题目
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)
题目中文翻译
在本题中必须展示所有运算过程。
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
解法一
思路
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。
答题过程
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Since z=x+iy, its complex conjugate is
z∗=x−iy.
Therefore,
===(z−2i)(z∗−2i)[x+i(y−2)]×[x−i(y+2)]x2−ix(y+2)+ix(y−2)+(y−2)(y+2)x2+y2−4−4xi.
Comparing real and imaginary parts with 21−12i gives
x2+y2−4=21
and
−4x=−12.
Hence x=3, and so
32+y2=25.
Thus y2=16, giving y=±4. Therefore,
z=3+4iorz=3−4i.
解法二
思路
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。
答题过程
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Expanding in terms of z and z∗,
==(z−2i)(z∗−2i)zz∗−2i(z+z∗)−4x2+y2−4xi−4.
Hence, by comparing real and imaginary parts,
x2+y2−4=−4x=21,−12.
It follows that x=3 and
y2=25−9=16.
Therefore,
z=3±4i.
解法三
思路
展开
把乘积先写成含 zz∗ 和 z+z∗ 的形式。两者都是实数,比较实部、虚部后可求出这两个共轭数的积与和,再建立以它们为根的二次方程。
答题过程
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Expanding the given equation gives
=zz∗−2i(z+z∗)−421−12i.
Both zz∗ and z+z∗ are real. Comparing real and imaginary parts,
zz∗−4=−2(z+z∗)=21,−12.
Thus zz∗=25 and z+z∗=6. The numbers z and z∗ have sum 6 and product 25, so each satisfies
t2−6t+25=0.
In particular,
z2−6z+25=(z−3)2+16=z−3=0,0,±4i.
Hence,
z=3±4i,
giving x=3 and y=±4.