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FP1 2024 Jan Q2

A Level / Edexcel / FP1

IAL 2024 Jan Paper · Question 2

题目

Problem

 f(z)= 2z3+pz2+qz−41\begin{align*} &\,f(z)\\ =&\,2z^3+pz^2+qz-41 \end{align*}

where pp and qq are integers.

The complex number 5−4i5 - 4\mathrm{i} is a root of the equation f(z)=0f(z) = 0

(a) Write down another complex root of this equation.

(1)

(b) Solve the equation f(z)=0f(z) = 0 completely.

(4)

(c) Determine the value of pp and the value of qq.

(2)

When plotted on an Argand diagram, the points representing the roots of the equation f(z)=0f(z) = 0 form the vertices of a triangle.

(d) Determine the area of this triangle.

(2)
题目中文翻译

已知

 f(z)= 2z3+pz2+qz−41\begin{align*} &\,f(z)\\ =&\,2z^3+pz^2+qz-41 \end{align*}

其中 pp 和 qq 为整数。

复数 5−4i5 - 4\mathrm{i} 是方程 f(z)=0f(z) = 0 的一个根。

(a) 直接写出该方程的另一个复数根。

(1)

(b) 完全解方程 f(z)=0f(z) = 0。

(4)

(c) 确定 pp 和 qq 的值。

(2)

当在 Argand 图上绘制时，方程 f(z)=0f(z) = 0 的根所对应的点构成一个三角形的顶点。

(d) 确定该三角形的面积。

(2)

解答

(a)

解法一

思路

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多项式系数均为实数，因此非实复根必成共轭对出现。

答题过程

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Since ff has real coefficients and 5−4i5-4\mathrm{i} is a root, its complex conjugate is also a root. Hence another root is

5+4i.\boxed{5+4\mathrm{i}}.

(b)

解法一

思路

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先由共轭复根构造实系数二次因式。再利用首项系数与常数项确定剩余的一次因式，即可列出全部三个根。

答题过程

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The conjugate pair gives the factor

 (z−(5−4i)) ×(z−(5+4i))= (z−5)2−(4i)2= z2−10z+25+16= z2−10z+41.\begin{align*} &\,\big(z-(5-4\mathrm{i})\big)\\ &\,\hspace{2pt}\times\big(z-(5+4\mathrm{i})\big)\\ =&\,(z-5)^2-(4\mathrm{i})^2\\ =&\,z^2-10z+25+16\\ =&\,z^2-10z+41. \end{align*}

Since the leading coefficient of ff is 2, the remaining linear factor has the form 2z+a2z+a. Comparing constant terms gives

41a=−41,41a=-41,

so a=−1a=-1. Therefore,

 f(z)= (z2−10z+41) ×(2z−1).\begin{align*} &\,f(z)\\ =&\,(z^2-10z+41)\\ &\,\hspace{2pt}\times(2z-1). \end{align*}

Therefore,

z1= 5−4i,z2= 5+4i,z3= 12.\begin{align*} z_1=&\,5-4\mathrm{i},\\ z_2=&\,5+4\mathrm{i},\\ z_3=&\,\frac12. \end{align*}

解法二

思路

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利用三次多项式根积公式直接求第三根。两个共轭根的乘积是模长平方，而三个根的乘积由常数项和首项系数给出。

答题过程

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Let the third root be rr. The product of the conjugate roots is

 (5−4i)(5+4i)= 25+16= 41.\begin{align*} &\,(5-4\mathrm{i})(5+4\mathrm{i})\\ =&\,25+16\\ =&\,41. \end{align*}

For the cubic 2z3+pz2+qz−412z^3+pz^2+qz-41, the product of all three roots is

−−412=412.-\frac{-41}{2}=\frac{41}{2}.

Hence

41r=412,41r=\frac{41}{2},

so r=1/2r=1/2. Thus

z1= 5−4i,z2= 5+4i,z3= 12.\begin{align*} z_1=&\,5-4\mathrm{i},\\ z_2=&\,5+4\mathrm{i},\\ z_3=&\,\frac12. \end{align*}

解法三

思路

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先把复根代入 f(z)=0f(z)=0，比较实部和虚部，联立求出整数 p,qp,q；再分解所得三次多项式。这样也同时得到 (c) 的答案。

答题过程

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Let w=5+4iw=5+4\mathrm{i}. Then

 w2= 52+2(5)(4i)+(4i)2= 25+40i−16= 9+40i.\begin{align*} &\,w^2\\ =&\,5^2+2(5)(4\mathrm{i})+(4\mathrm{i})^2\\ =&\,25+40\mathrm{i}-16\\ =&\,9+40\mathrm{i}. \end{align*}

Hence

 w3= (9+40i)(5+4i)= 45+36i +200i+160i2= 45−160+(36+200)i= −115+236i.\begin{align*} &\,w^3\\ =&\,(9+40\mathrm{i})(5+4\mathrm{i})\\ =&\,45+36\mathrm{i}\\ &\,\hspace{2pt}+200\mathrm{i}+160\mathrm{i}^2\\ =&\,45-160+(36+200)\mathrm{i}\\ =&\,-115+236\mathrm{i}. \end{align*}

Therefore,

2w3= 2(−115+236i)= −230+472i.\begin{align*} 2w^3=&\,2(-115+236\mathrm{i})\\ =&\,-230+472\mathrm{i}. \end{align*}

