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FP2 2023 June Q2

A Level / Edexcel / FP2

IAL 2023 June Paper · Question 2

题目

Problem

2.

The complex number z1z_1 is defined as

z1= (cos⁡5π12+isin⁡5π12)4(cos⁡π3−isin⁡π3)3\begin{align*} z_1 =&\, \frac{\bigg(\cos \frac{5\pi}{12} + \mathrm{i}\sin \frac{5\pi}{12}\bigg)^4} {\bigg(\cos \frac{\pi}{3} - \mathrm{i}\sin \frac{\pi}{3}\bigg)^3} \end{align*}

(a) Without using your calculator show that

z1= cos⁡2π3+isin⁡2π3\begin{align*} z_1 =&\, \cos \frac{2\pi}{3} + \mathrm{i}\sin \frac{2\pi}{3} \end{align*}
(4)

(b) Shade, on a single Argand diagram, the region RR defined by

∣z−z1∣⩽1and0⩽arg⁡(z−z1)⩽3π4\begin{align*} |z - z_1| \leqslant 1 \qquad \text{and} \qquad 0 \leqslant \arg(z - z_1) \leqslant \frac{3\pi}{4} \end{align*}
(4)

Given that the complex number zz lies in RR

(c) determine the smallest possible positive value of arg⁡z\arg z

(2)
题目中文翻译

复数 z1z_1 定义为

z1= (cos⁡5π12+isin⁡5π12)4(cos⁡π3−isin⁡π3)3. \begin{align*} z_1=&\,\frac{\big(\cos\frac{5\pi}{12}+\mathrm{i}\sin\frac{5\pi}{12}\big)^4} {\big(\cos\frac{\pi}{3}-\mathrm{i}\sin\frac{\pi}{3}\big)^3}. \end{align*}

(a) 不使用计算器，证明

z1= cos⁡2π3+isin⁡2π3. \begin{align*} z_1=&\,\cos\frac{2\pi}{3}+\mathrm{i}\sin\frac{2\pi}{3}. \end{align*}
(4)

(b) 在同一张 Argand 图上，用阴影表示满足以下条件的区域 RR：

∣z−z1∣⩽1,0⩽arg⁡(z−z1)⩽3π4. \begin{align*} |z-z_1|\leqslant1,\qquad 0\leqslant\arg(z-z_1)\leqslant\frac{3\pi}{4}. \end{align*}
(4)

已知复数 zz 位于 RR 内。

(c) 求 arg⁡z\arg z 可能的最小正值。

(2)

解答

(a)

解法一

思路

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复数已经写成 cos⁡θ+isin⁡θ\cos\theta+\mathrm{i}\sin\theta 的形式，所以直接用 De Moivre’s theorem。注意分母中间是减号，要先写成角度为 −π3-\frac{\pi}{3} 的形式。

答题过程

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For the numerator,

(cos⁡5π12+isin⁡5π12)4= cos⁡(4⋅5π12)+isin⁡(4⋅5π12)= cos⁡5π3+isin⁡5π3\begin{align*} \bigg(\cos \frac{5\pi}{12} +\mathrm{i}\sin \frac{5\pi}{12}\bigg)^4 =&\, \cos\bigg(4\cdot\frac{5\pi}{12}\bigg) +\mathrm{i}\sin\bigg(4\cdot\frac{5\pi}{12}\bigg)\\[4mm] =&\, \cos\frac{5\pi}{3}+\mathrm{i}\sin\frac{5\pi}{3} \end{align*}

For the denominator,

cos⁡π3−isin⁡π3= cos⁡(−π3)+isin⁡(−π3)\begin{align*} \cos\frac{\pi}{3}-\mathrm{i}\sin\frac{\pi}{3} =&\, \cos\bigg(-\frac{\pi}{3}\bigg) +\mathrm{i}\sin\bigg(-\frac{\pi}{3}\bigg) \end{align*}

Hence

(cos⁡π3−isin⁡π3)3= cos⁡(−π)+isin⁡(−π)\begin{align*} \bigg(\cos\frac{\pi}{3}-\mathrm{i}\sin\frac{\pi}{3}\bigg)^3 =&\, \cos(-\pi)+\mathrm{i}\sin(-\pi) \end{align*}

