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IAL 2023 June Q5

A Level / Edexcel / FP2

IAL 2023 June Paper · Question 5

题目

Problem

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=z+1z−3z≠3\begin{align*} w ={}& \frac{z + 1}{z - 3} \qquad z \neq 3 \end{align*}

The straight line in the zz-plane with equation y=4xy = 4x is mapped by TT onto the circle CC in the ww-plane.

(a) Show that CC has equation

3u2+3v2−2u+v+k=0\begin{align*} 3u^2 + 3v^2 - 2u + v + k ={}& 0 \end{align*}

where kk is a constant to be determined.

(5)

(b) Hence determine

(i) the coordinates of the centre of CC

(ii) the radius of CC

(2)

解答

(a)

解法一

思路

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因为已知的是 zz-plane 中的直线 y=4xy=4x，而要求的是 ww-plane 中的圆，所以可以先把 zz 用 ww 表示，再把 w=u+ivw=u+\mathrm{i}v 代入，得到 x,yx,y 关于 u,vu,v 的表达式。

答题过程

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From

w=z+1z−3\begin{align*} w=\frac{z+1}{z-3} \end{align*}

make zz the subject:

w(z−3)=z+1wz−3w=z+1z(w−1)=3w+1z=3w+1w−1\begin{align*} w(z-3)={}&z+1\\[4mm] wz-3w={}&z+1\\[4mm] z(w-1)={}&3w+1\\[4mm] z={}&\frac{3w+1}{w-1} \end{align*}

Now let w=u+ivw=u+\mathrm{i}v:

z=3(u+iv)+1u+iv−1=(3u+1)+3iv(u−1)+iv\begin{align*} z ={}& \frac{3(u+\mathrm{i}v)+1}{u+\mathrm{i}v-1}\\[4mm] ={}& \frac{(3u+1)+3\mathrm{i}v}{(u-1)+\mathrm{i}v} \end{align*}

Rationalise the denominator:

z=((3u+1)+3iv)((u−1)−iv)(u−1)2+v2\begin{align*} z ={}& \frac{((3u+1)+3\mathrm{i}v)((u-1)-\mathrm{i}v)} {(u-1)^2+v^2} \end{align*}

Expand the numerator:

((3u+1)+3iv)((u−1)−iv)=(3u+1)(u−1)+3v2 +i(3v(u−1)−v(3u+1))=3u2−2u+3v2−1−4iv\begin{align*} ((3u+1)+3\mathrm{i}v)((u-1)-\mathrm{i}v) ={}& (3u+1)(u-1)+3v^2\\[4mm] &\,\hspace{2pt}+\mathrm{i}\left(3v(u-1)-v(3u+1)\right)\\[4mm] ={}& 3u^2-2u+3v^2-1-4\mathrm{i}v \end{align*}

Since z=x+iyz=x+\mathrm{i}y,

x=3u2−2u+3v2−1(u−1)2+v2y=−4v(u−1)2+v2\begin{align*} x={}& \frac{3u^2-2u+3v^2-1}{(u-1)^2+v^2}\\[4mm] y={}& \frac{-4v}{(u-1)^2+v^2} \end{align*}

The line in the zz-plane is y=4xy=4x, so

−4v(u−1)2+v2=4(3u2−2u+3v2−1(u−1)2+v2)\begin{align*} \frac{-4v}{(u-1)^2+v^2} ={}& 4\left( \frac{3u^2-2u+3v^2-1}{(u-1)^2+v^2} \right) \end{align*}

Multiplying by the common denominator,

−4v=12u2−8u+12v2−40=12u2+12v2−8u+4v−40=3u2+3v2−2u+v−1\begin{align*} -4v ={}& 12u^2-8u+12v^2-4\\[4mm] 0 ={}& 12u^2+12v^2-8u+4v-4\\[4mm] 0 ={}& 3u^2+3v^2-2u+v-1 \end{align*}

Therefore

3u2+3v2−2u+v+k=0\begin{align*} 3u^2+3v^2-2u+v+k=0 \end{align*}

where

k=−1\begin{align*} \boxed{k=-1} \end{align*}

解法二

思路

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另一条路线是不先反解 zz，而是直接把直线 y=4xy=4x 写成 z=x+4ixz=x+4\mathrm{i}x，代入变换式。这样得到 u,vu,v 关于参数 xx 的表达式，再代入目标圆的左边检验。

答题过程

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On the line y=4xy=4x,

z=x+4ix\begin{align*} z=x+4\mathrm{i}x \end{align*}

Substitute into the transformation:

w=x+4ix+1x+4ix−3=(x+1)+4ix(x−3)+4ix⋅(x−3)−4ix(x−3)−4ix\begin{align*} w ={}& \frac{x+4\mathrm{i}x+1}{x+4\mathrm{i}x-3}\\[4mm] ={}& \frac{(x+1)+4\mathrm{i}x}{(x-3)+4\mathrm{i}x} \cdot \frac{(x-3)-4\mathrm{i}x}{(x-3)-4\mathrm{i}x} \end{align*}

