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IAL 2023 June Q2

A Level / Edexcel / FP2

IAL 2023 June Paper · Question 2

题目

Problem

The complex number z1z_1 is defined as

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

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*} \left(\cos \frac{5\pi}{12} +\mathrm{i}\sin \frac{5\pi}{12}\right)^4 ={}& \cos\left(4\cdot\frac{5\pi}{12}\right) +\mathrm{i}\sin\left(4\cdot\frac{5\pi}{12}\right)\\[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\left(-\frac{\pi}{3}\right) +\mathrm{i}\sin\left(-\frac{\pi}{3}\right) \end{align*}

Hence

(cos⁡π3−isin⁡π3)3=cos⁡(−π)+isin⁡(−π)\begin{align*} \left(\cos\frac{\pi}{3}-\mathrm{i}\sin\frac{\pi}{3}\right)^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\left(\frac{5\pi}{3}-(-\pi)\right) +\mathrm{i}\sin\left(\frac{5\pi}{3}-(-\pi)\right)\\[4mm] ={}& \cos\frac{8\pi}{3}+\mathrm{i}\sin\frac{8\pi}{3}\\[4mm] ={}& \cos\left(\frac{8\pi}{3}-2\pi\right) +\mathrm{i}\sin\left(\frac{8\pi}{3}-2\pi\right)\\[4mm] ={}& \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=\left(-\frac12,\frac{\sqrt3}{2}\right);
  • 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] ={}& \left(-\frac12+\frac{\sqrt3}{2}\mathrm{i}\right)+1\\[4mm] ={}& \frac12+\frac{\sqrt3}{2}\mathrm{i} \end{align*}

Therefore

arg⁡z=arctan⁡(3212)=arctan⁡(3)=π3\begin{align*} \arg z ={}& \arctan\left(\frac{\frac{\sqrt3}{2}}{\frac12}\right)\\[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*}