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IAL 2025 June Q3

A Level / Edexcel / FP2

IAL 2025 June Paper · Question 3

题目

Problem

A complex number zz is represented by the point PP on an Argand diagram where ∣z∣=1|z|=1

(a) Sketch the locus of PP as zz varies.

(1)

The transformation TT from the zz-plane, where z=x+iyz=x+\mathrm{i}y, to the ww-plane, where w=u+ivw=u+\mathrm{i}v, is given by

w=9iz−iz+1z≠−1\begin{align*} w=\frac{9\mathrm{i}z-\mathrm{i}}{z+1} \qquad z\neq -1 \end{align*}

Given that the image under TT of the locus of PP in the zz-plane, where z≠−1z\neq -1, is the line ll in the ww-plane,

(b) determine, in simplest form, a Cartesian equation for ll

(5)

解答

(a)

解法一

思路

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∣z∣=1|z|=1 表示点到原点的距离恒为 11，所以轨迹是以原点为圆心、半径为 11 的圆。

答题过程

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The locus ∣z∣=1|z|=1 is a circle centered at the origin OO with radius 11 on the Argand diagram.

(b)

解法一

思路

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先把 zz 用 ww 表示。由于原轨迹是 ∣z∣=1|z|=1，代入后会得到一个“到两个固定点距离相等”的式子，所以像是两点连线的垂直平分线。

答题过程

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From

w=9iz−iz+1,\begin{align*} w=\frac{9\mathrm{i}z-\mathrm{i}}{z+1}, \end{align*}

make zz the subject:

w(z+1)=9iz−iwz+w=9iz−iz(w−9i)=−w−iz=−w−iw−9i\begin{align*} w(z+1) ={}& 9\mathrm{i}z-\mathrm{i}\\[2mm] wz+w ={}& 9\mathrm{i}z-\mathrm{i}\\[2mm] z(w-9\mathrm{i}) ={}& -w-\mathrm{i}\\[2mm] z ={}& \frac{-w-\mathrm{i}}{w-9\mathrm{i}} \end{align*}

Since ∣z∣=1|z|=1,

∣−w−iw−9i∣=1∣−w−i∣=∣w−9i∣∣w+i∣=∣w−9i∣\begin{align*} \left| \frac{-w-\mathrm{i}}{w-9\mathrm{i}} \right| ={}& 1\\[4mm] |-w-\mathrm{i}| ={}& |w-9\mathrm{i}|\\[2mm] |w+\mathrm{i}| ={}& |w-9\mathrm{i}| \end{align*}

This means that ww is equidistant from −i-\mathrm{i} and 9i9\mathrm{i}. Their midpoint is 4i4\mathrm{i}, and the segment joining them is vertical, so the perpendicular bisector is horizontal:

v=4\begin{align*} \boxed{v=4} \end{align*}

解法二

思路

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也可以完全代数化。令 w=u+ivw=u+\mathrm{i}v，把

∣w+i∣=∣w−9i∣\begin{align*} |w+\mathrm{i}|=|w-9\mathrm{i}| \end{align*}

两边平方后展开。

答题过程

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From the first method,

∣w+i∣=∣w−9i∣\begin{align*} |w+\mathrm{i}|=|w-9\mathrm{i}| \end{align*}

Let w=u+ivw=u+\mathrm{i}v. Then

∣w+i∣2=u2+(v+1)2\begin{align*} |w+\mathrm{i}|^2 ={}& u^2+(v+1)^2 \end{align*}

and

∣w−9i∣2=u2+(v−9)2\begin{align*} |w-9\mathrm{i}|^2 ={}& u^2+(v-9)^2 \end{align*}

Hence

u2+(v+1)2=u2+(v−9)2v2+2v+1=v2−18v+8120v=80v=4\begin{align*} u^2+(v+1)^2 ={}& u^2+(v-9)^2\\[3mm] v^2+2v+1 ={}& v^2-18v+81\\[3mm] 20v ={}& 80\\[2mm] v ={}& 4 \end{align*}

Therefore

v=4\begin{align*} \boxed{v=4} \end{align*}