题目

Problem

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

\begin{align*} w =\,& \frac{z}{z + 4\mathrm{i}} \qquad z \neq -4\mathrm{i}\\[2mm] \end{align*}

The circle with equation $|z| = 3$ is mapped by $T$ onto the circle $C$

Determine

(i) a Cartesian equation of $C$

(ii) the centre and radius of $C$

(8)

解答

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这题的核心是把 $z$ 用 $w$ 表示，再把 $|z|=3$ 转成 $w$ 平面里的条件。

令 $w=u+\mathrm{i}v$ ，最后把模长方程展开并配方，就能得到圆的直角坐标方程、圆心和半径。

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From

\begin{align*} w=\frac{z}{z+4\mathrm{i}}, \end{align*}

we have

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

Since $|z|=3$ ,

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

Let

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

Then

\begin{align*} 4\sqrt{u^2+v^2} =&\,3\sqrt{(1-u)^2+v^2}. \end{align*}

Squaring both sides gives

\begin{align*} 16(u^2+v^2)=&\,9\{(1-u)^2+v^2\}\\[2mm] 16u^2+16v^2 =&\,9(1-2u+u^2+v^2)\\[2mm] 16u^2+16v^2 =&\,9-18u+9u^2+9v^2\\[2mm] 7u^2+7v^2+18u-9=&\,0. \end{align*}

So a Cartesian equation of $C$ is

\begin{align*} \boxed{7u^2+7v^2+18u-9=0}. \end{align*}

Completing the square,

\begin{align*} 7u^2+18u+7v^2-9=&\,0\\[2mm] u^2+\frac{18}{7}u+v^2-\frac{9}{7}=&\,0\\[2mm] \left(u+\frac{9}{7}\right)^2+v^2 =&\,\frac{9}{7}+\frac{81}{49}\\[2mm] =&\,\frac{63+81}{49}\\[2mm] =&\,\frac{144}{49}. \end{align*}

Therefore the centre is

\begin{align*} \boxed{\left(-\frac{9}{7},0\right)} \end{align*}

and the radius is

\begin{align*} \boxed{\frac{12}{7}}. \end{align*}