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FP2 2022 Jan Q7

A Level / Edexcel / FP2

IAL 2022 Jan Paper · Question 7

题目

Problem

7. A transformation from the zz-plane to the ww-plane is given by

w= (1+i)z+2(1−i)z−i, z≠i\begin{align*} w=&\,\frac{(1+\mathrm{i})z+2(1-\mathrm{i})}{z-\mathrm{i}},\\[4mm] &\,z\ne\mathrm{i} \end{align*}

The transformation maps points on the imaginary axis in the zz-plane onto a line in the ww-plane.

(a) Find an equation for this line.

(2)

The transformation maps points on the real axis in the zz-plane onto a circle in the ww-plane.

(b) Find the centre and radius of this circle.

(6)
题目中文翻译

7. 从 zz 平面到 ww 平面的变换为

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

该变换将 zz 平面虚轴上的点映射到 ww 平面的一条直线上。

(a) 求这条直线的方程。

(2)

该变换将 zz 平面实轴上的点映射到 ww 平面的一个圆上。

(b) 求这个圆的圆心和半径。

(6)

解答

(a)

解法一

思路

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imaginary axis 上的点可以写成 z=iyz=\mathrm{i}y，其中 yy 是实数。代入 transformation 后，把 ww 写成 u+ivu+\mathrm{i}v，即可看出 uu 和 vv 的关系。

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= (1+i)(iy)+2(1−i)iy−i= −y+iy+2−2ii(y−1)= (2−y)+i(y−2)i(y−1).\begin{align*} w =&\,\frac{(1+\mathrm{i})(\mathrm{i}y)+2(1-\mathrm{i})} {\mathrm{i}y-\mathrm{i}}\\[4mm] =&\,\frac{-y+\mathrm{i}y+2-2\mathrm{i}} {\mathrm{i}(y-1)}\\[4mm] =&\,\frac{(2-y)+\mathrm{i}(y-2)} {\mathrm{i}(y-1)}. \end{align*}

Since

2−y=−(y−2),\begin{align*} 2-y=-(y-2), \end{align*}

the numerator is

(y−2)(−1+i).\begin{align*} (y-2)(-1+\mathrm{i}). \end{align*}

Thus

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

Therefore the real and imaginary parts are equal, so the line is

u=v.\begin{align*} u=v. \end{align*}

解法二

思路

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题目已经说明像轨迹是一条直线，因此求虚轴上两个不同点的像，就能确定这条直线。选 z=0z=0 和 z=2iz=2\mathrm{i}，避开变换没有定义的 z=iz=\mathrm{i}。

答题过程

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The image is given to be a line. When z=0z=0,

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

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

w= (1+i)2i+2(1−i)2i−i= −2+2i+2−2ii= 0.\begin{align*} w=&\,\frac{(1+\mathrm{i})2\mathrm{i}+2(1-\mathrm{i})} {2\mathrm{i}-\mathrm{i}}\\[4mm] =&\,\frac{-2+2\mathrm{i}+2-2\mathrm{i}}{\mathrm{i}}\\[4mm] =&\,0. \end{align*}

Writing w=u+ivw=u+\mathrm{i}v, the line passes through (2,2)(2,2) and (0,0)(0,0). Its gradient is

2−02−0=1.\begin{align*} \frac{2-0}{2-0}=1. \end{align*}

Hence the equation is

v=u.\begin{align*} v=u. \end{align*}

(b)

解法一

思路

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把 transformation 反过来，先把 zz 表示成 ww。real axis 的条件是 Im⁡(z)=0\operatorname{Im}(z)=0。把 w=u+ivw=u+\mathrm{i}v 代入后，令 zz 的虚部为 00，就能得到圆的 Cartesian equation。

答题过程

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Start with

w=(1+i)z+2(1−i)z−i.\begin{align*} w=\frac{(1+\mathrm{i})z+2(1-\mathrm{i})}{z-\mathrm{i}}. \end{align*}

Then

w(z−i)=(1+i)z+2(1−i).\begin{align*} w(z-\mathrm{i})=(1+\mathrm{i})z+2(1-\mathrm{i}). \end{align*}

Collect the terms involving zz:

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

So

z=iw+2(1−i)w−1−i.\begin{align*} z=\frac{\mathrm{i}w+2(1-\mathrm{i})}{w-1-\mathrm{i}}. \end{align*}