When dividing complex numbers in modulus-argument form, subtract the arguments:

z1= cos⁡(5π3−(−π))+isin⁡(5π3−(−π))= cos⁡8π3+isin⁡8π3= cos⁡(8π3−2π)+isin⁡(8π3−2π)= cos⁡2π3+isin⁡2π3\begin{align*} z_1 =&\, \cos\bigg(\frac{5\pi}{3}-(-\pi)\bigg) +\mathrm{i}\sin\bigg(\frac{5\pi}{3}-(-\pi)\bigg)\\[4mm] =&\, \cos\frac{8\pi}{3}+\mathrm{i}\sin\frac{8\pi}{3}\\[4mm] =&\, \cos\bigg(\frac{8\pi}{3}-2\pi\bigg) +\mathrm{i}\sin\bigg(\frac{8\pi}{3}-2\pi\bigg)\\[4mm] =&\, \cos\frac{2\pi}{3}+\mathrm{i}\sin\frac{2\pi}{3} \end{align*}

解法二

思路

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用 de Moivre 定理计算分子、分母的幂后，注意分母恰好是 −1-1。再用精确三角值求商，最后写回题目要求的三角形式。

答题过程

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By de Moivre’s theorem, the denominator is

(cos⁡π3−isin⁡π3)3= cos⁡(−π)+isin⁡(−π)= −1.\begin{align*} \bigg(\cos\frac{\pi}{3}-\mathrm{i}\sin\frac{\pi}{3}\bigg)^3 =&\,\cos(-\pi)+\mathrm{i}\sin(-\pi)\\[2mm] =&\,-1. \end{align*}

The numerator is

(cos⁡5π12+isin⁡5π12)4= cos⁡5π3+isin⁡5π3= 12−32i.\begin{align*} \bigg(\cos\frac{5\pi}{12}+\mathrm{i}\sin\frac{5\pi}{12}\bigg)^4 =&\,\cos\frac{5\pi}{3}+\mathrm{i}\sin\frac{5\pi}{3}\\[2mm] =&\,\frac12-\frac{\sqrt3}{2}\mathrm{i}. \end{align*}

Hence

z1= 12−32i−1= −12+32i= cos⁡2π3+isin⁡2π3.\begin{align*} z_1=&\,\frac{\frac12-\frac{\sqrt3}{2}\mathrm{i}}{-1}\\[2mm] =&\,-\frac12+\frac{\sqrt3}{2}\mathrm{i}\\[2mm] =&\,\cos\frac{2\pi}{3}+\mathrm{i}\sin\frac{2\pi}{3}. \end{align*}

(b)

解法一

思路

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∣z−z1∣⩽1|z-z_1|\leqslant 1 表示以 z1z_1 为圆心、半径为 11 的圆内区域。arg⁡(z−z1)\arg(z-z_1) 是从圆心 z1z_1 出发看点 zz 的方向角，所以区域是圆内从水平向右射线逆时针扫到 3π4\frac{3\pi}{4} 的扇形。

答题过程

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First write the centre in Cartesian form:

z1= cos⁡2π3+isin⁡2π3= −12+32i\begin{align*} z_1 =&\, \cos\frac{2\pi}{3}+\mathrm{i}\sin\frac{2\pi}{3}\\[4mm] =&\, -\frac12+\frac{\sqrt3}{2}\mathrm{i} \end{align*}

On a single Argand diagram:

  • plot the centre z1=(−12,32)z_1=\bigg(-\frac12,\frac{\sqrt3}{2}\bigg);
  • draw the circle with centre z1z_1 and radius 11;
  • draw the ray from z1z_1 parallel to the positive real axis;
  • draw the ray from z1z_1 making angle 3π4\frac{3\pi}{4} anticlockwise from that ray;
  • shade the part inside the circle between these two rays.

(c)

解法一

思路

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要让 arg⁡z\arg z 最小，点 zz 要尽量靠近正实轴方向。扇形里最靠右的边界点是从圆心 z1z_1 沿水平向右走一个半径到圆周上的点。

答题过程

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The smallest positive argument occurs at the right-hand end of the horizontal radius of the circle:

z= z1+1= (−12+32i)+1= 12+32i\begin{align*} z =&\, z_1+1\\[4mm] =&\, \bigg(-\frac12+\frac{\sqrt3}{2}\mathrm{i}\bigg)+1\\[4mm] =&\, \frac12+\frac{\sqrt3}{2}\mathrm{i} \end{align*}

Therefore

arg⁡z= arctan⁡(3212)= arctan⁡(3)= π3\begin{align*} \arg z =&\, \arctan\bigg(\frac{\frac{\sqrt3}{2}}{\frac12}\bigg)\\[4mm] =&\, \arctan(\sqrt3)\\[4mm] =&\, \frac{\pi}{3} \end{align*}

So the smallest possible positive value of arg⁡z\arg z is

π3\begin{align*} \boxed{\frac{\pi}{3}} \end{align*}