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FP1 2023 Jan Q3

A Level / Edexcel / FP1

IAL 2023 Jan Paper · Question 3

题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

 f(z)= 4z3+pz2 −24z+108\begin{align*} &\,f(z)\\ =&\,4z^3+pz^2\\ &\,\hspace{2pt}-24z+108 \end{align*}

where pp is a constant.

Given that −3-3 is a root of the equation f(z)=0f(z) = 0

(a) determine the value of pp

(2)

(b) using algebra, solve f(z)=0f(z) = 0 completely, giving the roots in simplest form,

(4)

(c) determine the modulus of the complex roots of f(z)=0f(z) = 0

(2)

(d) show the roots of f(z)=0f(z) = 0 on a single Argand diagram.

(2)
题目中文翻译

本题必须展示所有解题步骤。 完全依赖计算器技术的解答不可接受。

 f(z)= 4z3+pz2 −24z+108\begin{align*} &\,f(z)\\ =&\,4z^3+pz^2\\ &\,\hspace{2pt}-24z+108 \end{align*}

其中 pp 为常数。

已知 −3-3 是方程 f(z)=0f(z) = 0 的一个根，

(a) 确定 pp 的值

(2)

(b) 用代数方法完全解方程 f(z)=0f(z) = 0，根以最简形式表示

(4)

(c) 确定 f(z)=0f(z) = 0 的复数根的模

(2)

(d) 在同一 Argand 图上表示 f(z)=0f(z) = 0 的根

(2)

解答

(a)

解法一

思路

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由因式定理，−3-3 是根意味着 f(−3)=0f(-3)=0。代入后得到关于 pp 的一次方程。

答题过程

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Since −3-3 is a root, f(−3)=0f(-3)=0. Substituting gives

 4(−3)3+p(−3)2 −24(−3)+108= 0.\begin{align*} &\,4(-3)^3+p(-3)^2\\ &\,\hspace{2pt}-24(-3)+108\\ =&\,0. \end{align*}

Therefore,

 −108+9p +72+108= 0, 9p+72= 0, 9p= −72.\begin{align*} &\,-108+9p\\ &\,\hspace{2pt}+72+108\\ =&\,0,\\ &\,9p+72\\ =&\,0,\\ &\,9p\\ =&\,-72. \end{align*}

Hence

p=−8.\boxed{p=-8}.

解法二

思路

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由于 z+3z+3 是因式，可把二次因式的首项和常数项先确定，只留下中间系数 bb。展开乘积后，先比较 zz 的系数求 bb，再比较 z2z^2 的系数求 pp。

答题过程

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Since z+3z+3 is a factor, write

f(z)= (z+3) ×(4z2+bz+36).\begin{align*} f(z)=&\,(z+3)\\ &\,\hspace{2pt}\times(4z^2+bz+36). \end{align*}

Expanding,

 f(z)= 4z3+bz2+36z +12z2+3bz+108= 4z3+(b+12)z2 +(3b+36)z+108.\begin{align*} &\,f(z)\\ =&\,4z^3+bz^2+36z\\ &\,\hspace{2pt}+12z^2+3bz+108\\ =&\,4z^3+(b+12)z^2\\ &\,\hspace{2pt}+(3b+36)z+108. \end{align*}

Comparing the coefficients of zz gives

3b+36= −24,3b= −60,b= −20.\begin{align*} 3b+36=&\,-24,\\ 3b=&\,-60,\\ b=&\,-20. \end{align*}

Comparing the coefficients of z2z^2,

p= b+12= −20+12= −8.\begin{align*} p=&\,b+12\\ =&\,-20+12\\ =&\,\boxed{-8}. \end{align*}

(b)

解法一

思路

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代入 p=−8p=-8 后，用多项式除法除去已知因式 z+3z+3。每一步用当前最高次项确定下一商项，乘回并相减，直到余数为零。最后用二次公式求出共轭复根，并列出全部三个根。

答题过程

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With p=−8p=-8,

f(z)= 4z3−8z2 −24z+108.\begin{align*} f(z)=&\,4z^3-8z^2\\ &\,\hspace{2pt}-24z+108. \end{align*}

Since z=−3z=-3 is a root, z+3z+3 is a factor.

Dividing the leading terms gives 4z3/z=4z24z^3/z=4z^2. Multiplying back,

4z2(z+3)=4z3+12z2.4z^2(z+3)=4z^3+12z^2.