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

iw+2(1−i)= i(u+iv)+2−2i= 2−v+i(u−2),\begin{align*} \mathrm{i}w+2(1-\mathrm{i}) =&\,\mathrm{i}(u+\mathrm{i}v)+2-2\mathrm{i}\\[2mm] =&\,2-v+\mathrm{i}(u-2), \end{align*}

and

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

Therefore

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

After multiplying by the conjugate of the denominator, write

z=ND,\begin{align*} z=\frac{N}{D}, \end{align*}

where

D= (u−1)2+(v−1)2,N= [2−v+i(u−2)] ×[u−1−i(v−1)].\begin{align*} D=&\,(u-1)^2+(v-1)^2,\\[4mm] N=&\,\big[2-v+\mathrm{i}(u-2)\big]\\[4mm] &\,\hspace{2pt}\times\big[u-1-\mathrm{i}(v-1)\big]. \end{align*}

Expanding the numerator gives

N= (2−v)(u−1)+(u−2)(v−1) +i[(u−1)(u−2) −(2−v)(v−1)].\begin{align*} N=&\,(2-v)(u-1)+(u-2)(v-1)\\[4mm] &\,\hspace{2pt}+\mathrm{i}\big[(u-1)(u-2)\\[4mm] &\,\hspace{4pt}-(2-v)(v-1)\big]. \end{align*}

Since DD is real and positive, Im⁡(z)=0\operatorname{Im}(z)=0 requires Im⁡(N)=0\operatorname{Im}(N)=0.

For points on the real axis,

Im⁡(z)=0.\begin{align*} \operatorname{Im}(z)=0. \end{align*}

So

(u−1)(u−2)−(2−v)(v−1)=0.\begin{align*} (u-1)(u-2)-(2-v)(v-1)=0. \end{align*}

Expand:

u2−3u+2−(3v−v2−2)=0.\begin{align*} u^2-3u+2-\Big(3v-v^2-2\Big)=0. \end{align*}

Hence

u2+v2−3u−3v+4=0.\begin{align*} u^2+v^2-3u-3v+4=0. \end{align*}

Complete the square:

 (u−32)2−94 +(v−32)2−94+4= 0.\begin{align*} &\,\Big(u-\frac32\Big)^2-\frac94\\[4mm] &\,\hspace{2pt}+\Big(v-\frac32\Big)^2-\frac94+4\\[4mm] =&\,0. \end{align*}

So

(u−32)2+(v−32)2= 12.\begin{align*} \Big(u-\frac32\Big)^2 +\Big(v-\frac32\Big)^2 =&\,\frac12. \end{align*}

Therefore the centre is

(32,32),\begin{align*} \Big(\frac32,\frac32\Big), \end{align*}

and the radius is

12=22.\begin{align*} \sqrt{\frac12}=\frac{\sqrt2}{2}. \end{align*}

解法二

思路

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也可以直接令 real axis 上的点为 z=xz=x，其中 xx 是实数。把 transformation 化成 u+ivu+\mathrm{i}v 后消去 xx。

答题过程

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On the real axis,

z=x.\begin{align*} z=x. \end{align*}

Substitute into the transformation:

w= (1+i)x+2(1−i)x−i.\begin{align*} w =&\,\frac{(1+\mathrm{i})x+2(1-\mathrm{i})}{x-\mathrm{i}}. \end{align*}

Multiply numerator and denominator by x+ix+\mathrm{i}:

w= [(1+i)x+2(1−i)](x+i)x2+1.\begin{align*} w =&\,\frac{\Big[(1+\mathrm{i})x+2(1-\mathrm{i})\Big](x+\mathrm{i})} {x^2+1}. \end{align*}

Expanding the numerator gives

w= x2+x+2x2+1+ix2−x+2x2+1.\begin{align*} w =&\,\frac{x^2+x+2} {x^2+1} +\mathrm{i}\frac{x^2-x+2} {x^2+1}. \end{align*}

So

u= x2+x+2x2+1,v= x2−x+2x2+1.\begin{align*} u=&\,\frac{x^2+x+2}{x^2+1},\\ v=&\,\frac{x^2-x+2}{x^2+1}. \end{align*}

