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FP1 2026 Jan A Q2

A Level / Edexcel / FP1

IAL 2026 Jan A Paper · Question 2

题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

 f(z)= z4−6z3+38z2 −94z+221\begin{align*} &\,f(z)\\ =&\,z^4-6z^3+38z^2\\ &\,\hspace{2pt}-94z+221 \end{align*}

(a) Given that z=2+3iz = 2 + 3\mathrm{i} is a root of the equation f(z)=0f(z) = 0, use algebra to find the three other roots of f(z)=0f(z) = 0.

(7)

(b) Show the four roots of f(z)=0f(z) = 0 on a single Argand diagram.

(2)
题目中文翻译

在本题中必须展示所有运算过程。

依赖计算器技术的解答不可接受。

 f(z)= z4−6z3+38z2 −94z+221\begin{align*} &\,f(z)\\ =&\,z^4-6z^3+38z^2\\ &\,\hspace{2pt}-94z+221 \end{align*}

(a) 已知 z=2+3iz = 2 + 3\mathrm{i} 是方程 f(z)=0f(z) = 0 的一个根，用代数方法求 f(z)=0f(z) = 0 的另外三个根。

(7)

(b) 在同一幅 Argand 图上画出 f(z)=0f(z) = 0 的四个根。

(2)

解答

(a)

解法一

思路

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多项式系数全为实数，所以非实根成共轭对出现。先由已知根得到另一个根 2−3i2-3\mathrm{i}，将对应的两个一次因式相乘，再用所得二次因式除原多项式。最后用配方法求第二个二次因式的根。

答题过程

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Since f(z)f(z) has real coefficients and 2+3i2+3\mathrm{i} is a root, 2−3i2-3\mathrm{i} is also a root.

The corresponding quadratic factor is

 [z−(2+3i)] ×[z−(2−3i)]= [(z−2)−3i] ×[(z−2)+3i]= (z−2)2+9= z2−4z+13.\begin{align*} &\,\big[z-(2+3\mathrm{i})\big]\\[4mm] &\,\hspace{2pt}\times\big[z-(2-3\mathrm{i})\big]\\[4mm] =&\,\big[(z-2)-3\mathrm{i}\big]\\[4mm] &\,\hspace{2pt}\times\big[(z-2)+3\mathrm{i}\big]\\[4mm] =&\,(z-2)^2+9\\ =&\,z^2-4z+13. \end{align*}

Write D(z)=z2−4z+13D(z)=z^2-4z+13. To divide f(z)f(z) by D(z)D(z), first subtract z2D(z)z^2D(z):

R1(z)= f(z)−z2D(z)= −2z3+25z2 −94z+221.\begin{align*} R_1(z)=&\,f(z)-z^2D(z)\\[4mm] =&\,-2z^3+25z^2\\[4mm] &\,\hspace{2pt}-94z+221. \end{align*}

Next subtract (−2z)D(z)(-2z)D(z):

 R2(z)= R1(z)−(−2z)D(z)= 17z2−68z+221= 17D(z).\begin{align*} &\,R_2(z)\\[4mm] =&\,R_1(z)-(-2z)D(z)\\[4mm] =&\,17z^2-68z+221\\[4mm] =&\,17D(z). \end{align*}

Subtracting 17D(z)17D(z) leaves zero remainder. The quotient is therefore z2−2z+17z^2-2z+17, giving

 f(z)= (z2−4z+13) ×(z2−2z+17).\begin{align*} &\,f(z)\\[4mm] =&\,(z^2-4z+13)\\[4mm] &\,\hspace{2pt}\times(z^2-2z+17). \end{align*}

For completeness, expanding the right-hand side gives

 (z2−4z+13) ×(z2−2z+17)= z4−2z3+17z2 −4z3+8z2−68z +13z2−26z+221= z4−6z3+38z2 −94z+221= f(z),\begin{align*} &\,(z^2-4z+13)\\[4mm] &\,\hspace{2pt}\times(z^2-2z+17)\\[4mm] =&\,z^4-2z^3+17z^2\\[4mm] &\,\hspace{2pt}-4z^3+8z^2-68z\\[4mm] &\,\hspace{4pt}+13z^2-26z+221\\[4mm] =&\,z^4-6z^3+38z^2\\[4mm] &\,\hspace{2pt}-94z+221\\[4mm] =&\,f(z), \end{align*}

as required. The remaining roots satisfy

z2−2z+17= 0(z−1)2+16= 0(z−1)2= −16.\begin{align*} z^2-2z+17=&\,0\\ (z-1)^2+16=&\,0\\ (z-1)^2=&\,-16. \end{align*}

Hence z=1±4iz=1\pm4\mathrm{i}. The other three roots are therefore

 2−3i, 1+4i, 1−4i.\begin{align*} &\,\boxed{2-3\mathrm{i}},\\[4mm] &\,\boxed{1+4\mathrm{i}},\\[4mm] &\,\boxed{1-4\mathrm{i}}. \end{align*}

(b)

解法一

思路

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在 Argand 图上，复数 a+bia+b\mathrm{i} 对应点 (a,b)(a,b)。因此把四个根的实部作为横坐标、虚部作为纵坐标，并利用共轭根关于实轴对称的性质标点即可。

答题过程

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On an Argand diagram, plot and label the points

(2,3)(2,−3)(1,4)(1,−4)\begin{array}{cc} (2,3)&(2,-3)\\[4pt] (1,4)&(1,-4) \end{array}

Thus 2±3i2\pm3\mathrm{i} lie on the vertical line Re⁡(z)=2\operatorname{Re}(z)=2, and 1±4i1\pm4\mathrm{i} lie on the vertical line Re⁡(z)=1\operatorname{Re}(z)=1. Each conjugate pair is symmetric about the real axis.