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IAL 2025 Jan Q8

A Level / Edexcel / FP2

IAL 2025 Jan Paper · Question 8

题目

Problem

A 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=(3−i)(z−2)z+2z≠−2\begin{align*} w=\frac{(\sqrt3-\mathrm{i})(z-2)}{z+2} \qquad z\neq -2 \end{align*}

(a) Show that the real axis in the zz-plane is mapped by TT onto the line with equation

v=−13u\begin{align*} v=-\frac1{\sqrt3}u \end{align*}

in the ww-plane.

(3)

(b) Show that the circle in the zz-plane with equation ∣z∣=2|z|=2 is mapped by TT onto a line in the ww-plane, stating clearly an equation for this line.

(5)

The region RR in the zz-plane is defined by

{z∈C:∣z∣<2}∩{z∈C:Im⁡z>0}\begin{align*} \{z\in\mathbb{C}:|z|<2\} \cap \{z\in\mathbb{C}:\operatorname{Im}z>0\} \end{align*}

(c) Determine the image of RR under TT, giving your answer in the form

{w∈C:α<arg⁡w<β}\begin{align*} \{w\in\mathbb{C}:\alpha<\arg w<\beta\} \end{align*}

where α\alpha and β\beta are rational multiples of π\pi

(5)

解答

(a)

解法一

思路

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实轴上的点可写成 z=xz=x，其中 xx 为实数。代入变换后，乘上的复数 3−i\sqrt3-\mathrm{i} 只是一个固定方向，而 x−2x+2\frac{x-2}{x+2} 是实数，所以 uu 和 vv 的比例很快能看出来。

答题过程

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On the real axis in the zz-plane,

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

where xx is real. Substitute into the transformation:

w=(3−i)(x−2)x+2=3(x−2x+2)−i(x−2x+2)\begin{align*} w ={}& \frac{(\sqrt3-\mathrm{i})(x-2)}{x+2}\\[3mm] ={}& \sqrt3\left(\frac{x-2}{x+2}\right) -\mathrm{i}\left(\frac{x-2}{x+2}\right) \end{align*}

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

u=3(x−2x+2),v=−(x−2x+2)\begin{align*} u ={}& \sqrt3\left(\frac{x-2}{x+2}\right),\\ v ={}& -\left(\frac{x-2}{x+2}\right) \end{align*}

Therefore

v=−13u\begin{align*} v ={}& -\frac1{\sqrt3}u \end{align*}

as required.

(b)

解法一

思路

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要把 ∣z∣=2|z|=2 转到 ww-plane，先把 zz 用 ww 表示，再代入 ∣z∣=2|z|=2。之后令 w=u+ivw=u+\mathrm{i}v，展开模长平方即可得到直线。

答题过程

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Start with

w=(3−i)(z−2)z+2\begin{align*} w ={}& \frac{(\sqrt3-\mathrm{i})(z-2)}{z+2} \end{align*}

Rearrange:

w(z+2)=(3−i)(z−2)wz+2w=(3−i)z−2(3−i)z(w−3+i)=−2w−2(3−i)\begin{align*} w(z+2) ={}& (\sqrt3-\mathrm{i})(z-2)\\[2mm] wz+2w ={}& (\sqrt3-\mathrm{i})z-2(\sqrt3-\mathrm{i})\\[2mm] z(w-\sqrt3+\mathrm{i}) ={}& -2w-2(\sqrt3-\mathrm{i}) \end{align*}

Hence

z=−2w−2(3−i)w−3+i\begin{align*} z ={}& \frac{-2w-2(\sqrt3-\mathrm{i})} {w-\sqrt3+\mathrm{i}} \end{align*}

Use ∣z∣=2|z|=2:

∣−2w−2(3−i)w−3+i∣=2\begin{align*} \left| \frac{-2w-2(\sqrt3-\mathrm{i})} {w-\sqrt3+\mathrm{i}} \right| ={}& 2 \end{align*}

Divide by 22:

∣w+3−i∣=∣w−3+i∣\begin{align*} \left|w+\sqrt3-\mathrm{i}\right| ={}& \left|w-\sqrt3+\mathrm{i}\right| \end{align*}