The denominator is

(x−3)2+16x2=17x2−6x+9\begin{align*} (x-3)^2+16x^2 = 17x^2-6x+9 \end{align*}

The numerator is

((x+1)+4ix)((x−3)−4ix)=(x+1)(x−3)+16x2 +i(4x(x−3)−4x(x+1))=17x2−2x−3−16ix\begin{align*} ((x+1)+4\mathrm{i}x)((x-3)-4\mathrm{i}x) ={}& (x+1)(x-3)+16x^2\\[4mm] &\,\hspace{2pt}+\mathrm{i}\left(4x(x-3)-4x(x+1)\right)\\[4mm] ={}& 17x^2-2x-3-16\mathrm{i}x \end{align*}

Hence

u=17x2−2x−317x2−6x+9v=−16x17x2−6x+9\begin{align*} u={}& \frac{17x^2-2x-3}{17x^2-6x+9}\\[4mm] v={}& \frac{-16x}{17x^2-6x+9} \end{align*}

Substituting these into 3u2+3v2−2u+v−13u^2+3v^2-2u+v-1 gives

3u2+3v2−2u+v−1=3(17x2−2x−3)2+3(−16x)2(17x2−6x+9)2 −2(17x2−2x−3)17x2−6x+9 −16x17x2−6x+9−1\begin{align*} 3u^2+3v^2-2u+v-1 ={}& \frac{ 3(17x^2-2x-3)^2+3(-16x)^2 }{(17x^2-6x+9)^2}\\[4mm] &\,\hspace{2pt}-\frac{2(17x^2-2x-3)}{17x^2-6x+9}\\[4mm] &\,\hspace{4pt}-\frac{16x}{17x^2-6x+9}-1 \end{align*}

Putting the terms over the common denominator (17x2−6x+9)2(17x^2-6x+9)^2 and simplifying gives

3u2+3v2−2u+v−1=3(17x2−2x−3)2+768x2(17x2−6x+9)2 −2(17x2−2x−3)(17x2−6x+9)(17x2−6x+9)2 −16x(17x2−6x+9)(17x2−6x+9)2 −(17x2−6x+9)2(17x2−6x+9)2=(289x4−204x3+342x2−108x+81)(17x2−6x+9)2 −(289x4−204x3+342x2−108x+81)(17x2−6x+9)2=0\begin{align*} 3u^2+3v^2-2u+v-1 ={}& \frac{ 3(17x^2-2x-3)^2+768x^2 }{(17x^2-6x+9)^2}\\[4mm] &\,\hspace{2pt}-\frac{ 2(17x^2-2x-3)(17x^2-6x+9) }{(17x^2-6x+9)^2}\\[4mm] &\,\hspace{4pt}-\frac{ 16x(17x^2-6x+9) }{(17x^2-6x+9)^2}\\[4mm] &\,\hspace{6pt}-\frac{ (17x^2-6x+9)^2 }{(17x^2-6x+9)^2}\\[4mm] ={}& \frac{ (289x^4-204x^3+342x^2-108x+81) }{(17x^2-6x+9)^2}\\[4mm] &\,\hspace{2pt}-\frac{ (289x^4-204x^3+342x^2-108x+81) }{(17x^2-6x+9)^2}\\[4mm] ={}& 0 \end{align*}

Hence

3u2+3v2−2u+v−1=0\begin{align*} 3u^2+3v^2-2u+v-1=0 \end{align*}

Therefore CC has equation

3u2+3v2−2u+v−1=0\begin{align*} 3u^2+3v^2-2u+v-1=0 \end{align*}

so

k=−1\begin{align*} \boxed{k=-1} \end{align*}

(b)

解法一

思路

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由 (a) 得到圆的方程后，除以 33 再配方。注意 vv 的一次项是 +13v+\frac13v，所以配方中心的 vv 坐标会是负数。

答题过程

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From part (a),

3u2+3v2−2u+v−1=0\begin{align*} 3u^2+3v^2-2u+v-1=0 \end{align*}

Divide by 33:

u2−23u+v2+13v−13=0\begin{align*} u^2-\frac23u+v^2+\frac13v-\frac13=0 \end{align*}

Complete the square:

(u−13)2−19+(v+16)2−136−13=0(u−13)2+(v+16)2=19+136+13=4+1+1236=1736\begin{align*} \left(u-\frac13\right)^2-\frac19 +\left(v+\frac16\right)^2-\frac1{36} -\frac13 ={}&0\\[4mm] \left(u-\frac13\right)^2 +\left(v+\frac16\right)^2 ={}& \frac19+\frac1{36}+\frac13\\[4mm] ={}& \frac{4+1+12}{36}\\[4mm] ={}& \frac{17}{36} \end{align*}

Therefore the centre is

(13,−16)\begin{align*} \boxed{\left(\frac13,-\frac16\right)} \end{align*}

and the radius is

176\begin{align*} \boxed{\frac{\sqrt{17}}{6}} \end{align*}