z-\omega \right)\left(z-{\omega }^2\right)\dots \left(z-{\omega }^{n-1}\right)$         A1

[3 marks]

METHOD 1

substitutes $z=1$into $\left(z-\omega \right)\left(z-{\omega }^2\right)\dots \left(z-{\omega }^{n-1}\right)\equiv z^{n-1}+ z^{n-2}+ \dots +z+1$           M1

$\left(1-\omega \right)\left(1-{\omega }^2\right)\dots \left(1-{\omega }^{n-1}\right)=n$         (A1)

takes modulus of both sides           M1

$\left|\left(1-\omega \right)\left(1-{\omega }^2\right)\dots \left(1-{\omega }^{n-1}\right)\right|=\left|n\right|$

$\left|1-\omega \right|\left|1-{\omega }^2\right|\dots \left|1-{\omega }^{n-1}\right|=n$                 A1

so ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2\times \dots \times {\text{P}}_0{\text{P}}_{n-1}=n$                 AG

Note: Award a maximum of M1A1FTM1A0 from part (e).

METHOD 2

$\left(1-\omega \right), \left(1-{\omega }^2\right), \dots , \left(1-{\omega }^{n-1}\right)$ are the roots of ${\left(1-v\right)}^{n-1}+{\left(1-v\right)}^{n-2}+ \dots + \left(1-v\right)+1=0$           M1

coefficient of $v^{n-1}$ is ${\left(-1\right)}^{n-1}$ and the coefficient of $1$ is $n$                 A1

product of the roots is $\frac{{\left(-1\right)}^{n-1}n}{{\left(-1\right)}^{n-1}}=n$                 A1

$\left|1-\omega \right|\left|1-{\omega }^2\right|\dots \left|1-{\omega }^{n-1}\right|=n$                 A1

so ${\text{P}}_0{\text{P}}_1\times {\text{P}}_0{\text{P}}_2\times \dots \times {\text{P}}_0{\text{P}}_{n-1}=n$                 AG

[4 marks]