题目

Problem

The transformation $T$ from the $z$ -plane, where $z = x + iy$ , to the $w$ -plane, where $w = u + iv$ , is given by

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

The transformation $T$ maps the circle $|z| = 2$ in the $z$ -plane onto a circle $C$ in the $w$ -plane.

Find

(i) the centre of $C$ ,

(ii) the radius of $C$ .

(8)

解答

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要找 $w$ -plane 里的圆，先把 $z$ 用 $w$ 表示，再代入原来的圆 $|z|=2$ 。之后令 $w=u+\mathrm{i}v$ ，把模长方程转成 Cartesian equation。

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Starting from

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

we rearrange:

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

Since the original locus is $|z|=2$ ,

\begin{align*} \left|\frac{2+\mathrm{i}w}{w-1}\right|=2. \end{align*}

Let

\begin{align*} w=u+\mathrm{i}v. \end{align*}

Then

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

and

\begin{align*} w-1=(u-1)+\mathrm{i}v. \end{align*}

\begin{align*} \frac{|2+\mathrm{i}w|}{|w-1|}=2. \end{align*}

Squaring both sides gives

\begin{align*} (2-v)^2+u^2 =&\,4\big((u-1)^2+v^2\big). \end{align*}

Expand and collect terms:

\begin{align*} 4-4v+v^2+u^2 =&\,4u^2-8u+4+4v^2\\[4mm] 3u^2+3v^2-8u+4v =&\,0. \end{align*}

Divide by $3$ :

\begin{align*} u^2+v^2-\frac83u+\frac43v=0. \end{align*}

Complete the square:

\begin{align*} \left(u-\frac43\right)^2 +\left(v+\frac23\right)^2 =&\,\frac{16}{9}+\frac49\\[4mm] =&\,\frac{20}{9}. \end{align*}

Therefore, the centre of $C$ is

\begin{align*} \left(\frac43,-\frac23\right), \end{align*}

and the radius is

\begin{align*} \sqrt{\frac{20}{9}}=\frac{2\sqrt5}{3}. \end{align*}