Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2023 Nov Paper 31 Q4

A Level / CIE / P3

CIE 9709 2023 Nov Paper 31 Paper · Question 4

题目

Problem

The complex number uu is defined by u=3+2ia−5iu=\dfrac{3+2\mathrm{i}}{a-5\mathrm{i}}, where aa is real.

(a) Express uu in the Cartesian form x+iyx+\mathrm{i}y, where xx and yy are in terms of aa.

(3)

(b) Given that arg⁡u=14π\arg u=\frac14\pi, find the value of aa.

(2)
题目中文翻译

复数 uu 定义为 u=3+2ia−5iu=\dfrac{3+2\mathrm{i}}{a-5\mathrm{i}},其中 aa 为实数。

(a) 将 uu 表示为直角坐标形式 x+iyx+\mathrm{i}y,其中 xx 和 yy 均用 aa 表示。

(3)

(b) 已知 arg⁡u=14π\arg u=\frac14\pi,求 aa 的值。

(2)
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

(a)

解法一

思路

展开

分母含复数,用共轭 a+5ia+5\mathrm{i} 化为实分母,再分开实部和虚部。

答题过程

展开

Multiplying numerator and denominator by a+5ia+5\mathrm{i},

u= (3+2i)(a+5i)(a−5i)(a+5i)= 3a+15i+2ai+10i2a2+25= 3a−10a2+25+i2a+15a2+25.\begin{align*} u=&\,\frac{(3+2\mathrm{i})(a+5\mathrm{i})} {(a-5\mathrm{i})(a+5\mathrm{i})}\\[4mm] =&\,\frac{3a+15\mathrm{i}+2a\mathrm{i}+10\mathrm{i}^2}{a^2+25}\\[4mm] =&\,\boxed{\frac{3a-10}{a^2+25} +\mathrm{i}\frac{2a+15}{a^2+25}}. \end{align*}

解法二

思路

展开

也可以设商为 x+iyx+\mathrm{i}y,清分母后比较实部和虚部,解二元一次方程。

答题过程

展开

Let u=x+iyu=x+\mathrm{i}y. Then

3+2i= (a−5i)(x+iy)= (ax+5y)+i(ay−5x).\begin{align*} 3+2\mathrm{i}=&\,(a-5\mathrm{i})(x+\mathrm{i}y)\\[4mm] =&\,(ax+5y)+\mathrm{i}(ay-5x). \end{align*}

Thus

ax+5y= 3,−5x+ay= 2.\begin{align*} ax+5y=&\,3,\\[4mm] -5x+ay=&\,2. \end{align*}

Multiplying the first by aa and subtracting five times the second,

(a2+25)x=3a−10.(a^2+25)x=3a-10.

Multiplying the second by aa and adding five times the first,

(a2+25)y=2a+15.(a^2+25)y=2a+15.

Since a2+25>0a^2+25>0,

u=3a−10a2+25+i2a+15a2+25.\boxed{u=\frac{3a-10}{a^2+25} +\mathrm{i}\frac{2a+15}{a^2+25}}.

(b)

解法一

思路

展开

辐角为 π/4\pi/4 不仅要求实部等于虚部,还要求两者都为正。先解等式,再检查象限。

答题过程

展开

For arg⁡u=π/4\arg u=\pi/4, the real and imaginary parts must be equal and positive. From part (a),

3a−10a2+25=2a+15a2+25.\frac{3a-10}{a^2+25}=\frac{2a+15}{a^2+25}.

Thus

3a−10=2a+15,a=25.3a-10=2a+15,\qquad a=25.

At a=25a=25,

Re⁡u=Im⁡u=65650=110>0,\operatorname{Re}u=\operatorname{Im}u=\frac{65}{650}=\frac1{10}>0,

so the argument is indeed π/4\pi/4. Therefore a=25\boxed{a=25}.