Then

u−1= x+1x2+1,\begin{align*} u-1 =&\,\frac{x+1}{x^2+1}, \end{align*}

and

v−1= 1−xx2+1.\begin{align*} v-1 =&\,\frac{1-x}{x^2+1}. \end{align*}

Hence

(u−1)2+(v−1)2= (x+1)2+(1−x)2(x2+1)2= 2x2+2(x2+1)2= 2x2+1.\begin{align*} (u-1)^2+(v-1)^2 =&\,\frac{(x+1)^2+(1-x)^2} {(x^2+1)^2}\\[2mm] =&\,\frac{2x^2+2}{(x^2+1)^2}\\[2mm] =&\,\frac{2}{x^2+1}. \end{align*}

Also,

u+v−2= 2x2+4x2+1−2= 2x2+1.\begin{align*} u+v-2 =&\,\frac{2x^2+4}{x^2+1}-2\\[2mm] =&\,\frac{2}{x^2+1}. \end{align*}

Therefore

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

Expand:

u2+v2−3u−3v+4=0.\begin{align*} u^2+v^2-3u-3v+4=0. \end{align*}

Completing the square,

(u−32)2+(v−32)2= 12.\begin{align*} \Big(u-\frac32\Big)^2 +\Big(v-\frac32\Big)^2 =&\,\frac12. \end{align*}

Hence the centre is

(32,32),\begin{align*} \Big(\frac32,\frac32\Big), \end{align*}

and the radius is

22.\begin{align*} \frac{\sqrt2}{2}. \end{align*}

解法三

思路

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题目已经说明实轴的像轨迹是圆。取实轴上 z=0,1,−1z=0,1,-1 三点，求出三个像点后，利用两条弦的垂直平分线求圆心，再用圆心到其中一个点的距离求半径。

答题过程

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The image of the real axis is given to be a circle. From part (a), z=0z=0 maps to w=2+2iw=2+2\mathrm{i}.

When z=1z=1,

w= (1+i)+2(1−i)1−i= 3−i1−i= (3−i)(1+i)2= 4+2i2= 2+i.\begin{align*} w=&\,\frac{(1+\mathrm{i})+2(1-\mathrm{i})}{1-\mathrm{i}}\\[4mm] =&\,\frac{3-\mathrm{i}}{1-\mathrm{i}}\\[4mm] =&\,\frac{(3-\mathrm{i})(1+\mathrm{i})}{2}\\[4mm] =&\,\frac{4+2\mathrm{i}}{2}\\[4mm] =&\,2+\mathrm{i}. \end{align*}

When z=−1z=-1,

w= −(1+i)+2(1−i)−1−i= 1−3i−1−i= (1−3i)(−1+i)2= 2+4i2= 1+2i.\begin{align*} w=&\,\frac{-(1+\mathrm{i})+2(1-\mathrm{i})}{-1-\mathrm{i}}\\[4mm] =&\,\frac{1-3\mathrm{i}}{-1-\mathrm{i}}\\[4mm] =&\,\frac{(1-3\mathrm{i})(-1+\mathrm{i})}{2}\\[4mm] =&\,\frac{2+4\mathrm{i}}{2}\\[4mm] =&\,1+2\mathrm{i}. \end{align*}

Thus three points on the image circle are

A= (2,2),B= (2,1),C= (1,2).\begin{align*} A=&\,(2,2),\\[4mm] B=&\,(2,1),\\[4mm] C=&\,(1,2). \end{align*}

The chord ABAB is vertical, with midpoint (2,32)(2,\frac32). Its perpendicular bisector is

v=32.\begin{align*} v=\frac32. \end{align*}

The chord ACAC is horizontal, with midpoint (32,2)(\frac32,2). Its perpendicular bisector is

u=32.\begin{align*} u=\frac32. \end{align*}

Their intersection gives the centre

(32,32).\begin{align*} \Big(\frac32,\frac32\Big). \end{align*}

Using the distance from the centre to AA, the radius is

r= (2−32)2+(2−32)2= 14+14= 22.\begin{align*} r=&\,\sqrt{\Big(2-\frac32\Big)^2+\Big(2-\frac32\Big)^2}\\[4mm] =&\,\sqrt{\frac14+\frac14}\\[4mm] =&\,\frac{\sqrt2}{2}. \end{align*}