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IAL 2021 Oct Q3

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 3

题目

Problem

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−iz≠i\begin{align*} w = \frac{(2 + \mathrm{i}) z + 4}{z - \mathrm{i}} \qquad z \neq \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)

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 \quad\Longrightarrow\quad 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 =&\,(2+\mathrm{i})z+4,\\[2mm] wz-(2+\mathrm{i})z =&\,\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*}

Multiply numerator and denominator by the conjugate of the denominator:

z= (4−v+iu)(u−2−i(v−1))(u−2)2+(v−1)2.\begin{align*} z =&\,\frac{\left(4-v+\mathrm{i}u\right) \left(u-2-\mathrm{i}(v-1)\right)} {(u-2)^2+(v-1)^2}. \end{align*}

The denominator is real, so the real part of zz comes from the real part of the numerator. Expand the numerator:

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

Since zz lies on the imaginary axis,

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

Therefore

(4−v)(u−2)+u(v−1)=0.\begin{align*} (4-v)(u-2)+u(v-1)=0. \end{align*}

Expand:

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

Hence

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