# IBDP Maths analysis and approaches Topic: AHL 1.14 :Conjugate roots of polynomial equations HL Paper 1

### Question:

In the following Argand diagram, the points Z1 , O and Z2 are the vertices of triangle Z1OZ2 described anticlockwise.

The point Z1 represents the complex number z1 = r1e , where r1 > 0. The point Z2 represents the complex number z2 = r2e , where r2 > 0.

Angles α, θ are measured anticlockwise from the positive direction of the real axis such that 0 ≤ α, θ < 2π and 0 < α – θ < π.

(a) Show that z1z2 * = r1r2e i(α – θ) where z2* is the complex conjugate of z2 .

Ans:

Note: Accept working in modulus-argument form

(b) Given that Re (z1 z2*) = 0, show that Z1OZ2 is a right-angled triangle.
In parts (c), (d) and (e), consider the case where Z1OZ2 is an equilateral triangle.

Ans:

(c) (i) Express z1 in terms of z2 .
(ii) Hence show that z12 + z22 = z1 z2 .

Let z1 and z2 be the distinct roots of the equation z2 + az + b = 0 where z ∈ C and a , b ∈ R.

Ans:

Note: Accept working in either modulus-argument form to obtain $$z_{1} = z_{2}\left ( cos\frac{\pi }{3} + isin\frac{\pi }{3} \right )$$ or in Cartesian form to obtain $$z_{1} = z_{2}\left ( \frac{1}{2} + \frac{\sqrt{3}}{2} \right )i$$

(d) Use the result from part (c)(ii) to show that a2 – 3b = 0.

Consider the equation z2 + az + 12 = 0, where z ∈ C and a ∈ R.

Ans:

(e) Given that 0 < α – θ < π, deduce that only one equilateral triangle Z1OZ2 can be formed from the point O and the roots of this equation.

Ans:

so (for 0 <  α – θ < π), only one equilateral triangle can be formed from point O and the two roots of this equation

### Question

Consider the equation $$\frac{2z}{3-z}$$ = i where z = x + i y and x , y ∈ R.

Find the value of x and the value of y .  [Maximum mark: 5]

Ans:

substituting z =  x + iy and *z = x – iy

$$\frac{2(x+iy)}{3-(x-iy)}$$=i

2x+2iy = -y + i (3-x)

equate real and imaginary:

y= -2x AND 2y= 3-x

solving simultaneously: x= -1, y=2 (z=-1+2i)

### Question

Given that $${z_1} = 2$$ and $${z_2} = 1 + \sqrt 3 {\text{i}}$$ are roots of the cubic equation $${z^3} + b{z^2} + cz + d = 0$$ where b, c, $$d \in \mathbb{R}$$,

(a)     write down the third root, $${z_3}$$, of the equation;

Ans: $$1 – \sqrt 3 {\text{i}}$$     A1

(b)     find the values of b, c and d ;

Ans:

EITHER

$$\left( {z – (1 + \sqrt 3 {\text{i)}}} \right)\left( {z – (1 – \sqrt 3 {\text{i)}}} \right) = {z^2} – 2z + 4$$     (M1)A1

$$p(z) = (z – 2)({z^2} – 2z + 4)$$     (M1)

$$= {z^3} – 4{z^2} + 8z – 8$$     A1

therefore $$b = – 4,{\text{ }}c = 8,{\text{ }}d = – 8$$

OR

relating coefficients of cubic equations to roots

$$– b = 2 + 1 + \sqrt 3 {\text{i}} + 1 – \sqrt 3 {\text{i}} = 4$$     M1

$$c = 2(1 + \sqrt 3 {\text{i}}) + 2(1 – \sqrt 3 {\text{i}}) + (1 + \sqrt 3 {\text{i}})(1 – \sqrt 3 {\text{i}}) = 8$$

$$– d = 2(1 + \sqrt 3 {\text{i}})(1 – \sqrt 3 {\text{i}}) = 8$$

$$b = – 4,{\text{ }}c = 8,{\text{ }}d = – 8$$     A1A1A1

(c)     write $${z_2}$$ and $${z_3}$$ in the form $$r{{\text{e}}^{{\text{i}}\theta }}$$.

Ans: $${z_2} = 2{{\text{e}}^{\frac{{{\text{i}}\pi }}{3}}},{\text{ }}{z_3} = 2{{\text{e}}^{ – \frac{{{\text{i}}\pi }}{3}}}$$     A1A1A1

Note: Award A1 for modulus,

A1 for each argument.

