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IAL 2020 May FP1 Q3

A Level / Edexcel / FP1

IAL 2020 May Paper · Question 3

题目

Problem

3. f(z)=z4+az3+bz2+cz+df(z) = z^4 + az^3 + bz^2 + cz + d

where aa, bb, cc and dd are integers.

The complex numbers 3+i3 + i and 12i-1 - 2i are roots of the equation f(z)=0f(z) = 0

(a) Write down the other roots of this equation.

(2)

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

(2)

(c) Determine the values of aa, bb, cc and dd.

(5)
题目中文翻译
  1. f(z)=z4+az3+bz2+cz+df(z) = z^4 + az^3 + bz^2 + cz + d

其中 aabbccdd 是整数。

复数 3+i3 + i12i-1 - 2i 是方程 f(z)=0f(z) = 0 的根。

(a) 写出该方程的其他根。

(b) 在同一幅 Argand 图上画出方程 f(z)=0f(z) = 0 的所有根。

(c) 求 aabbccdd 的值。

解答

(a)

解法一

思路

展开

多项式的系数都是实数,因此每个非实复根的共轭复数也必定是根。

答题过程

展开

Since f(z)f(z) has real coefficients, non-real roots occur in conjugate pairs. Therefore, the other roots are

3iand1+2i.\boxed{3-\mathrm{i}\quad\text{and}\quad-1+2\mathrm{i}}.

(b)

解法一

思路

展开

在 Argand 图上,复数 x+yix+y\mathrm{i} 对应点 (x,y)(x,y)。因此标出 (3,1)(3,1)(3,1)(3,-1)(1,2)(-1,2)(1,2)(-1,-2);两对点均关于实轴对称。

答题过程

展开

The four roots are represented by the points

(3,1),(3,1),(3,1),\qquad(3,-1),

and

(1,2),(1,2)(-1,2),\qquad(-1,-2)

on the Argand diagram.

(c)

解法一

思路

展开

先把每对共轭复根组成一个实系数二次因式,再将两个二次因式相乘。展开所得四次多项式后,与题目给出的系数逐项比较。

答题过程

展开

The conjugate pair 3±i3\pm\mathrm{i} gives the factor

[z(3+i)][z(3i)]=(z3)2+1=z26z+10.\begin{align*} &\,\big[z-(3+\mathrm{i})\big] \big[z-(3-\mathrm{i})\big] \\ =&\,(z-3)^2+1 \\ =&\,z^2-6z+10. \end{align*}

Similarly, the pair 1±2i-1\pm2\mathrm{i} gives

[z(1+2i)][z(12i)]=(z+1)2+4=z2+2z+5.\begin{align*} &\,\big[z-(-1+2\mathrm{i})\big] \big[z-(-1-2\mathrm{i})\big] \\ =&\,(z+1)^2+4 \\ =&\,z^2+2z+5. \end{align*}

Therefore,

f(z)=(z26z+10)(z2+2z+5)=z44z3+3z210z+50.\begin{align*} f(z) =&\,(z^2-6z+10)(z^2+2z+5) \\ =&\,z^4-4z^3+3z^2-10z+50. \end{align*}

Hence

a=4,b=3,c=10,d=50.\boxed{a=-4,\quad b=3,\quad c=-10,\quad d=50}.

解法二

思路

展开

使用韦达定理。将两对共轭根分别看成一组,先求每组的和与积;这样二阶、三阶对称和都可由两组的和与积快速得到,不必列出所有组合逐项计算。

答题过程

展开

Let the roots be r1,r2,r3,r4r_1,r_2,r_3,r_4, where

{r1,r2}={3+i,3i}\{r_1,r_2\}=\{3+\mathrm{i},3-\mathrm{i}\}

and

{r3,r4}={1+2i,12i}.\{r_3,r_4\}=\{-1+2\mathrm{i},-1-2\mathrm{i}\}.

Then

r1+r2=6,r1r2=10,r_1+r_2=6, \qquad r_1r_2=10,

and

r3+r4=2,r3r4=5.r_3+r_4=-2, \qquad r_3r_4=5.

The sum of all four roots is

6+(2)=4,6+(-2)=4,

so a=4a=-4.

The sum of the six products of two distinct roots is

r1r2+r3r4+(r1+r2)(r3+r4)=10+5+6(2)=3,\begin{align*} &\,r_1r_2+r_3r_4 +(r_1+r_2)(r_3+r_4) \\ =&\,10+5+6(-2) \\ =&\,3, \end{align*}

so b=3b=3.

The sum of the four products of three distinct roots is

r1r2(r3+r4)+r3r4(r1+r2)=10(2)+5(6)=10,\begin{align*} &\,r_1r_2(r_3+r_4) +r_3r_4(r_1+r_2) \\ =&\,10(-2)+5(6) \\ =&\,10, \end{align*}

so c=10c=-10.

Finally,

r1r2r3r4=10(5)=50,r_1r_2r_3r_4=10(5)=50,

so d=50d=50. Therefore,

a=4,b=3,c=10,d=50.\boxed{a=-4,\quad b=3,\quad c=-10,\quad d=50}.