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FP2 2020 Oct Q4

A Level / Edexcel / FP2

IAL 2020 Oct Paper · Question 4

题目

Problem

4.

(a) Express the complex number 183−18i18\sqrt3-18\mathrm{i} in the form

 r(cos⁡θ+isin⁡θ) −π<θ⩽π\begin{align*} &\,r(\cos\theta+\mathrm{i}\sin\theta)\\[3mm] &\,-\pi<\theta\leqslant\pi \end{align*}
(3)

(b) Solve the equation

z4=183−18i\begin{align*} z^4=18\sqrt3-18\mathrm{i} \end{align*}

giving your answers in the form reiθre^{\mathrm{i}\theta} where −π<θ⩽π-\pi<\theta\leqslant \pi

(5)
题目中文翻译

(a) 将复数 183−18i18\sqrt3-18\mathrm{i} 写成

 r(cos⁡θ+isin⁡θ) −π<θ⩽π\begin{align*} &\,r(\cos\theta+\mathrm{i}\sin\theta)\\[3mm] &\,-\pi<\theta\leqslant\pi \end{align*}

的形式。

(3)

(b) 解方程

z4=183−18i\begin{align*} z^4=18\sqrt3-18\mathrm{i} \end{align*}

将答案写成 reiθre^{\mathrm{i}\theta} 的形式，其中 −π<θ⩽π-\pi<\theta\leqslant\pi。

(5)

解答

(a)

解法一

思路

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先求模长，再找辐角。实部为正、虚部为负，所以复数在第四象限，辐角应为负角。

答题过程

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The modulus is

r= (183)2+(−18)2= 972+324= 36.\begin{align*} r=&\,\sqrt{(18\sqrt3)^2+(-18)^2}\\[4mm] =&\,\sqrt{972+324}\\[4mm] =&\,36. \end{align*}

For the argument,

tan⁡θ= −18183= −13.\begin{align*} \tan\theta=&\,\frac{-18}{18\sqrt3}\\[4mm] =&\,-\frac1{\sqrt3}. \end{align*}

Since the complex number is in the fourth quadrant,

θ= −π6.\begin{align*} \theta=&\,-\frac{\pi}{6}. \end{align*}

Therefore

 183−18i= 36[cos⁡(−π6) +isin⁡(−π6)].\begin{align*} &\,18\sqrt3-18\mathrm{i}\\[4mm] =&\,36\Big[\cos\big(-\frac{\pi}{6}\big)\\[2mm] &\,\hspace{2pt}+\mathrm{i}\sin\big(-\frac{\pi}{6}\big)\Big]. \end{align*}

(b)

解法一

思路

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用 De Moivre 定理分别比较模长和辐角。四次根的模长是 6\sqrt6；辐角先加上整周角，再除以 44。最后把四个根逐一写成指数形式，并将辐角调整到题目指定范围。

答题过程

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Write z=ρ(cos⁡ϕ+isin⁡ϕ)z=\rho(\cos\phi+\mathrm{i}\sin\phi). From part (a), De Moivre’s theorem gives

ρ4= 36,4ϕ= −π6+2kπ.\begin{align*} \rho^4=&\,36,\\[2mm] 4\phi=&\,-\frac{\pi}{6}+2k\pi. \end{align*}

Hence

ρ= 361/4=6,ϕ= −π/6+2kπ4= −π24+kπ2,\begin{align*} \rho=&\,36^{1/4}=\sqrt6,\\[4mm] \phi=&\,\frac{-\pi/6+2k\pi}{4}\\[4mm] =&\,-\frac{\pi}{24}+\frac{k\pi}{2}, \end{align*}

where k=0,1,2,3k=0,1,2,3. The four arguments are

ϕ0= −π24,ϕ1= 11π24,ϕ2= 23π24,ϕ3= 35π24.\begin{align*} \phi_0=&\,-\frac{\pi}{24},\\[2mm] \phi_1=&\,\frac{11\pi}{24},\\[2mm] \phi_2=&\,\frac{23\pi}{24},\\[2mm] \phi_3=&\,\frac{35\pi}{24}. \end{align*}

The last angle is greater than π\pi, so subtract 2π2\pi:

35π24−2π= −13π24.\begin{align*} \frac{35\pi}{24}-2\pi=&\,-\frac{13\pi}{24}. \end{align*}

Therefore the four roots, with arguments in (−π,π](-\pi,\pi], are

z0= 6e−π24i,z1= 6e11π24i,z2= 6e23π24i,z3= 6e−13π24i.\begin{align*} z_0=&\,\sqrt6e^{-\frac{\pi}{24}\mathrm{i}},\\[4mm] z_1=&\,\sqrt6e^{\frac{11\pi}{24}\mathrm{i}},\\[4mm] z_2=&\,\sqrt6e^{\frac{23\pi}{24}\mathrm{i}},\\[4mm] z_3=&\,\sqrt6e^{-\frac{13\pi}{24}\mathrm{i}}. \end{align*}

解法二

思路

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先取辐角为 −π24-\frac{\pi}{24} 的一个根，再连续逆时针旋转 π2\frac{\pi}{2} 得到另外三个根。因为乘以 i\mathrm{i} 后的四次方不变，这四个不同的数都是所求的根。

答题过程

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One fourth root of 36e−π6i36e^{-\frac{\pi}{6}\mathrm{i}} is

z0= 6e−π24i,\begin{align*} z_0=&\,\sqrt6e^{-\frac{\pi}{24}\mathrm{i}}, \end{align*}

since its modulus to the fourth power is 3636 and its argument multiplied by 44 is −π/6-\pi/6.

For any root zz,

(iz)4= i4z4=z4.\begin{align*} (\mathrm{i}z)^4=&\,\mathrm{i}^4z^4=z^4. \end{align*}

Thus rotation through π/2\pi/2 produces another root. Repeating this rotation gives

z1= iz0= 6e11π24i,z2= iz1= 6e23π24i,z3= iz2= 6e35π24i= 6e−13π24i.\begin{align*} z_1=&\,\mathrm{i}z_0\\[2mm] =&\,\sqrt6e^{\frac{11\pi}{24}\mathrm{i}},\\[4mm] z_2=&\,\mathrm{i}z_1\\[2mm] =&\,\sqrt6e^{\frac{23\pi}{24}\mathrm{i}},\\[4mm] z_3=&\,\mathrm{i}z_2\\[2mm] =&\,\sqrt6e^{\frac{35\pi}{24}\mathrm{i}}\\[2mm] =&\,\sqrt6e^{-\frac{13\pi}{24}\mathrm{i}}. \end{align*}

In the final line, 35π/2435\pi/24 has been reduced by 2π2\pi. The four distinct roots, all with arguments in (−π,π](-\pi,\pi], are

z0= 6e−π24i,z1= 6e11π24i,z2= 6e23π24i,z3= 6e−13π24i.\begin{align*} z_0=&\,\sqrt6e^{-\frac{\pi}{24}\mathrm{i}},\\[4mm] z_1=&\,\sqrt6e^{\frac{11\pi}{24}\mathrm{i}},\\[4mm] z_2=&\,\sqrt6e^{\frac{23\pi}{24}\mathrm{i}},\\[4mm] z_3=&\,\sqrt6e^{-\frac{13\pi}{24}\mathrm{i}}. \end{align*}