Subtracting this from f(z)f(z) leaves

 (−8−12)z2−24z +108= −20z2−24z+108.\begin{align*} &\,(-8-12)z^2-24z\\ &\,\hspace{2pt}+108\\ =&\,-20z^2-24z+108. \end{align*}

The next quotient term is −20z2/z=−20z-20z^2/z=-20z. Multiplying back,

 (−20z)(z+3)= −20z2−60z.\begin{align*} &\,(-20z)(z+3)\\ =&\,-20z^2-60z. \end{align*}

Subtracting leaves

 (−24+60)z+108= 36z+108.\begin{align*} &\,(-24+60)z+108\\ =&\,36z+108. \end{align*}

The final quotient term is 36z/z=3636z/z=36, and

36(z+3)=36z+108.36(z+3)=36z+108.

The remainder is therefore zero, so

 f(z)= (z+3) ×(4z2−20z+36).\begin{align*} &\,f(z)\\ =&\,(z+3)\\ &\,\hspace{2pt}\times(4z^2-20z+36). \end{align*}

For the quadratic factor, the discriminant is

Δ= (−20)2−4(4)(36)= 400−576= −176.\begin{align*} \Delta=&\,(-20)^2-4(4)(36)\\ =&\,400-576\\ =&\,-176. \end{align*}

Hence

z= 20±Δ8= 20±−1768= 20±4i118= 5±i112.\begin{align*} z=&\,\frac{20\pm\sqrt{\Delta}}8\\ =&\,\frac{20\pm\sqrt{-176}}8\\ =&\,\frac{20\pm4\mathrm{i}\sqrt{11}}8\\ =&\,\frac{5\pm\mathrm{i}\sqrt{11}}2. \end{align*}

Thus all three roots are

z1= −3,z2= 5+i112,z3= 5−i112.\begin{align*} z_1=&\,\boxed{-3},\\ z_2=&\,\boxed{\frac{5+\mathrm{i}\sqrt{11}}2},\\ z_3=&\,\boxed{\frac{5-\mathrm{i}\sqrt{11}}2}. \end{align*}

解法二

思路

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沿用前面的因式分解，但用配方法代替二次公式。先把二次方程除以 4，再补成平方，即可求出两个复根。

答题过程

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Using the factorisation from Method 1, one root is z=−3z=-3. The other roots satisfy

4z2−20z+36=0.4z^2-20z+36=0.

Dividing by 44 and completing the square,

z2−5z+9= 0,z2−5z= −9,z2−5z+254= −9+254,(z−52)2= −114.\begin{align*} z^2-5z+9=&\,0,\\ z^2-5z=&\,-9,\\ z^2-5z+\frac{25}4=&\,-9+\frac{25}4,\\ \Big(z-\frac52\Big)^2=&\,-\frac{11}4. \end{align*}

Thus

z−52= ±i112,z= 5±i112.\begin{align*} z-\frac52=&\,\pm\frac{\mathrm{i}\sqrt{11}}2,\\ z=&\,\frac{5\pm\mathrm{i}\sqrt{11}}2. \end{align*}

All three roots are therefore

z1= −3,z2= 5+i112,z3= 5−i112.\begin{align*} z_1=&\,\boxed{-3},\\ z_2=&\,\boxed{\frac{5+\mathrm{i}\sqrt{11}}2},\\ z_3=&\,\boxed{\frac{5-\mathrm{i}\sqrt{11}}2}. \end{align*}

(c)

解法一

思路

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复数的模由实部与虚部的平方和开方得到。两个复根互为共轭，因此模相同，只需计算其中一个。

答题过程

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For either complex root,

 ∣5±i112∣= (52)2+(112)2= 25+114= 9= 3.\begin{align*} &\,\Big\vert\frac{5\pm\mathrm{i}\sqrt{11}}2\Big\vert\\ =&\,\sqrt{ \Big(\frac52\Big)^2 +\Big(\frac{\sqrt{11}}2\Big)^2 }\\ =&\,\sqrt{\frac{25+11}4}\\ =&\,\sqrt9\\ =&\,\boxed{3}. \end{align*}

解法二

思路

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两个复根是共轭数，其乘积等于模的平方；另一方面，它们的乘积可直接由二次因式的常数项与首项系数得到。

答题过程

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The product of the roots of

4z2−20z+36=04z^2-20z+36=0

is

364=9.\frac{36}{4}=9.

Since the roots are a conjugate pair, their product is the square of their common modulus. Hence

∣z∣2=9,|z|^2=9,

so

∣z∣=3.\boxed{|z|=3}.

(d)

解法一

思路

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在 Argand 图上，横轴表示实部、纵轴表示虚部。实根位于负实轴；两个共轭根实部均为 5/25/2，虚部分别为 ±11/2\pm\sqrt{11}/2，应关于实轴对称。

答题过程

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Plot the three points

(−3,0),(-3,0), (52,112),\Big(\frac52,\frac{\sqrt{11}}2\Big),

and

(52,−112)\Big(\frac52,-\frac{\sqrt{11}}2\Big)

on a single Argand diagram, with the conjugate pair symmetric about the real axis.