题目

Problem

The transformation $T$ from the $z$ -plane to the $w$ -plane is given by

\begin{align*} w=\frac{z-3\mathrm{i}}{z+2\mathrm{i}} \qquad z\ne -2\mathrm{i} \end{align*}

The circle with equation $|z|=1$ in the $z$ -plane is mapped by $T$ onto the circle $C$ in the $w$ -plane.

Determine

(i) the centre of $C$ ,

(ii) the radius of $C$ .

(7)

解答

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先把变换式改写成 $z$ 关于 $w$ 的表达式。因为原来的轨迹是 $|z|=1$ ，所以把 $z$ 的表达式代入模长条件。再令 $w=u+\mathrm{i}v$ ，整理成圆的标准方程。

### 答题过程

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Starting from

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

Since $|z|=1$ ,

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

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

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

\begin{align*} (2u+3)^2+(2v)^2 =&\,(1-u)^2+(-v)^2\\[4mm] 4u^2+12u+9+4v^2 =&\,1-2u+u^2+v^2\\[4mm] 3u^2+3v^2+14u+8 =&\,0\\[4mm] u^2+v^2+\frac{14}{3}u+\frac83 =&\,0 \end{align*}

Complete the square:

\begin{align*} \left(u+\frac73\right)^2+v^2 =&\,\frac{49}{9}-\frac{8}{3}\\[4mm] =&\,\frac{25}{9} \end{align*}

Therefore the centre of $C$ is

\begin{align*} \left(-\frac73,0\right) \end{align*}

and the radius is

\begin{align*} \frac53 \end{align*}