(a)     Use de Moivre’s theorem to find the roots of the equation ${z^4} = 1 - {\text{i}}$ .

(b)     Draw these roots on an Argand diagram.

(c)     If ${{\text{z}}_1}$ is the root in the first quadrant and ${{\text{z}}_2}$ is the root in the second quadrant, find $\frac{{{{\text{z}}_2}}}{{{{\text{z}}_1}}}$ in the form a + ib .

(a)     Expand and simplify $(x - 1)({x^4} + {x^3} + {x^2} + x + 1)$ .

(b)     Given that b is a root of the equation ${z^5} - 1 = 0$ which does not lie on the real axis in the Argand diagram, show that $1 + b + {b^2} + {b^3} + {b^4} = 0$ .

(c)     If $u = b + {b^4}$ and $v = {b^2} + {b^3}$ show that

(i)     u + v = uv = −1;

(ii)     $u - v = \sqrt 5$ , given that $u - v > 0$ .

(a)     $z = {(1 - {\text{i}})^{\frac{1}{4}}}$

Let $1 - {\text{i}} = r(\cos \theta  + {\text{i}}\sin \theta )$

$\Rightarrow r = \sqrt 2$     A1

$\theta = - \frac{\pi }{4}$     A1

$z = {\left( {\sqrt 2 \left( {\cos \left( { - \frac{\pi }{4}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{4}} \right)} \right)} \right)^{\frac{1}{4}}}$     M1

$= {\left( {\sqrt 2 \left( {\cos \left( { - \frac{\pi }{4} + 2n\pi } \right) + {\text{i}}\sin \left( { - \frac{\pi }{4} + 2n\pi } \right)} \right)} \right)^{\frac{1}{4}}}$

$= {2^{\frac{1}{8}}}\left( {\cos \left( { - \frac{\pi }{{16}} + \frac{{n\pi }}{2}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{{16}} + \frac{{n\pi }}{2}} \right)} \right)$     M1

$= {2^{\frac{1}{8}}}\left( {\cos \left( { - \frac{\pi }{{16}}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{{16}}} \right)} \right)$

Note: Award M1 above for this line if the candidate has forgotten to add $2\pi$ and no other solution given.

$= {2^{\frac{1}{8}}}\left( {\cos \left( {\frac{{7\pi }}{{16}}} \right) + {\text{i}}\sin \left( {\frac{{7\pi }}{{16}}} \right)} \right)$

$= {2^{\frac{1}{8}}}\left( {\cos \left( {\frac{{15\pi }}{{16}}} \right) + {\text{i}}\sin \left( {\frac{{15\pi }}{{16}}} \right)} \right)$

$= {2^{\frac{1}{8}}}\left( {\cos \left( { - \frac{{9\pi }}{{16}}} \right) + {\text{i}}\sin \left( { - \frac{{9\pi }}{{16}}} \right)} \right)$     A2

Note: Award A1 for 2 correct answers. Accept any equivalent form.

[6 marks]

(b)

    A2

Note: Award A1 for roots being shown equidistant from the origin and one in each quadrant.

A1 for correct angular positions. It is not necessary to see written evidence of angle, but must agree with the diagram.

[2 marks]

(c)     $\frac{{{z_2}}}{{{z_1}}} = \frac{{{2^{\frac{1}{8}}}\left( {\left( {\cos \frac{{15\pi }}{{16}}} \right) + {\text{i}}\sin \left( {\frac{{15\pi }}{{16}}} \right)} \right)}}{{{2^{\frac{1}{8}}}\left( {\left( {\cos \frac{{7\pi }}{{16}}} \right) + {\text{i}}\sin \left( {\frac{{7\pi }}{{16}}} \right)} \right)}}$     M1A1

$= \cos \frac{\pi }{2} + {\text{i}}\sin \frac{\pi }{2}$     (A1)

$= {\text{i}}$     A1     N2

${\text{(}} \Rightarrow a = 0,{\text{ }}b = 1)$

[4 marks] 

(a)     $(x - 1)({x^4} + {x^3} + {x^2} + x + 1)$

$= {x^5} + {x^4} + {x^3} + {x^2} + x - {x^4} - {x^3} - {x^2} - x - 1$     (M1)

$= {x^5} - 1$     A1

[2 marks]

(b)     b is a root

$f(b) = 0$

${b^5} = 1$     M1

${b^5} - 1 = 0$     A1

$(b - 1)({b^4} + {b^3} + {b^2} + b + 1) = 0$

$b \ne 1$     R1

$1 + b + {b^2} + {b^3} + {b^4} = 0$ as shown.     AG

[3 marks]

(c)     (i)     $u + v = {b^4} + {b^3} + {b^2} + b = - 1$     A1

$uv = (b + {b^4})({b^2} + {b^3}) = {b^3} + {b^4} + {b^6} + {b^7}$     A1

Now ${b^5} = 1$     (A1)

Hence $uv = {b^3} + {b^4} + b + {b^2} = - 1$     A1

Hence $u + v = uv = - 1$     AG

(ii)     ${(u - v)^2} = ({u^2} + {v^2}) - 2uv$     (M1)

$= \left( {{{(u + v)}^2} - 2uv} \right) - 2uv\,\,\,\,\,\left( { = {{(u + v)}^2} - 4uv} \right)$     (M1)A1

Given $u - v > 0$ 

$u - v = \sqrt {{{(u + v)}^2} - 4uv}$

$= \sqrt {{{( - 1)}^2} - 4( - 1)}$

$= \sqrt {1 + 4}$     A1

$= \sqrt 5$     AG

Note: Award A0 unless an indicator is given that $u - v = - \sqrt 5$ is invalid.

[8 marks]

Total [13 marks]

The response to Part A was disappointing. Many candidates did not know that they had to apply de Moivre’s theorem and did not appreciate that they needed to find four roots.

Part B started well for most candidates, but in part (b) many candidates did not appreciate the significance of b not lying on the real axis. A majority of candidates started (c) (i) and many fully correct answers were seen. Part (c) (ii) proved unsuccessful for all but the very best candidates.