The relations R and S are defined on quadratic polynomials P of the form

\[P(z) = {z^2} + az + b{\text{ , where }}a{\text{ , }}b \in \mathbb{R}{\text{ , }}z \in \mathbb{C}{\text{ .}}\]

(a)     The relation R is defined by \({P_1}R{P_2}\) if and only if the sum of the two zeros of \({P_1}\) is equal to the sum of the two zeros of \({P_2}\) .

(i)     Show that R is an equivalence relation.

(ii)     Determine the equivalence class containing \({z^2} - 4z + 5\) .

(b)     The relation S is defined by \({P_1}S{P_2}\) if and only if \({P_1}\) and \({P_2}\) have at least one zero in common. Determine whether or not S is transitive.

(a)     (i)     R is reflexive, i.e. PRP because the sum of the zeroes of P is equal to the sum of the zeros of P     R1

R is symmetric, i.e. \({P_1}R{P_2} \Rightarrow {P_2}R{P_1}\) because the sums of the zeros of \({P_1}\) and \({P_2}\) are equal implies that the sums of the zeros of \({P_2}\) and \({P_1}\) are equal     R1

suppose that \({P_1}R{P_2}\) and \({P_2}R{P_3}\)     M1

it follows that \({P_1}R{P_3}\) so R is transitive, because the sum of the zeros of \({P_1}\) is equal to the sum of the zeros of \({P_2}\) which in turn is equal to the sum of the zeros of \({P_3}\) , which implies that the sum of the zeros of \({P_1}\) is equal to the sum of the zeros of \({P_3}\)     R1

the three requirements for an equivalence relation are therefore satisfied     AG

(ii)     the zeros of \({z^2} - 4z + 5\) are \(2 \pm {\text{i}}\) , for which the sum is 4     M1A1

\({{\text{z}}^2} + az + b\) has zeros of \(\frac{{ - a \pm \sqrt {{a^2} - 4b} }}{2}\) , so the sum is –a     (M1)

Note: Accept use of the result (although not in the syllabus) that the sum of roots is minus the coefficient of z.

hence – a = 4 and so a = – 4     A1

the equivalence class is \({z^2} - 4z + k{\text{ , }}(k \in \mathbb{R})\)     A1

[9 marks]

(b)     for example, \((z - 1)(z - 2)S(z - 1)(z - 3)\) and

\((z - 1)(z - 3)S(z - 3)(z - 4)\) but \((z - 1)(z - 2)S(z - 3)(z - 4)\) is not true     M1A1

so S is not transitive     A1

[3 marks]

Total [12 marks]

Most candidates were able to show, in (a), that R is an equivalence relation although few were able to identify the required equivalence class. In (b), the explanation that S is not transitive was often unconvincing.