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IAL 2022 Jan Q7

A Level / Edexcel / FP2

IAL 2022 Jan Paper · Question 7

题目

Problem

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

w=(1+i)z+2(1−i)z−iz≠i\begin{align*} w = \frac{(1 + \mathrm{i}) z + 2(1 - \mathrm{i})}{z - \mathrm{i}} \qquad z \neq \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)

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*}

(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*}

Multiply by the conjugate of the denominator:

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

The numerator expands to

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

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-\left(3v-v^2-2\right)=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*} \left(u-\frac32\right)^2-\frac94 +\left(v-\frac32\right)^2-\frac94 +4=0. \end{align*}

So

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

Therefore the centre is

(32,32),\begin{align*} \left(\frac32,\frac32\right), \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{\left[(1+\mathrm{i})x+2(1-\mathrm{i})\right](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*} \left(u-\frac32\right)^2 +\left(v-\frac32\right)^2 =&\,\frac12. \end{align*}

Hence the centre is

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

and the radius is

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