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FP2 2021 Oct Q3

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 3

题目

Problem

3. A transformation maps points from the zz-plane, where z=x+iyz = x + \mathrm{i} y, to the ww-plane, where w=u+ivw = u + \mathrm{i} v. The transformation is given by

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

The transformation maps the imaginary axis in the zz-plane onto the line ll in the ww-plane.

Determine a Cartesian equation of ll, giving your answer in the form au+bv+c=0au + bv + c = 0 where aa, bb and cc are integers to be found.

(6)
题目中文翻译

3. 某变换将 zz 平面（其中 z=x+iyz=x+\mathrm{i}y）上的点映到 ww 平面（其中 w=u+ivw=u+\mathrm{i}v）上。变换定义为

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

该变换将 zz 平面上的虚轴映为 ww 平面上的直线 ll。

求 ll 的直角坐标方程，将答案写成 au+bv+c=0au+bv+c=0 的形式，其中 aa、bb 和 cc 是待求的整数。

(6)

解答

解法一

思路

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imaginary axis 上的点满足 z=iyz=\mathrm{i}y，其中 yy 是实数。把它直接代入 transformation，就能把 uu 和 vv 都表示成 yy 的函数。然后消去 yy，得到 ww-plane 中的直线方程。

答题过程

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On the imaginary axis in the zz-plane,

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

Substitute this into the transformation:

w= (2+i)(iy)+4iy−i= 2iy+i2y+4i(y−1)= 4−y+2iyi(y−1).\begin{align*} w =&\,\frac{(2+\mathrm{i})(\mathrm{i}y)+4} {\mathrm{i}y-\mathrm{i}}\\[4mm] =&\,\frac{2\mathrm{i}y+\mathrm{i}^2y+4} {\mathrm{i}(y-1)}\\[4mm] =&\,\frac{4-y+2\mathrm{i}y} {\mathrm{i}(y-1)}. \end{align*}

Multiply numerator and denominator by −i-\mathrm{i}:

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

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

u= 2yy−1,v= y−4y−1.\begin{align*} u=&\,\frac{2y}{y-1},\\[2mm] v=&\,\frac{y-4}{y-1}. \end{align*}

From

u=2yy−1,\begin{align*} u=\frac{2y}{y-1}, \end{align*}

we get

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

So

uy−u= 2y,y(u−2)= u.\begin{align*} uy-u=&\,2y,\\[4mm] y(u-2)=&\,u. \end{align*}

Hence

y=uu−2.\begin{align*} y=\frac{u}{u-2}. \end{align*}

Now use

v=y−4y−1.\begin{align*} v=\frac{y-4}{y-1}. \end{align*}

Substitute y=uu−2y=\dfrac{u}{u-2}:

v= uu−2−4uu−2−1= u−4(u−2)u−(u−2)= −3u+82.\begin{align*} v =&\,\frac{\frac{u}{u-2}-4} {\frac{u}{u-2}-1}\\[4mm] =&\,\frac{u-4(u-2)} {u-(u-2)}\\[4mm] =&\,\frac{-3u+8}{2}. \end{align*}

Therefore

2v=−3u+8.\begin{align*} 2v=-3u+8. \end{align*}

So the Cartesian equation of ll is

3u+2v−8=0.\begin{align*} 3u+2v-8=0. \end{align*}

解法二

思路

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也可以先把 transformation 反过来，把 zz 表示成 ww。因为 imaginary axis 的条件是 Re⁡(z)=0\operatorname{Re}(z)=0，所以把反解后的 zz 写成 u,vu,v 的形式，再令实部为 00。

答题过程

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Start with

w=(2+i)z+4z−i.\begin{align*} w=\frac{(2+\mathrm{i})z+4}{z-\mathrm{i}}. \end{align*}

Then

w(z−i)=(2+i)z+4.\begin{align*} w(z-\mathrm{i})=(2+\mathrm{i})z+4. \end{align*}

Expand and collect the terms involving zz:

