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FP2 2021 June Q2

A Level / Edexcel / FP2

IAL 2021 June Paper · Question 2

题目

Problem

2. 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+2z−i, z≠i.\begin{align*} w=&\,\frac{z+2}{z-\mathrm{i}},\\[4mm] &\,z\ne\mathrm{i}. \end{align*}

The transformation TT maps the circle ∣z∣=2|z| = 2 in the zz-plane onto a circle CC in the ww-plane.

Find

(i) the centre of CC,

(ii) the radius of CC.

(8)
题目中文翻译

2. 变换 TT 将 zz 平面（其中 z=x+iyz=x+\mathrm{i}y）映到 ww 平面（其中 w=u+ivw=u+\mathrm{i}v），其定义为

w= z+2z−i, z≠i.\begin{align*} w=&\,\frac{z+2}{z-\mathrm{i}},\\[4mm] &\,z\ne\mathrm{i}. \end{align*}

变换 TT 将 zz 平面上的圆 ∣z∣=2|z|=2 映为 ww 平面上的圆 CC。

求

(i) CC 的圆心；

(ii) CC 的半径。

(8)

解答

解法一

思路

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要找 ww-plane 里的圆，先把 zz 用 ww 表示，再代入原来的圆 ∣z∣=2|z|=2。之后令 w=u+ivw=u+\mathrm{i}v，把模长方程转成 Cartesian equation。

答题过程

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Starting from

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

we rearrange:

w(z−i)= z+2wz−iw= z+2z(w−1)= 2+iwz= 2+iww−1.\begin{align*} w(z-\mathrm{i})=&\,z+2\\[4mm] wz-\mathrm{i}w=&\,z+2\\[4mm] z(w-1)=&\,2+\mathrm{i}w\\[4mm] z=&\,\frac{2+\mathrm{i}w}{w-1}. \end{align*}

Since the original locus is ∣z∣=2|z|=2,

∣2+iww−1∣=2.\begin{align*} \Big|\frac{2+\mathrm{i}w}{w-1}\Big|=2. \end{align*}

Let

w=u+iv.\begin{align*} w=u+\mathrm{i}v. \end{align*}

Then

2+iw= 2+i(u+iv)= (2−v)+iu,\begin{align*} 2+\mathrm{i}w =&\,2+\mathrm{i}(u+\mathrm{i}v)\\[4mm] =&\,(2-v)+\mathrm{i}u, \end{align*}

and

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

So

∣2+iw∣∣w−1∣=2.\begin{align*} \frac{|2+\mathrm{i}w|}{|w-1|}=2. \end{align*}

Squaring both sides gives

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

Expand and collect terms:

 4−4v+v2+u2= 4u2−8u+4+4v2.\begin{align*} &\,4-4v+v^2+u^2\\[4mm] =&\,4u^2-8u+4+4v^2. \end{align*}

Collecting terms,

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

Divide by 33:

u2+v2−83u+43v=0.\begin{align*} u^2+v^2-\frac83u+\frac43v=0. \end{align*}

Complete the square:

 (u−43)2 +(v+23)2= 169+49= 209.\begin{align*} &\,\big(u-\frac43\big)^2\\[4mm] &\,\hspace{2pt}+\big(v+\frac23\big)^2\\[4mm] =&\,\frac{16}{9}+\frac49\\[4mm] =&\,\frac{20}{9}. \end{align*}

Therefore, the centre of CC is

(43,−23),\begin{align*} \big(\frac43,-\frac23\big), \end{align*}

and the radius is

209=253.\begin{align*} \sqrt{\frac{20}{9}}=\frac{2\sqrt5}{3}. \end{align*}

解法二

思路

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也可以找像圆的一条直径的两个端点，再用中点及两点间距离求圆心和半径。

关键不是“直径一定会映成直径”。这里虚轴穿过原圆圆心，且包含使分母为零的点 i\mathrm{i}，所以虚轴的像是一条直线。这类分式线性变换保持交角，原圆与虚轴正交，因此像圆也与这条像直线正交；两个交点的像才是像圆的直径端点。

答题过程

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The imaginary axis meets ∣z∣=2|z|=2 orthogonally at z=2iz=2\mathrm{i} and z=−2iz=-2\mathrm{i}.

To find its image, put z=tiz=t\mathrm{i}, where tt is real and t≠1t\ne1. Then

w= 1+2+iz−i= 1+1−2it−1.\begin{align*} w=&\,1+\frac{2+\mathrm{i}}{z-\mathrm{i}}\\[4mm] =&\,1+\frac{1-2\mathrm{i}}{t-1}. \end{align*}

Thus its image is a straight line. Since the transformation preserves angles, this line meets CC orthogonally and hence passes through the centre of CC.

Therefore the images of 2i2\mathrm{i} and −2i-2\mathrm{i} are the endpoints of a diameter of CC.

For z=2iz=2\mathrm{i},

w1= 2i+2i= 2−2i.\begin{align*} w_1=&\,\frac{2\mathrm{i}+2}{\mathrm{i}}\\[4mm] =&\,2-2\mathrm{i}. \end{align*}

For z=−2iz=-2\mathrm{i},

w2= −2i+2−3i= 23+23i.\begin{align*} w_2=&\,\frac{-2\mathrm{i}+2}{-3\mathrm{i}}\\[4mm] =&\,\frac23+\frac23\mathrm{i}. \end{align*}

The centre is the midpoint of these endpoints. Its complex coordinate is

c= w1+w22= 12(83−43i)= 43−23i.\begin{align*} c=&\,\frac{w_1+w_2}{2}\\[4mm] =&\,\frac12\bigg(\frac83-\frac43\mathrm{i}\bigg)\\[4mm] =&\,\frac43-\frac23\mathrm{i}. \end{align*}

Hence the centre is

(43,−23).\bigg(\frac43,-\frac23\bigg).

The radius is half the diameter:

R= 12∣w1−w2∣= 12∣43−83i∣= 12169+649= 253.\begin{align*} R=&\,\frac12|w_1-w_2|\\[4mm] =&\,\frac12\Big|\frac43-\frac83\mathrm{i}\Big|\\[4mm] =&\,\frac12\sqrt{\frac{16}{9}+\frac{64}{9}}\\[4mm] =&\,\frac{2\sqrt5}{3}. \end{align*}