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 \neq -2\mathrm{i}\\[2mm] \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 $z$ 用 $w$ 表示，再把 $|z|=1$ 转成 $w$ 平面里的模长方程。

令 $w=u+\mathrm{i}v$ 后，把模长平方展开，就会得到圆的方程；最后配方读出圆心和半径。

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Given

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

Rearrange to make $z$ the subject:

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

Let

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

Then

\begin{align*} 2w+3=(2u+3)+2\mathrm{i}v \end{align*}

and

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

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

Divide by $3$ :

\begin{align*} u^2+v^2+\frac{14}{3}u+\frac{8}{3}=0. \end{align*}

Complete the square:

\begin{align*} \left(u+\frac{7}{3}\right)^2+v^2 =&\,\frac{49}{9}-\frac{8}{3}\\[2mm] =&\,\frac{49}{9}-\frac{24}{9}\\[2mm] =&\,\frac{25}{9}. \end{align*}

Therefore the centre of $C$ is

\begin{align*} \boxed{\left(-\frac{7}{3},0\right)} \end{align*}

and the radius of $C$ is

\begin{align*} \boxed{\frac{5}{3}}. \end{align*}