Question

Problem

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

Given that $z = x + iy$ , where $x$ and $y$ are real numbers, solve the equation

$(z - 2i)(z^* - 2i) = 21 - 12i$

where $z^*$ is the complex conjugate of $z$ .

(6)

中文翻译

在本题中必须展示所有运算过程。完全依赖计算器技术的解答不可接受。

已知 $z = x + iy$ ，其中 $x$ 和 $y$ 是实数，解方程

$(z - 2i)(z^* - 2i) = 21 - 12i$

其中 $z^*$ 是 $z$ 的共轭复数。

(6)

解答

展开

先写出 $z^*=x-iy$ ，再把左边展开。比较等式两边的实部和虚部，可以分别得到关于 $x$ 、 $y$ 的方程。

展开

Since $z=x+iy$ , its complex conjugate is

z^*=x-iy.

Therefore,

\begin{align*} (z-2i)(z^*-2i) =&\,[x+i(y-2)][x-i(y+2)]\\ =&\,x^2-ix(y+2)+ix(y-2)+(y-2)(y+2)\\ =&\,x^2+y^2-4-4xi. \end{align*}

Comparing real and imaginary parts with $21-12i$ gives

x^2+y^2-4=21

and

-4x=-12.

Hence $x=3$ , and so

3^2+y^2=25.

Thus $y^2=16$ , giving $y=\pm4$ . Therefore,

\boxed{z=3+4i\quad\text{or}\quad z=3-4i}.

展开

先利用 $zz^*=|z|^2$ 和 $z+z^*=2x$ 化简乘积，再比较实部与虚部。这种写法可以避免逐项展开两个括号。

展开

Expanding in terms of $z$ and $z^*$ ,

\begin{align*} (z-2i)(z^*-2i) =&\,zz^*-2i(z+z^*)-4\\ =&\,x^2+y^2-4xi-4. \end{align*}

Hence, by comparing real and imaginary parts,

x^2+y^2-4=21, \qquad -4x=-12.

It follows that $x=3$ and

y^2=25-9=16.

Therefore,

\boxed{z=3\pm4i}.