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FP1 2026 Jan Q7

A Level / Edexcel / FP1

IAL 2026 Jan Paper · Question 7

题目

Problem

z=3+k3 i2+3 iz = \dfrac{3 + k\sqrt{3}\,\mathrm{i}}{2 + \sqrt{3}\,\mathrm{i}}

where k∈Zk \in \mathbb{Z}.

(a) Express zz in the form a+bia+b\mathrm{i} where aa and bb are real functions of kk.

(3)

Given that arg⁡z=π6\arg z = \dfrac{\pi}{6}

(b) determine the exact value of ∣z∣|z|.

(5)
题目中文翻译 z=3+k3 i2+3 iz = \dfrac{3 + k\sqrt{3}\,\mathrm{i}}{2 + \sqrt{3}\,\mathrm{i}}

其中 k∈Zk \in \mathbb{Z}。

(a) 将 zz 表示为 a+bia+b\mathrm{i} 的形式，其中 aa 和 bb 是 kk 的实函数。

(3)

已知 arg⁡z=π6\arg z = \dfrac{\pi}{6}

(b) 确定 ∣z∣|z| 的精确值。

(5)

解答

(a)

解法一

思路

展开

分子、分母同乘分母的共轭复数 2−3 i2-\sqrt3\,\mathrm{i}，使分母变成实数 7。展开分子后分别合并实部与虚部。

答题过程

展开

Multiplying the numerator and denominator by the complex conjugate of the denominator,

 z= 3+k3 i2+3 i ×2−3 i2−3 i.\begin{align*} &\,z\\ =&\,\frac{3+k\sqrt3\,\mathrm{i}} {2+\sqrt3\,\mathrm{i}}\\ &\,\hspace{2pt}\times \frac{2-\sqrt3\,\mathrm{i}} {2-\sqrt3\,\mathrm{i}}. \end{align*}

The denominator is 4+3=74+3=7. Expanding the numerator gives

 (3+k3 i) ×(2−3 i)= 6−33 i +2k3 i+3k= 6+3k +3(2k−3) i.\begin{align*} &\,(3+k\sqrt3\,\mathrm{i})\\ &\,\hspace{2pt}\times(2-\sqrt3\,\mathrm{i})\\ =&\,6-3\sqrt3\,\mathrm{i}\\ &\,\hspace{2pt}+2k\sqrt3\,\mathrm{i}+3k\\ =&\,6+3k\\ &\,\hspace{2pt}+\sqrt3(2k-3)\,\mathrm{i}. \end{align*}

Therefore,

 z= 6+3k7 +3(2k−3)7 i.\begin{align*} &\,z\\ =&\,\frac{6+3k}{7}\\ &\,\hspace{2pt}+\frac{\sqrt3(2k-3)}{7}\,\mathrm{i}. \end{align*}

Thus

a= 6+3k7,b= 3(2k−3)7.\begin{align*} a=&\,\frac{6+3k}{7},\\ b=&\,\frac{\sqrt3(2k-3)}{7}. \end{align*}

(b)

解法一

思路

展开

由 arg⁡z=π6\arg z=\frac\pi6 可知 ba=tan⁡π6=13\frac ba=\tan\frac\pi6=\frac1{\sqrt3}。代入 (a) 的实部、虚部求出 kk，再写出 zz 并计算模长。

答题过程

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Since arg⁡z=π6\arg z=\dfrac\pi6,

Im⁡(z)Re⁡(z)=tan⁡π6=13.\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)} =\tan\frac\pi6 =\frac1{\sqrt3}.

Using part (a),

3(2k−3)6+3k=13.\frac{\sqrt3(2k-3)}{6+3k} =\frac1{\sqrt3}.

Therefore,

3(2k−3)= 6+3k6k−9= 6+3kk= 5.\begin{align*} 3(2k-3)=&\,6+3k\\ 6k-9=&\,6+3k\\ k=&\,5. \end{align*}

Substituting k=5k=5 into the result from part (a),

z=3+3 i.z=3+\sqrt3\,\mathrm{i}.

Hence,

∣z∣= 32+(3)2= 12= 23.\begin{align*} |z| =&\,\sqrt{3^2+(\sqrt3)^2}\\ =&\,\sqrt{12}\\ =&\,\boxed{2\sqrt3}. \end{align*}

解法二

思路

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也可以直接把 zz 写成模辐角形式。设 ∣z∣=R|z|=R，利用已知辐角写出实部、虚部，再与 (a) 的结果分别比较并消去 kk。

答题过程

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Let ∣z∣=R|z|=R. Since arg⁡z=π6\arg z=\dfrac\pi6,

 z= R(cos⁡π6+isin⁡π6)= 3R2+R2 i.\begin{align*} &\,z\\ =&\,R\bigg(\cos\frac\pi6 +\mathrm{i}\sin\frac\pi6\bigg)\\ =&\,\frac{\sqrt3R}{2}+\frac R2\,\mathrm{i}. \end{align*}

Comparing this with the result from part (a) gives

6+3k7=3R2\frac{6+3k}{7}=\frac{\sqrt3R}{2}

and

3(2k−3)7=R2.\frac{\sqrt3(2k-3)}{7}=\frac R2.

The first equation gives

k=73R6−2.k=\frac{7\sqrt3R}{6}-2.

Substituting this into the second equation,

 3[2(73R6−2) −3]= 7R2\begin{align*} &\,\sqrt3\Big[ 2\Big(\frac{7\sqrt3R}{6}-2\Big)\\ &\,\hspace{2pt}-3\Big]\\ =&\,\frac{7R}{2}\\ \end{align*}

The left-hand side simplifies as follows:

 3(73R3−4−3)= 3(73R3−7)= 7R−73.\begin{align*} &\,\sqrt3\Big(\frac{7\sqrt3R}{3}-4-3\Big)\\ =&\,\sqrt3\Big(\frac{7\sqrt3R}{3}-7\Big)\\ =&\,7R-7\sqrt3. \end{align*}

Therefore,

7R−73= 7R2,7R2= 73,R= 23.\begin{align*} 7R-7\sqrt3=&\,\frac{7R}{2},\\ \frac{7R}{2}=&\,7\sqrt3,\\ R=&\,2\sqrt3. \end{align*}

Hence,

∣z∣=R=23.\boxed{|z|=R=2\sqrt3}.