IAL 2024 June Q1 | 國際數學站

The complex number $z = x + iy$ satisfies the equation

\begin{align*} |z - 3 - 4\mathrm{i}| = |z + 1 + \mathrm{i}| \end{align*}

(a) Determine an equation for the locus of $z$ giving your answer in the form $ax + by + c = 0$ where $a$ , $b$ and $c$ are integers.

(3)

(b) Shade, on an Argand diagram, the region defined by

\begin{align*} |z - 3 - 4\mathrm{i}| \leqslant |z + 1 + \mathrm{i}| \end{align*}

You do not need to determine the coordinates of any intercepts on the coordinate axes.

(1)

解答

(a)

Let $z = x + \mathrm{i}y$ . Substitute this into the given equation:

\begin{align*} |x + \mathrm{i}y - 3 - 4\mathrm{i}| = &\,|x + \mathrm{i}y + 1 + \mathrm{i}|\\[4mm] |(x - 3) + \mathrm{i}(y - 4)| = &\,|(x + 1) + \mathrm{i}(y + 1)| \end{align*}

Squaring both sides and using the definition of the modulus:

\begin{align*} (x - 3)^2 + (y - 4)^2 = &\,(x + 1)^2 + (y + 1)^2\\[4mm] x^2 - 6x + 9 + y^2 - 8y + 16 = &\,x^2 + 2x + 1 + y^2 + 2y + 1\\[4mm] -6x - 8y + 25 = &\,2x + 2y + 2\\[4mm] 8x + 10y - 23 = &\,0 \end{align*}

(b)

The locus from part (a) is the perpendicular bisector of the line segment joining $(3, 4)$ and $(-1, -1)$ . The inequality $|z - 3 - 4\mathrm{i}| \leqslant |z + 1 + \mathrm{i}|$ represents the region of points that are closer to $3 + 4\mathrm{i}$ than to $-1 - \mathrm{i}$ (or equidistant). This is the half-plane containing the point $(3, 4)$ , bounded by the line $8x + 10y - 23 = 0$ .