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IAL 2026 Jan Q6

A Level / Edexcel / FP2

IAL 2026 Jan Paper · Question 6

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+23z+4,z≠−43\begin{align*} w=\frac{z+2}{3z+4}, \qquad z\ne-\frac43 \end{align*}

The circle CC in the zz-plane has equation

(x+1)2+(y−1)2=1\begin{align*} (x+1)^2+(y-1)^2=1 \end{align*}

Given that TT maps CC to the circle DD in the ww-plane, determine the equation of DD, giving your answer in the form

u2+v2+au+bv+c=0\begin{align*} u^2+v^2+au+bv+c=0 \end{align*}

where aa, bb and cc are real numbers.

(7)

解答

解法一

思路

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先把原来的圆写成复数模长形式，再由 w=z+23z+4w=\dfrac{z+2}{3z+4} 解出 zz。代回圆的方程后，把 w=u+ivw=u+\mathrm{i}v 展开并比较模长即可。

答题过程

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The circle

(x+1)2+(y−1)2=1\begin{align*} (x+1)^2+(y-1)^2=1 \end{align*}

can be written as

∣z−(−1+i)∣=1\begin{align*} |z-(-1+\mathrm{i})|=1 \end{align*}

so

∣z+1−i∣=1\begin{align*} |z+1-\mathrm{i}|=1 \end{align*}

From

w=z+23z+4\begin{align*} w=\frac{z+2}{3z+4} \end{align*}

we get

w(3z+4)=z+23wz+4w=z+2z(3w−1)=2−4w\begin{align*} w(3z+4) ={}& z+2\\[3mm] 3wz+4w ={}& z+2\\[3mm] z(3w-1) ={}& 2-4w \end{align*}

Hence

z=2−4w3w−1\begin{align*} z=\frac{2-4w}{3w-1} \end{align*}

Substitute this into ∣z+1−i∣=1|z+1-\mathrm{i}|=1:

∣2−4w3w−1+1−i∣=1\begin{align*} \left| \frac{2-4w}{3w-1}+1-\mathrm{i} \right| =1 \end{align*}

Therefore

∣2−4w+(1−i)(3w−1)3w−1∣=1∣1−w−3iw+i∣=∣3w−1∣\begin{align*} \left| \frac{2-4w+(1-\mathrm{i})(3w-1)} {3w-1} \right| ={}&1\\[3mm] |1-w-3\mathrm{i}w+\mathrm{i}| ={}& |3w-1| \end{align*}

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

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

Also

3w−1=(3u−1)+3iv\begin{align*} 3w-1=(3u-1)+3\mathrm{i}v \end{align*}

Hence

(1−u+3v)2+(1−v−3u)2=(3u−1)2+9v2\begin{align*} (1-u+3v)^2+(1-v-3u)^2 = (3u-1)^2+9v^2 \end{align*}

Expanding,

1+u2+9v2−2u+6v−6uv+1+v2+9u2−2v−6u+6uv= 9u2−6u+1+9v2\begin{align*} 1+u^2+9v^2-2u+6v-6uv &+1+v^2+9u^2-2v-6u+6uv\\[3mm] =&\, 9u^2-6u+1+9v^2 \end{align*}

So

u2+v2−2u+4v+1=0\begin{align*} u^2+v^2-2u+4v+1=0 \end{align*}

Therefore the equation of DD is

u2+v2−2u+4v+1=0\begin{align*} \boxed{u^2+v^2-2u+4v+1=0} \end{align*}

解法二

思路

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也可以把圆上的三个点映射到 ww 平面，再用一般圆方程求出 a,b,ca,b,c。这种方法计算直接，但要小心选择的点必须都在原圆上。

答题过程

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Choose three points on

(x+1)2+(y−1)2=1\begin{align*} (x+1)^2+(y-1)^2=1 \end{align*}

For example,

z=−1,z=−1+2i,z=i\begin{align*} z=-1,\qquad z=-1+2\mathrm{i},\qquad z=\mathrm{i} \end{align*}

Under

w=z+23z+4\begin{align*} w=\frac{z+2}{3z+4} \end{align*}

these become

z=−1⟹w=1z=−1+2i⟹w=1+2i1+6i=13−4i37z=i⟹w=2+i4+3i=11−2i25\begin{align*} z=-1 &\quad\Longrightarrow\quad w=1 \\[3mm] z=-1+2\mathrm{i} &\quad\Longrightarrow\quad w=\frac{1+2\mathrm{i}}{1+6\mathrm{i}} =\frac{13-4\mathrm{i}}{37} \\[3mm] z=\mathrm{i} &\quad\Longrightarrow\quad w=\frac{2+\mathrm{i}}{4+3\mathrm{i}} =\frac{11-2\mathrm{i}}{25} \end{align*}

Let the image circle be

u2+v2+au+bv+c=0\begin{align*} u^2+v^2+au+bv+c=0 \end{align*}

Using (u,v)=(1,0)(u,v)=(1,0),

1+a+c=0\begin{align*} 1+a+c=0 \end{align*}

Using (u,v)=(1337,−437)\left(u,v\right)=\left(\dfrac{13}{37},-\dfrac{4}{37}\right),

1851369+13a37−4b37+c=0\begin{align*} \frac{185}{1369} +\frac{13a}{37} -\frac{4b}{37} +c =0 \end{align*}

Using (u,v)=(1125,−225)\left(u,v\right)=\left(\dfrac{11}{25},-\dfrac{2}{25}\right),

125625+11a25−2b25+c=0\begin{align*} \frac{125}{625} +\frac{11a}{25} -\frac{2b}{25} +c =0 \end{align*}

Solving these simultaneous equations gives

a=−2,b=4,c=1\begin{align*} a=-2,\qquad b=4,\qquad c=1 \end{align*}

Therefore

u2+v2−2u+4v+1=0\begin{align*} \boxed{u^2+v^2-2u+4v+1=0} \end{align*}