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CIE 9709 2025 November Paper 35 Q5

A Level / CIE / P3

CIE 9709 2025 Nov Paper 35 Paper · Question 5

题目

Problem

Find the complex numbers, zz, which satisfy the equation

zz∗+5iz+2−10i=0.zz^*+5\mathrm{i}z+2-10\mathrm{i}=0.

Give your answers in the form x+iyx+\mathrm{i}y, where xx and yy are real.

(5)
题目中文翻译

求满足下列方程的复数 zz:

zz∗+5iz+2−10i=0.zz^*+5\mathrm{i}z+2-10\mathrm{i}=0.

答案写成 x+iyx+\mathrm{i}y 的形式,其中 xx 和 yy 为实数。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

展开

利用共轭乘积为实数。比较实部与虚部后,先解线性方程,再解二次方程。

答题过程

展开

Put z=x+yiz=x+y\mathrm{i}, so z∗=x−yiz^*=x-y\mathrm{i} and zz∗=x2+y2zz^*=x^2+y^2. The equation becomes

x2+y2+5xi−5y+2−10i=0.x^2+y^2+5x\mathrm{i}-5y+2-10\mathrm{i}=0.

Equating parts,

x2+y2−5y+2= 0,5x−10= 0.\begin{align*} x^2+y^2-5y+2=&\,0,\\[4mm] 5x-10=&\,0. \end{align*}

Thus x=2x=2. Substituting,

y2−5y+6=0⟹(y−2)(y−3)=0.y^2-5y+6=0 \quad\Longrightarrow\quad (y-2)(y-3)=0.

Therefore

z=2+2iorz=2+3i.\boxed{z=2+2\mathrm{i}\quad\text{or}\quad z=2+3\mathrm{i}}.