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

So

z=iw+4w−2−i.\begin{align*} z=\frac{\mathrm{i}w+4}{w-2-\mathrm{i}}. \end{align*}

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

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

For compact notation, define the real quantities

a= 4−v,b= u−2,c= v−1.\begin{align*} a=&\,4-v,\\[4mm] b=&\,u-2,\\[4mm] c=&\,v-1. \end{align*}

Then

z= a+iub+ic.\begin{align*} z=&\,\frac{a+\mathrm{i}u}{b+\mathrm{i}c}. \end{align*}

Multiply numerator and denominator by b−icb-\mathrm{i}c:

z= Nb2+c2,\begin{align*} z=&\,\frac{N}{b^2+c^2}, \end{align*}

where the numerator is

N= (a+iu)(b−ic)= ab−iac +iub+uc= (ab+uc) +i(ub−ac).\begin{align*} N=&\,(a+\mathrm{i}u)(b-\mathrm{i}c)\\[4mm] =&\,ab-\mathrm{i}ac\\[4mm] &\,\hspace{2pt}+\mathrm{i}ub+uc\\[4mm] =&\,(ab+uc)\\[4mm] &\,\hspace{2pt}+\mathrm{i}(ub-ac). \end{align*}

Since the denominator is real and zz lies on the imaginary axis,

Re⁡(z)= 0,ab+uc= 0.\begin{align*} \operatorname{Re}(z)=&\,0,\\[4mm] ab+uc=&\,0. \end{align*}

Substitute the definitions of aa, bb and cc:

 (4−v)(u−2) +u(v−1)= 0.\begin{align*} &\,(4-v)(u-2)\\[4mm] &\,\hspace{2pt}+u(v-1)\\[4mm] =&\,0. \end{align*}

Expanding and simplifying gives

 4u−8−uv+2v +uv−u= 0.\begin{align*} &\,4u-8-uv+2v\\[4mm] &\,\hspace{2pt}+uv-u\\[4mm] =&\,0. \end{align*}

Hence

3u+2v−8= 0.\begin{align*} 3u+2v-8=&\,0. \end{align*}

解法三

思路

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题目已经说明虚轴的像是一条直线，因此只需选虚轴上两个不同的点，求出它们的像，再写出通过这两个像点的直线方程。取 z=0z=0 与 z=2iz=2\mathrm{i} 可以避开变换中不允许的 z=iz=\mathrm{i}，计算也较简单。

答题过程

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The image is given to be a line. We find the images of two distinct points on the imaginary axis.

When z=0z=0,

w= 4−i= 4i.\begin{align*} w=&\,\frac4{-\mathrm{i}}\\[4mm] =&\,4\mathrm{i}. \end{align*}

So one point on ll is (0,4)(0,4).

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

w= (2+i)(2i)+42i−i= 4i−2+4i= 2+4ii= 4−2i.\begin{align*} w=&\,\frac{(2+\mathrm{i})(2\mathrm{i})+4}{2\mathrm{i}-\mathrm{i}}\\[4mm] =&\,\frac{4\mathrm{i}-2+4}{\mathrm{i}}\\[4mm] =&\,\frac{2+4\mathrm{i}}{\mathrm{i}}\\[4mm] =&\,4-2\mathrm{i}. \end{align*}

So another point on ll is (4,−2)(4,-2).

The gradient of the line through these points is

m= −2−44−0= −32.\begin{align*} m=&\,\frac{-2-4}{4-0}\\[4mm] =&\,-\frac32. \end{align*}

Using the point (0,4)(0,4),

v−4= −32(u−0),2v−8= −3u.\begin{align*} v-4=&\,-\frac32(u-0),\\[4mm] 2v-8=&\,-3u. \end{align*}

Hence the required Cartesian equation is

3u+2v−8= 0.\begin{align*} 3u+2v-8=&\,0. \end{align*}