题目

Problem

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

\begin{align*} w =&\, \frac{z - \mathrm{i}}{z + 1} \qquad z \neq -1 \end{align*}

Given that $T$ maps the imaginary axis in the $z$ -plane to the circle $C$ in the $w$ -plane, determine

(i) the coordinates of the centre of $C$

(ii) the radius of $C$

(7)

解答

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虚轴上的点满足 $\operatorname{Re}(z)=0$ 。先把 $z$ 用 $w$ 表示，再令 $w=u+\mathrm{i}v$ ，取 $z$ 的实部等于 $0$ ，就能得到 $w$ -plane 中的圆方程。

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From

\begin{align*} w=\frac{z-\mathrm{i}}{z+1}, \end{align*}

make $z$ the subject:

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

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

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

Rationalise the denominator:

\begin{align*} z =&\, \frac{(u+\mathrm{i}(v+1))((1-u)+\mathrm{i}v)} {(1-u)^2+v^2} \end{align*}

The real part of the numerator is

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

Since the original locus is the imaginary axis, $\operatorname{Re}(z)=0$ . Hence

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

Complete the square:

\begin{align*} \left(u-\frac12\right)^2-\frac14 +\left(v+\frac12\right)^2-\frac14 =&\,0\\[4mm] \left(u-\frac12\right)^2 +\left(v+\frac12\right)^2 =&\, \frac12 \end{align*}

Therefore the centre is

\begin{align*} \boxed{\left(\frac12,-\frac12\right)} \end{align*}

and the radius is

\begin{align*} \boxed{\sqrt{\frac12}=\frac{1}{\sqrt2}} \end{align*}

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虚轴可以写成 $z=\mathrm{i}y$ 。直接代入变换，得到 $u,v$ 关于参数 $y$ 的式子，再消去 $y$ ，也能得到同一个圆。

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On the imaginary axis,

\begin{align*} z=\mathrm{i}y \end{align*}

Substitute into the transformation:

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

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

\begin{align*} u=&\,\frac{y^2-y}{1+y^2}\\[2mm] v=&\,\frac{y-1}{1+y^2} \end{align*}

From these two equations,

\begin{align*} u-yv =&\, \frac{y^2-y}{1+y^2} -y\frac{y-1}{1+y^2}\\[4mm] =&\,0 \end{align*}

so $u=yv$ . Substitute $y=\frac{u}{v}$ into $v=\frac{y-1}{1+y^2}$ :

\begin{align*} v =&\, \frac{\frac{u}{v}-1}{1+\frac{u^2}{v^2}}\\[4mm] v(u^2+v^2) =&\, uv-v^2 \end{align*}

If $v=0$ , then $y=1$ and $u=0$ , which also lies on the circle found below. For the remaining points, divide by $v$ :

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

Hence

\begin{align*} \left(u-\frac12\right)^2+\left(v+\frac12\right)^2 =&\, \frac12 \end{align*}

So the centre is $\left(\frac12,-\frac12\right)$ and the radius is $\frac1{\sqrt2}$ .