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FP2 2024 June Q6

A Level / Edexcel / FP2

IAL 2024 June Paper · Question 6

题目

Problem

6.

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

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

Given that TT maps the imaginary axis in the zz-plane to the circle CC in the ww-plane, determine

(i) the coordinates of the centre of CC

(ii) the radius of CC

(7)
题目中文翻译

从 zz 平面到 ww 平面的变换 TT 定义为

w= z−iz+1,z≠−1.\begin{align*} w=&\,\frac{z-\mathrm{i}}{z+1}, \qquad z\ne-1. \end{align*}

已知 TT 将 zz 平面的虚轴映射为 ww 平面的圆 CC，求：

(i) 圆 CC 的圆心坐标；

(ii) 圆 CC 的半径。

(7)

解答

(i)

解法一

思路

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虚轴上的点满足 Re⁡(z)=0\operatorname{Re}(z)=0。先把 zz 用 ww 表示，再令 w=u+ivw=u+\mathrm{i}v，取 zz 的实部等于 00，就能得到 ww-plane 中的圆方程。

答题过程

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From

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

make zz the subject:

w(z+1)= z−iwz+w= z−iz(w−1)= −w−iz= w+i1−w\begin{align*} w(z+1)=&\,z-\mathrm{i}\\[2mm] wz+w=&\,z-\mathrm{i}\\[2mm] z(w-1)=&\,-w-\mathrm{i}\\[2mm] z=&\,\frac{w+\mathrm{i}}{1-w} \end{align*}

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

z= u+iv+i1−u−iv= u+i(v+1)(1−u)−iv\begin{align*} z =&\, \frac{u+\mathrm{i}v+\mathrm{i}}{1-u-\mathrm{i}v}\\[4mm] =&\, \frac{u+\mathrm{i}(v+1)}{(1-u)-\mathrm{i}v} \end{align*}

Rationalise the denominator:

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

The real part of the numerator is

u(1−u)−v(v+1)= u−u2−v2−v\begin{align*} u(1-u)-v(v+1) =&\, u-u^2-v^2-v \end{align*}

Since the original locus is the imaginary axis, Re⁡(z)=0\operatorname{Re}(z)=0. Hence

u−u2−v2−v= 0u2−u+v2+v= 0\begin{align*} u-u^2-v^2-v=&\,0\\[2mm] u^2-u+v^2+v=&\,0 \end{align*}

Complete the square:

(u−12)2−14+(v+12)2−14= 0(u−12)2+(v+12)2= 12\begin{align*} \bigg(u-\frac12\bigg)^2-\frac14 +\bigg(v+\frac12\bigg)^2-\frac14 =&\,0\\[4mm] \bigg(u-\frac12\bigg)^2 +\bigg(v+\frac12\bigg)^2 =&\, \frac12 \end{align*}

Therefore the centre is

(12,−12)\begin{align*} \boxed{\bigg(\frac12,-\frac12\bigg)} \end{align*}

解法二

思路

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虚轴可以写成 z=iyz=\mathrm{i}y。直接代入变换，得到 u,vu,v 关于参数 yy 的式子，再消去 yy，也能得到同一个圆。

答题过程

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On the imaginary axis,

z=iy\begin{align*} z=\mathrm{i}y \end{align*}

Substitute into the transformation:

w= iy−iiy+1= i(y−1)1+iy⋅1−iy1−iy= y2−y+i(y−1)1+y2\begin{align*} w =&\, \frac{\mathrm{i}y-\mathrm{i}}{\mathrm{i}y+1}\\[4mm] =&\, \frac{\mathrm{i}(y-1)}{1+\mathrm{i}y} \cdot \frac{1-\mathrm{i}y}{1-\mathrm{i}y}\\[4mm] =&\, \frac{y^2-y+\mathrm{i}(y-1)}{1+y^2} \end{align*}

