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CIE 9709 2025 November Paper 33 Q7

A Level / CIE / P3

CIE 9709 2025 Nov Paper 33 Paper · Question 7

题目

Problem

Solve the equation 5z2−i−zz∗+20+8i=0\frac{5z}{2-\mathrm{i}}-zz^*+20+8\mathrm{i}=0. Give your answers in the form x+iyx+\mathrm{i}y, where xx and yy are real.

(6)
题目中文翻译

解方程 5z2−i−zz∗+20+8i=0\frac{5z}{2-\mathrm{i}}-zz^*+20+8\mathrm{i}=0。答案写成 x+iyx+\mathrm{i}y 的形式，其中 xx 和 yy 为实数。

(6)
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

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先把复分母实数化，并利用 zz∗=x2+y2zz^*=x^2+y^2 为实数。虚部给线性关系，代入实部得到二次方程。

答题过程

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Since

52−i=5(2+i)5=2+i,\frac5{2-\mathrm{i}}=\frac{5(2+\mathrm{i})}{5}=2+\mathrm{i},

putting z=x+yiz=x+y\mathrm{i} gives

(2+i)(x+yi)−(x2+y2)+20+8i=0.(2+\mathrm{i})(x+y\mathrm{i})-(x^2+y^2)+20+8\mathrm{i}=0.

Equating real and imaginary parts,

2x−y−x2−y2+20= 0,x+2y+8= 0.\begin{align*} 2x-y-x^2-y^2+20=&\,0,\\[4mm] x+2y+8=&\,0. \end{align*}

Thus x=−2y−8x=-2y-8. Substituting in the real equation and simplifying,

 2(−2y−8)−y−(−2y−8)2−y2+20= −5y2−37y−60=0.\begin{align*} &\,2(-2y-8)-y-(-2y-8)^2-y^2+20\\[4mm] =&\,-5y^2-37y-60=0. \end{align*}

Multiplying by −1-1 and factorising,

(y+5)(5y+12)=0.(y+5)(5y+12)=0.

If y=−5y=-5, then x=2x=2. If y=−12/5y=-12/5, then x=−16/5x=-16/5.

Therefore

z= 2−5i,orz= −165−125i.\begin{align*} z=&\,2-5\mathrm{i},\\[4mm] \text{or}\quad z=&\,-\frac{16}{5}-\frac{12}{5}\mathrm{i}. \end{align*}