Challenging +1.8 This is a Further Maths locus problem requiring algebraic manipulation of complex numbers, rationalization of a complex fraction, separation of real and imaginary parts, and recognition that the resulting equation represents a circle. While the steps are systematic, the problem demands careful algebra across multiple stages and geometric interpretation, making it significantly harder than standard A-level questions but still within established Further Maths techniques.
C is the locus of numbers, \(z\), for which \(\text{Im}\left(\frac{z + 7i}{z - 24}\right) = \frac{1}{4}\).
By writing \(z = x + iy\) give a complete description of the shape of C on an Argand diagram. [7]
Completing both squares with half signed coefficients of x and y
\((x - 38)^2 + (y + 48)^2 = 2500\)
A1
So the shape of C is a circle...
E1
...centre \(38 - 48i\), radius 50
E1
Or (38, -48)
$\frac{z + 7i}{z - 24} = \frac{x + iy + 7i}{x - 24 + iy} \times \frac{x - 24 - iy}{x - 24 - iy}$ | M1 | Substituting $z = x + iy$ into $\frac{z + 7i}{z - 24}$ and multiplying top and bottom by conjugate of bottom
$\text{Im}\frac{z + 7i}{z - 24} = \frac{-xy + (y + 7)(x - 24)}{(x - 24)^2 + y^2} = \frac{1}{4}$ | M1 | Multiplying out (ignore errors in real part) and equating imaginary part to 1/4 (without i unless later cancelled or recovered)
$28x - 96y - 672 = x^2 - 48x + 576 + y^2$ | M1 | Multiplying out to get horizontal equation with no xy term and no double brackets
$0 = (x - 38)^2 - 1444 + (y + 48)^2 - 2304 + 1248$ | M1 | Completing both squares with half signed coefficients of x and y
$(x - 38)^2 + (y + 48)^2 = 2500$ | A1 |
So the shape of C is a circle... | E1 |
...centre $38 - 48i$, radius 50 | E1 | Or (38, -48)
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C is the locus of numbers, $z$, for which $\text{Im}\left(\frac{z + 7i}{z - 24}\right) = \frac{1}{4}$.
By writing $z = x + iy$ give a complete description of the shape of C on an Argand diagram. [7]
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q7 [7]}}