Let w=u+ivw=u+\mathrm{i}v. Then

∣w+3−i∣2=(u+3)2+(v−1)2\begin{align*} \left|w+\sqrt3-\mathrm{i}\right|^2 ={}& (u+\sqrt3)^2+(v-1)^2 \end{align*}

and

∣w−3+i∣2=(u−3)2+(v+1)2\begin{align*} \left|w-\sqrt3+\mathrm{i}\right|^2 ={}& (u-\sqrt3)^2+(v+1)^2 \end{align*}

So

(u+3)2+(v−1)2=(u−3)2+(v+1)2u2+23u+3+v2−2v+1=u2−23u+3+v2+2v+143u=4v\begin{align*} (u+\sqrt3)^2+(v-1)^2 ={}& (u-\sqrt3)^2+(v+1)^2\\[3mm] u^2+2\sqrt3u+3+v^2-2v+1 ={}& u^2-2\sqrt3u+3+v^2+2v+1\\[3mm] 4\sqrt3u ={}& 4v \end{align*}

Therefore the image is the line

v=3u\begin{align*} \boxed{v=\sqrt3u} \end{align*}

解法二

思路

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从

∣w+3−i∣=∣w−3+i∣\begin{align*} \left|w+\sqrt3-\mathrm{i}\right| = \left|w-\sqrt3+\mathrm{i}\right| \end{align*}

可以看出，ww 到两个固定点 (−3,1)(-\sqrt3,1) 和 (3,−1)(\sqrt3,-1) 的距离相等，所以轨迹是这两点连线的垂直平分线。

答题过程

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As in the first method,

z=−2w−2(3−i)w−3+i\begin{align*} z ={}& \frac{-2w-2(\sqrt3-\mathrm{i})} {w-\sqrt3+\mathrm{i}} \end{align*}

Using ∣z∣=2|z|=2 gives

∣w+3−i∣=∣w−3+i∣\begin{align*} \left|w+\sqrt3-\mathrm{i}\right| ={}& \left|w-\sqrt3+\mathrm{i}\right| \end{align*}

This means that ww is equidistant from

−3+iand3−i\begin{align*} -\sqrt3+\mathrm{i} \qquad \text{and} \qquad \sqrt3-\mathrm{i} \end{align*}

So the locus is the perpendicular bisector of the segment joining

(−3,1)and(3,−1)\begin{align*} (-\sqrt3,1) \qquad \text{and} \qquad (\sqrt3,-1) \end{align*}

The midpoint is (0,0)(0,0), and the slope of the segment is

−1−13−(−3)=−13\begin{align*} \frac{-1-1}{\sqrt3-(-\sqrt3)} ={}& -\frac1{\sqrt3} \end{align*}

Thus the perpendicular slope is 3\sqrt3. Since the perpendicular bisector passes through the origin,

v=3u\begin{align*} \boxed{v=\sqrt3u} \end{align*}

(c)

解法一

思路

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RR 是上半圆内部，所以边界来自两部分：实轴和圆 ∣z∣=2|z|=2。由 (a) 和 (b)，它们分别映到两条过原点的直线。最后用一个区域内的点，例如 z=iz=\mathrm{i}，判断夹角是哪一侧。

答题过程

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From part (b), the circle ∣z∣=2|z|=2 maps to

v=3u\begin{align*} v=\sqrt3u \end{align*}

This line has argument

arg⁡w=π3\begin{align*} \arg w=\frac{\pi}{3} \end{align*}

From part (a), the real axis maps to

v=−13u\begin{align*} v=-\frac1{\sqrt3}u \end{align*}

This line has possible arguments

−π6or5π6\begin{align*} -\frac{\pi}{6} \qquad \text{or} \qquad \frac{5\pi}{6} \end{align*}

To choose the correct side of the region, use a point inside RR, for example z=iz=\mathrm{i}:

w=(3−i)(i−2)i+2=4−335+i3+435\begin{align*} w ={}& \frac{(\sqrt3-\mathrm{i})(\mathrm{i}-2)}{\mathrm{i}+2}\\[4mm] ={}& \frac{4-3\sqrt3}{5} +\mathrm{i}\frac{3+4\sqrt3}{5} \end{align*}

This point lies in the second quadrant in the ww-plane, so the image of RR lies between the rays with arguments π3\frac{\pi}{3} and 5π6\frac{5\pi}{6}.

Therefore

{w∈C:π3<arg⁡w<5π6}\begin{align*} \boxed{ \{w\in\mathbb{C}:\frac{\pi}{3}<\arg w<\frac{5\pi}{6}\} } \end{align*}