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IAL 2023 Jan Q6

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 6

题目

Problem

A complex number zz is represented by the point PP in an Argand diagram.

Given that

∣z−2i∣= ∣z−3∣\begin{align*} |z - 2\mathrm{i}| =\,& |z - 3| \end{align*}

(a) sketch the locus of PP. You do not need to find the coordinates of any intercepts.

(2)

The transformation TT from the zz-plane to the ww-plane is given by

w= izz−2iz≠2i\begin{align*} w =\,& \frac{\mathrm{i}z}{z - 2\mathrm{i}} \qquad z \neq 2\mathrm{i} \end{align*}

Given that TT maps ∣z−2i∣=∣z−3∣|z - 2\mathrm{i}| = |z - 3| to a circle CC in the ww-plane,

(b) find the equation of CC, giving your answer in the form

∣w−(p+qi)∣= r\begin{align*} |w - (p + q\mathrm{i})| =\,& r \end{align*}

where pp, qq and rr are real numbers to be determined.

(6)

解答

(a)

解法一

思路

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∣z−2i∣=∣z−3∣|z-2\mathrm{i}|=|z-3| 表示点 PP 到 2i2\mathrm{i} 和 33 的距离相等，所以轨迹是连接 (0,2)(0,2) 与 (3,0)(3,0) 线段的垂直平分线。

答题过程

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The equation

∣z−2i∣=∣z−3∣\begin{align*} |z-2\mathrm{i}|=|z-3| \end{align*}

means that PP is equidistant from the points representing 2i2\mathrm{i} and 33.

So the locus is the perpendicular bisector of the line segment joining

(0,2)and(3,0).\begin{align*} (0,2)\quad\text{and}\quad (3,0). \end{align*}

The sketch should be a straight line with positive gradient, not passing through the origin, lying through quadrants I, III and IV.

(b)

解法一

思路

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把 zz 用 ww 表示，再直接代入原来的模长关系。这样可以避免先求直线方程再代入，整体更短。

答题过程

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Given

w=izz−2i.\begin{align*} w=\frac{\mathrm{i}z}{z-2\mathrm{i}}. \end{align*}

Rearrange:

w(z−2i)= izwz−2iw= izz(w−i)= 2iwz= 2iww−i.\begin{align*} w(z-2\mathrm{i})=&\,\mathrm{i}z\\[2mm] wz-2\mathrm{i}w=&\,\mathrm{i}z\\[2mm] z(w-\mathrm{i})=&\,2\mathrm{i}w\\[2mm] z=&\,\frac{2\mathrm{i}w}{w-\mathrm{i}}. \end{align*}

The original locus is

∣z−2i∣=∣z−3∣.\begin{align*} |z-2\mathrm{i}|=|z-3|. \end{align*}

Substitute z=2iww−iz=\dfrac{2\mathrm{i}w}{w-\mathrm{i}}:

∣2iww−i−2i∣= ∣2iww−i−3∣.\begin{align*} \left|\frac{2\mathrm{i}w}{w-\mathrm{i}}-2\mathrm{i}\right| =&\,\left|\frac{2\mathrm{i}w}{w-\mathrm{i}}-3\right|. \end{align*}

Since w≠iw\neq \mathrm{i}, multiply both sides by ∣w−i∣|w-\mathrm{i}|:

∣2iw−2i(w−i)∣= ∣2iw−3(w−i)∣∣−2∣= ∣(2i−3)w+3i∣.\begin{align*} |2\mathrm{i}w-2\mathrm{i}(w-\mathrm{i})| =&\,|2\mathrm{i}w-3(w-\mathrm{i})|\\[2mm] |-2|=&\,|(2\mathrm{i}-3)w+3\mathrm{i}|. \end{align*}

So

2= ∣(2i−3)w+3i∣= ∣2i−3∣∣w+3i2i−3∣.\begin{align*} 2=&\,|(2\mathrm{i}-3)w+3\mathrm{i}|\\[2mm] =&\,|2\mathrm{i}-3| \left|w+\frac{3\mathrm{i}}{2\mathrm{i}-3}\right|. \end{align*}

Now

∣2i−3∣=13.\begin{align*} |2\mathrm{i}-3|=\sqrt{13}. \end{align*}

Also,

3i2i−3= 3i(−3−2i)(2i−3)(−3−2i)= −9i−6i213= 6−9i13.\begin{align*} \frac{3\mathrm{i}}{2\mathrm{i}-3} =&\,\frac{3\mathrm{i}(-3-2\mathrm{i})}{(2\mathrm{i}-3)(-3-2\mathrm{i})}\\[2mm] =&\,\frac{-9\mathrm{i}-6\mathrm{i}^2}{13}\\[2mm] =&\,\frac{6-9\mathrm{i}}{13}. \end{align*}

Therefore

2= 13∣w+613−913i∣.\begin{align*} 2=&\,\sqrt{13}\left|w+\frac{6}{13}-\frac{9}{13}\mathrm{i}\right|. \end{align*}

Hence

∣w−(−613+913i)∣= 213= 21313.\begin{align*} \left|w-\left(-\frac{6}{13}+\frac{9}{13}\mathrm{i}\right)\right| =&\,\frac{2}{\sqrt{13}}\\[2mm] =&\,\frac{2\sqrt{13}}{13}. \end{align*}

So

∣w−(−613+913i)∣=21313.\begin{align*} \boxed{ \left|w-\left(-\frac{6}{13}+\frac{9}{13}\mathrm{i}\right)\right| =\frac{2\sqrt{13}}{13} }. \end{align*}