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 + p\mathrm{i}}{\mathrm{i}z + 3} \qquad z \neq 3\mathrm{i} \qquad p \in \mathbb{Z} \end{align*}

The point representing $\mathrm{i}(1 + \sqrt{3})$ is invariant under $T$ .

Determine the value of $p$ .

(3)

解答

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The point representing $\mathrm{i}(1 + \sqrt{3})$ is invariant under $T$ 这句，说明 $\mathrm{i}(1 + \sqrt{3})$ 是不动点。

不动点的意思是这个点经过 transformation 后仍然回到自己，所以这里要令

\begin{align*} z = w = \mathrm{i}(1+\sqrt{3}). \end{align*}

然后代入 transformation，解出整数 $p$ 。

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Since the point is invariant under $T$ ,

\begin{align*} z = w = \mathrm{i}(1+\sqrt{3}). \end{align*}

Substituting into the transformation gives

\begin{align*} \mathrm{i}(1+\sqrt{3}) =&\,\frac{\mathrm{i}(1+\sqrt{3})+p\mathrm{i}}{\mathrm{i}\big(\mathrm{i}(1+\sqrt{3})\big)+3}\\[4mm] =&\,\frac{\mathrm{i}(1+\sqrt{3}+p)}{2-\sqrt{3}}. \end{align*}

Therefore,

\begin{align*} \mathrm{i}(1+\sqrt{3})(2-\sqrt{3}) =&\,\mathrm{i}(1+\sqrt{3}+p). \end{align*}

Cancelling $\mathrm{i}$ ,

\begin{align*} (1+\sqrt{3})(2-\sqrt{3}) =&\,1+\sqrt{3}+p\\[4mm] 2-\sqrt{3}+2\sqrt{3}-3 =&\,1+\sqrt{3}+p\\[4mm] -1+\sqrt{3} =&\,1+\sqrt{3}+p. \end{align*}

Hence,

\begin{align*} p=-2. \end{align*}