Substitution into f(w)=0f(w)=0 gives

 f(w)= −230+472i +p(9+40i) +q(5+4i)−41= (−271+9p+5q) +(472+40p+4q)i.\begin{align*} &\,f(w)\\ =&\,-230+472\mathrm{i}\\ &\,\hspace{2pt}+p(9+40\mathrm{i})\\ &\,\hspace{4pt}+q(5+4\mathrm{i})-41\\ =&\,(-271+9p+5q)\\ &\,\hspace{2pt}+(472+40p+4q)\mathrm{i}. \end{align*}

Equating real and imaginary parts gives

9p+5q= 271,40p+4q= −472.\begin{align*} 9p+5q=&\,271,\\ 40p+4q=&\,-472. \end{align*}

Dividing the second equation by 4 gives

10p+q=−118,10p+q=-118,

so q=−118−10pq=-118-10p. Substituting into the first equation,

 9p+5(−118−10p)= 271,\begin{align*} &\,9p+5(-118-10p)\\ =&\,271, \end{align*}

and hence

9p−590−50p= 271,−41p= 861,p= −21.\begin{align*} 9p-590-50p=&\,271,\\ -41p=&\,861,\\ p=&\,-21. \end{align*}

It follows that

q= −118−10(−21)= −118+210= 92.\begin{align*} q=&\,-118-10(-21)\\ =&\,-118+210\\ =&\,92. \end{align*}

Therefore,

 f(z)= 2z3−21z2+92z−41= (z2−10z+41) ×(2z−1).\begin{align*} &\,f(z)\\ =&\,2z^3-21z^2+92z-41\\ =&\,(z^2-10z+41)\\ &\,\hspace{2pt}\times(2z-1). \end{align*}

Hence

z1= 5−4i,z2= 5+4i,z3= 12.\begin{align*} z_1=&\,5-4\mathrm{i},\\ z_2=&\,5+4\mathrm{i},\\ z_3=&\,\frac12. \end{align*}

(c)

解法一

思路

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承接 (b) 解法一，把已经确定的两个因式相乘，再比较 z2z^2 与 zz 的系数。

答题过程

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From part (b),

 f(z)= (z2−10z+41) ×(2z−1)= 2z3−z2−20z2 +10z+82z−41= 2z3−21z2+92z−41.\begin{align*} &\,f(z)\\ =&\,(z^2-10z+41)\\ &\,\hspace{2pt}\times(2z-1)\\ =&\,2z^3-z^2-20z^2\\ &\,\hspace{2pt}+10z+82z-41\\ =&\,2z^3-21z^2+92z-41. \end{align*}

Comparing this with

 f(z)= 2z3+pz2+qz−41.\begin{align*} &\,f(z)\\ =&\,2z^3+pz^2+qz-41. \end{align*}

gives

p=−21,q=92.\boxed{p=-21,\qquad q=92}.

解法二

思路

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利用三次方程的根与系数关系：三根之和确定 pp，两两乘积之和确定 qq。

答题过程

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The sum of the roots is

 (5−4i)+(5+4i) +12= 10+12= 212.\begin{align*} &\,(5-4\mathrm{i})+(5+4\mathrm{i})\\ &\,\hspace{2pt}+\frac12\\ =&\,10+\frac12\\ =&\,\frac{21}{2}. \end{align*}

Hence

−p2=212,-\frac p2=\frac{21}{2},

so p=−21p=-21.

The sum of the products of the roots taken two at a time is

 (5−4i)(5+4i) +12(5−4i) +12(5+4i)= 41+5= 46.\begin{align*} &\,(5-4\mathrm{i})(5+4\mathrm{i})\\ &\,\hspace{2pt}+\frac12(5-4\mathrm{i})\\ &\,\hspace{4pt}+\frac12(5+4\mathrm{i})\\ =&\,41+5\\ =&\,46. \end{align*}

Therefore,

q2=46,\frac q2=46,

so

p=−21,q=92.\boxed{p=-21,\qquad q=92}.

解法三

思路

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直接使用 (b) 解法三中代入复根所得的实部、虚部方程，联立求解 p,qp,q。

答题过程

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Substitution of 5+4i5+4\mathrm{i} into f(z)=0f(z)=0 gives

9p+5q= 271,40p+4q= −472.\begin{align*} 9p+5q=&\,271,\\ 40p+4q=&\,-472. \end{align*}

Dividing the second equation by 4 gives

10p+q=−118,10p+q=-118,

so q=−118−10pq=-118-10p. Substituting into the first equation,

 9p+5(−118−10p)= 271,\begin{align*} &\,9p+5(-118-10p)\\ =&\,271, \end{align*}

and hence

9p−590−50p= 271,−41p= 861,p= −21.\begin{align*} 9p-590-50p=&\,271,\\ -41p=&\,861,\\ p=&\,-21. \end{align*}

It follows that

q= −118−10(−21)= −118+210= 92.\begin{align*} q=&\,-118-10(-21)\\ =&\,-118+210\\ =&\,92. \end{align*}

Therefore,

p=−21,q=92.\boxed{p=-21,\qquad q=92}.

(d)

解法一

思路

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在 Argand 图上，共轭根对应点 (5,±4)(5,\pm4)，第三个根对应点 (1/2,0)(1/2,0)。取竖直线段为底，其长度为 8；第三点到直线 x=5x=5 的水平距离为 9/29/2。

答题过程

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The points representing the roots are

(5,−4),(5,4),(12,0).\begin{gathered} (5,-4),\\ (5,4),\\ \Big(\frac12,0\Big). \end{gathered}

The vertical base has length 8, and the perpendicular height is

5−12=92.5-\frac12=\frac92.

Therefore, the area of the triangle is

 12×8×92= 18.\begin{align*} &\,\frac12\times8\times\frac92\\ =&\,18. \end{align*}

The area of the triangle is 18 square units.