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CIE 9709 2023 March Paper 32 Q4

A Level / CIE / P3

CIE 9709 2023 March Paper 32 Paper · Question 4

题目

Problem

Solve the equation

5z1+2i−zz∗+30+10i=0,\frac{5z}{1+2\mathrm{i}}-zz^*+30+10\mathrm{i}=0,

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

(5)
题目中文翻译

解方程

5z1+2i−zz∗+30+10i=0,\frac{5z}{1+2\mathrm{i}}-zz^*+30+10\mathrm{i}=0,

将答案表示为 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

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先用共轭消去常数分母,再设 z=x+yiz=x+y\mathrm{i},利用 zz∗=x2+y2zz^*=x^2+y^2 比较实部、虚部。

答题过程

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Since

51+2i=5(1−2i)1+4=1−2i,\frac5{1+2\mathrm{i}} =\frac{5(1-2\mathrm{i})}{1+4} =1-2\mathrm{i},

write z=x+yiz=x+y\mathrm{i} and zz∗=x2+y2zz^*=x^2+y^2. The equation becomes

(x+yi)(1−2i)−(x2+y2)+30+10i=0.(x+y\mathrm{i})(1-2\mathrm{i}) -(x^2+y^2)+30+10\mathrm{i}=0.

Equating parts,

x+2y−x2−y2+30= 0,−2x+y+10= 0.\begin{align*} x+2y-x^2-y^2+30=&\,0,\\[4mm] -2x+y+10=&\,0. \end{align*}

The second gives y=2x−10y=2x-10. Substituting into the first,

x+2(2x−10)−x2−(2x−10)2+30=0.x+2(2x-10)-x^2-(2x-10)^2+30=0.

Hence

−5x2+45x−90= 0,(x−3)(x−6)= 0.\begin{align*} -5x^2+45x-90=&\,0,\\[4mm] (x-3)(x-6)=&\,0. \end{align*}

If x=3x=3, y=−4y=-4; if x=6x=6, y=2y=2. Thus

z=3−4iorz=6+2i.\boxed{z=3-4\mathrm{i}\quad\text{or}\quad z=6+2\mathrm{i}}.