A2 June 2021 Paper 2 Q8
8 The complex number \(z\) satisfies the equations
\[|z^* - 1 - 2\mathrm{i}| = |z - 3|\]and
\[|z - a| = 3\]where \(a\) is real.
Show that \(a\) must lie in the interval \(\left[1 - s\sqrt{t},\ 1 + s\sqrt{t}\right]\), where \(s\) and \(t\) are prime numbers. [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Expresses \(z\) and \(z^*\) in terms of \(x\) and \(y\) and substitutes these in the first equation | M1 | 1.1a |
| Deduces a correct linear Cartesian equation for first equation | A1 | 2.2a |
| Obtains correct Cartesian equation for second equation or states that this represents a circle centre \((a, 0)\) radius 3 | B1 | 1.1b |
| Solves their Cartesian equations simultaneously | M1 | 1.1a |
| States that the discriminant is non-negative or creates an inequality based on this | E1 | 2.4 |
| Completes a rigorous argument to obtain the correct range of values for \(a\) | R1 | 2.1 |
| (6 marks) |
Typical solution
Let \(z = x + \mathrm{i}y\)
Then
\[\begin{aligned} |x - \mathrm{i}y - 1 - 2\mathrm{i}| &= |x + \mathrm{i}y - 3| \\ (x - 1)^2 + (y + 2)^2 &= (x - 3)^2 + y^2 \\ -2x + 1 + 4y + 4 &= -6x + 9 \\ y &= 1 - x \end{aligned}\]and
\[(x - a)^2 + y^2 = 9\]Solving simultaneously,
\[(x - a)^2 + (1 - x)^2 = 9\]\[2x^2 + (-2a - 2)x + (a^2 - 8) = 0\]\(\Delta \geqslant 0\) so
\[\begin{aligned} (2a + 2)^2 - 4(2)(a^2 - 8) &\geqslant 0 \\ a^2 - 2a - 17 &\leqslant 0 \\ (a - 1)^2 &\leqslant 18 \end{aligned}\]So \(1 - \sqrt{18} \leqslant a \leqslant 1 + \sqrt{18}\)
And \(a\) must lie in the interval
\[\left[1 - 3\sqrt{2},\ 1 + 3\sqrt{2}\right]\]