[8 marks]

### Question

Let $$z = \cos \theta + i\sin \theta$$.

a. Use de Moivre’s theorem to find the value of $${\left( {\cos \left( {\frac{\pi }{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{3}} \right)} \right)^3}$$.[2]

Ans:

$${\left( {\cos \left( {\frac{\pi }{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{3}} \right)} \right)^3} = \cos \pi + {\text{i}}\sin \pi$$    M1

$$= – 1$$    A1

[2 marks]

b. Use mathematical induction to prove that

${(\cos \theta – {\text{i}}\sin \theta )^n} = \cos n\theta – {\text{i}}\sin n\theta {\text{ for }}n \in {\mathbb{Z}^ + }.$[6]

Ans:

show the expression is true for $$n = 1$$     R1

assume true for $$n = k,{\text{ }}{(\cos \theta – {\text{i}}\sin \theta )^k} = \cos k\theta – {\text{i}}\sin k\theta$$     M1

Note:     Do not accept “let $$n = k$$” or “assume $$n = k$$”, assumption of truth must be present.

$${(\cos \theta – {\text{i}}\sin \theta )^{k + 1}} = {(\cos \theta – {\text{i}}\sin \theta )^k}(\cos \theta – {\text{i}}\sin \theta )$$

$$= (\cos k\theta – {\text{i}}\sin k\theta )(\cos \theta – {\text{i}}\sin \theta )$$    M1

$$= \cos k\theta \cos \theta – \sin k\theta \sin \theta – {\text{i}}(\cos k\theta \sin \theta + \sin k\theta \cos \theta )$$    A1

Note:     Award A1 for any correct expansion.

$$= \cos \left( {(k + 1)\theta } \right) – {\text{i}}\sin \left( {(k + 1)\theta } \right)$$    A1

therefore if true for $$n = k$$ true for $$n = k + 1$$, true for $$n = 1$$, so true for all $$n( \in {\mathbb{Z}^ + })$$     R1

Note:     To award the final R mark the first 4 marks must be awarded.

[6 marks]

c. Find an expression in terms of $$\theta$$ for $${(z)^n} + {(z{\text{*}})^n},{\text{ }}n \in {\mathbb{Z}^ + }$$ where $$z{\text{*}}$$ is the complex conjugate of $$z$$.[2]

Ans:

$${(z)^n} + {(z{\text{*}})^n} = {(\cos \theta + {\text{i}}\sin \theta )^n} + {(\cos \theta – {\text{i}}\sin \theta )^n}$$

$$= \cos n\theta + {\text{i}}\sin n\theta + \cos n\theta – {\text{i}}\sin n\theta = 2\cos (n\theta )$$    (M1)A1

[2 marks]

d. (i)     Show that $$zz{\text{*}} = 1$$.]

Ans:

$$zz* = (\cos \theta + {\text{i}}\sin \theta )(\cos \theta – {\text{i}}\sin \theta )$$

$$= {\cos ^2}\theta + {\sin ^2}\theta$$    A1

$$= 1$$    AG

Note:     Allow justification starting with $$|z| = 1$$.

(ii)     Write down the binomial expansion of $${(z + z{\text{*}})^3}$$ in terms of $$z$$ and $$z{\text{*}}$$.

Ans: $${(z + z{\text{*}})^3} = {z^3} + 3{z^2}z{\text{*}} + 3z{({z^*})^2} + (z{\text{*}})3\left( { = {z^3} + 3z + 3z{\text{*}} + {{(z{\text{*}})}^3}} \right)$$     A1

(iii)     Hence show that $$\cos 3\theta = 4{\cos ^3}\theta – 3\cos \theta$$.[5]

Ans:

$${(z + z{\text{*}})^3} = {(2\cos \theta )^3}$$     A1

$${z^3} + 3z + 3z{\text{*}} + {(z{\text{*}})^3} = 2\cos 3\theta + 6\cos \theta$$    M1A1

$$\cos 3\theta = 4{\cos ^3}\theta – 3\cos \theta$$   AG

Note:     M1 is for using $$zz{\text{*}} = 1$$, this might be seen in d(ii).

[5 marks]

e. Hence solve $$4{\cos ^3}\theta – 2{\cos ^2}\theta – 3\cos \theta + 1 = 0$$ for $$0 \leqslant \theta < \pi$$.[6]

Ans:

$$4{\cos ^3}\theta – 2{\cos ^2}\theta – 3\cos \theta + 1 = 0$$

$$4{\cos ^3}\theta – 3\cos \theta = 2{\cos ^2}\theta – 1$$

$$\cos (3\theta ) = \cos (2\theta )$$    A1A1

Note:     A1 for $$\cos (3\theta )$$ and A1 for $$\cos (2\theta )$$.

$$\theta = 0$$    A1

or $$3\theta = 2\pi – 2\theta {\text{ }}({\text{or }}3\theta = 4\pi – 2\theta )$$     M1

$$\theta = \frac{{2\pi }}{5},{\text{ }}\frac{{4\pi }}{5}$$    A1A1

Note:     Do not accept solutions via factor theorem or other methods that do not follow “hence”.

[6 marks]

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