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FP1 2026 Jan A Q4

A Level / Edexcel / FP1

IAL 2026 Jan A Paper · Question 4

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+iyz = x + \mathrm{i}y, where xx and yy are real numbers, solve the equation

 (z−2i)(z∗−2i)= 21−12i\begin{align*} &\,(z-2\mathrm{i})(z^*-2\mathrm{i})\\ =&\,21-12\mathrm{i} \end{align*}

where z∗z^* is the complex conjugate of zz.

(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+iyz = x + \mathrm{i}y,其中 xx 和 yy 是实数,解方程

 (z−2i)(z∗−2i)= 21−12i\begin{align*} &\,(z-2\mathrm{i})(z^*-2\mathrm{i})\\ =&\,21-12\mathrm{i} \end{align*}

其中 z∗z^* 是 zz 的共轭复数。

(6)

解答

解法一

思路

展开

先写出 z∗=x−iyz^*=x-\mathrm{i}y,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 xx、yy 的方程。

答题过程

展开

Since z=x+iyz=x+\mathrm{i}y, its complex conjugate is

z∗=x−iy.z^*=x-\mathrm{i}y.

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.\begin{align*} &\,(z-2\mathrm{i})(z^*-2\mathrm{i})\\[4mm] =&\,\big[x+\mathrm{i}(y-2)\big]\\[4mm] &\,\hspace{2pt}\times\big[x-\mathrm{i}(y+2)\big]\\[4mm] =&\,x^2-\mathrm{i}x(y+2)\\[4mm] &\,\hspace{2pt}+\mathrm{i}x(y-2)\\[4mm] &\,\hspace{4pt}+(y-2)(y+2)\\[4mm] =&\,x^2+y^2-4-4x\mathrm{i}. \end{align*}

Comparing real and imaginary parts with 21−12i21-12\mathrm{i} gives

x2+y2−4=21x^2+y^2-4=21

and

−4x=−12.-4x=-12.

Hence x=3x=3, and so

32+y2=25.3^2+y^2=25.

Thus y2=16y^2=16, giving y=±4y=\pm4. Therefore,

 z=3+4i orz=3−4i.\begin{align*} &\,\boxed{z=3+4\mathrm{i}}\\[4mm] &\,\text{or}\quad\boxed{z=3-4\mathrm{i}}. \end{align*}

解法二

思路

展开

先利用 zz∗=∣z∣2zz^*=|z|^2 和 z+z∗=2xz+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

展开

Expanding in terms of zz and z∗z^*,

 (z−2i)(z∗−2i)= zz∗−2i(z+z∗)−4= x2+y2−4xi−4.\begin{align*} &\,(z-2\mathrm{i})(z^*-2\mathrm{i})\\[4mm] =&\,zz^*-2\mathrm{i}(z+z^*)-4\\[4mm] =&\,x^2+y^2-4x\mathrm{i}-4. \end{align*}

Hence, by comparing real and imaginary parts,

x2+y2−4= 21,−4x= −12.\begin{align*} x^2+y^2-4=&\,21,\\[4mm] -4x=&\,-12. \end{align*}

It follows that x=3x=3 and

y2=25−9=16.y^2=25-9=16.

Therefore,

z=3±4i.\boxed{z=3\pm4\mathrm{i}}.

解法三

思路

展开

把乘积先写成含 zz∗zz^* 和 z+z∗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

展开

Expanding the given equation gives

 zz∗−2i(z+z∗)−4= 21−12i.\begin{align*} &\,zz^*-2\mathrm{i}(z+z^*)-4\\[4mm] =&\,21-12\mathrm{i}. \end{align*}

Both zz∗zz^* and z+z∗z+z^* are real. Comparing real and imaginary parts,

zz∗−4= 21,−2(z+z∗)= −12.\begin{align*} zz^*-4=&\,21,\\[4mm] -2(z+z^*)=&\,-12. \end{align*}

Thus zz∗=25zz^*=25 and z+z∗=6z+z^*=6. The numbers zz and z∗z^* have sum 6 and product 25, so each satisfies

t2−6t+25=0.t^2-6t+25=0.

In particular,

z2−6z+25= 0,(z−3)2+16= 0,z−3= ±4i.\begin{align*} z^2-6z+25=&\,0,\\[4mm] (z-3)^2+16=&\,0,\\[4mm] z-3=&\,\pm4\mathrm{i}. \end{align*}

Hence,

z=3±4i,\boxed{z=3\pm4\mathrm{i}},

giving x=3\boxed{x=3} and y=±4\boxed{y=\pm4}.