Since w=u+ivw=u+\mathrm{i}v,

u= y2−y1+y2v= y−11+y2\begin{align*} u=&\,\frac{y^2-y}{1+y^2}\\[2mm] v=&\,\frac{y-1}{1+y^2} \end{align*}

From these two equations,

u−yv= y2−y1+y2−yy−11+y2= 0\begin{align*} u-yv =&\, \frac{y^2-y}{1+y^2} -y\frac{y-1}{1+y^2}\\[4mm] =&\,0 \end{align*}

so u=yvu=yv. Substitute y=uvy=\frac{u}{v} into v=y−11+y2v=\frac{y-1}{1+y^2}:

v= uv−11+u2v2v(u2+v2)= uv−v2\begin{align*} v =&\, \frac{\frac{u}{v}-1}{1+\frac{u^2}{v^2}}\\[4mm] v(u^2+v^2) =&\, uv-v^2 \end{align*}

If v=0v=0, then y=1y=1 and u=0u=0, which also lies on the circle found below. For the remaining points, divide by vv:

u2+v2= u−vu2−u+v2+v= 0\begin{align*} u^2+v^2 =&\, u-v\\[2mm] u^2-u+v^2+v =&\, 0 \end{align*}

Hence

(u−12)2+(v+12)2= 12\begin{align*} \bigg(u-\frac12\bigg)^2+\bigg(v+\frac12\bigg)^2 =&\, \frac12 \end{align*}

So the centre is (12,−12)\big(\frac12,-\frac12\big).

解法三

思路

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已知像轨迹是圆，可以取虚轴上三个容易代入的点，求出它们的像，再用两条弦的垂直平分线找圆心。取 z=0,i,−iz=0,\mathrm{i},-\mathrm{i}，三个像点的坐标都很简单。

答题过程

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For three points on the imaginary axis,

z=0⇒w=−iz=i⇒w=0\begin{align*} z=0\quad&\Rightarrow\quad w=-\mathrm{i}\\[2mm] z=\mathrm{i}\quad&\Rightarrow\quad w=0 \end{align*}

For z=−iz=-\mathrm{i},

w= −2i1−i= −2i(1+i)2= 1−i\begin{align*} w=&\,\frac{-2\mathrm{i}}{1-\mathrm{i}}\\[2mm] =&\,\frac{-2\mathrm{i}(1+\mathrm{i})}{2}\\[2mm] =&\,1-\mathrm{i} \end{align*}

Thus CC passes through A=(0,−1)A=(0,-1), B=(0,0)B=(0,0) and D=(1,−1)D=(1,-1).

The chord ABAB is vertical with midpoint (0,−12)(0,-\frac12), so its perpendicular bisector is

v=−12\begin{align*} v=-\frac12 \end{align*}

The chord ADAD is horizontal with midpoint (12,−1)(\frac12,-1), so its perpendicular bisector is

u=12\begin{align*} u=\frac12 \end{align*}

Their intersection gives the centre of CC:

(12,−12)\begin{align*} \boxed{\bigg(\frac12,-\frac12\bigg)} \end{align*}

(ii)

解法一

思路

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由 (i) 前两种解法的圆方程直接读出半径平方；若用第三种解法，则求圆心到任一像点的距离，结果相同。

答题过程

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From the circle equation in part (i),

(u−12)2+(v+12)2= 12\begin{align*} \bigg(u-\frac12\bigg)^2+\bigg(v+\frac12\bigg)^2 =&\,\frac12 \end{align*}

the radius is

r= 12=12\begin{align*} r=&\,\sqrt{\frac12}=\boxed{\frac1{\sqrt2}} \end{align*}

Equivalently, using the image point B=(0,0)B=(0,0) from the third method,

r= (0−12)2+(0+12)2= 12\begin{align*} r=&\,\sqrt{\bigg(0-\frac12\bigg)^2+\bigg(0+\frac12\bigg)^2}\\[2mm] =&\,\frac1{\sqrt2} \end{align*}