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FP1 2025 Jan Q4

A Level / Edexcel / FP1

IAL 2025 Jan Paper · Question 4

题目

Problem

 f(z)= 6z3+Az2 +Bz+C\begin{align*} &\,f(z)\\ =&\,6z^3+Az^2\\ &\,\hspace{2pt}+Bz+C \end{align*}

where A,BA, B and CC are integers.

Given that z=23+173iz = \frac{2}{3} + \frac{\sqrt{17}}{3}\mathrm{i} is a root of the equation f(z)=0f(z) = 0

(a) write down the other complex root of the equation f(z)=0f(z) = 0

(1)

Given that z=−32z = -\frac{3}{2} is also a root of the equation f(z)=0f(z) = 0

(b) determine the value of AA, the value of BB and the value of CC.

(5)

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

(2)
题目中文翻译

已知多项式

 f(z)= 6z3+Az2 +Bz+C\begin{align*} &\,f(z)\\ =&\,6z^3+Az^2\\ &\,\hspace{2pt}+Bz+C \end{align*}

其中 A,B,CA, B, C 均为整数。

已知 z=23+173iz = \frac{2}{3} + \frac{\sqrt{17}}{3}\mathrm{i} 是方程 f(z)=0f(z) = 0 的一个根。

(a) 写出方程 f(z)=0f(z) = 0 的另一个复数根。

(1)

已知 z=−32z = -\frac{3}{2} 也是方程 f(z)=0f(z) = 0 的一个根。

(b) 确定 AA 的值、BB 的值以及 CC 的值。

(5)

(c) 在同一个阿甘图（Argand diagram）上表示出方程 f(z)=0f(z) = 0 的所有根。

(2)

解答

(a)

解法一

思路

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多项式的系数都是整数，因此是实系数多项式。非实复根必成共轭对，所以把已知根的虚部变号即可。

答题过程

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Since ff has real coefficients, non-real roots occur in conjugate pairs. Therefore, the other complex root is

z=23−173i.\boxed{z=\frac23-\frac{\sqrt{17}}{3}\mathrm{i}}.

(b)

解法一

思路

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先把共轭根对应的两个一次因式相乘，得到实系数二次因式；再乘以实根对应的因式。由于原多项式首项系数为 6，最后将三个首一因式的乘积乘以 6，并比较系数。

答题过程

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The conjugate pair gives the quadratic factor

 (z−23−173i) ×(z−23+173i)= (z−23)2+179= z2−43z+49+179= z2−43z+73.\begin{align*} &\,\big(z-\tfrac23-\tfrac{\sqrt{17}}3\mathrm{i}\big)\\ &\,\hspace{2pt}\times\big(z-\tfrac23+\tfrac{\sqrt{17}}3\mathrm{i}\big)\\ =&\,\big(z-\tfrac23\big)^2+\frac{17}{9}\\ =&\,z^2-\frac43z+\frac49+\frac{17}{9}\\ =&\,z^2-\frac43z+\frac73. \end{align*}

The third root is −32-\frac32, so, using the leading coefficient 66,

 f(z)= 6(z+32) ×(z2−43z+73)= 6[z3+(−43+32)z2 +(73−2)z+72]= 6(z3+16z2 +13z+72)= 6z3+z2+2z+21.\begin{align*} &\,f(z)\\ =&\,6\big(z+\tfrac32\big)\\ &\,\hspace{2pt}\times\big(z^2-\tfrac43z+\tfrac73\big)\\ =&\,6\Big[z^3+\big(-\tfrac43+\tfrac32\big)z^2\\ &\,\hspace{2pt}+\big(\tfrac73-2\big)z+\tfrac72\Big]\\ =&\,6\big(z^3+\tfrac16z^2\\ &\,\hspace{2pt}+\tfrac13z+\tfrac72\big)\\ =&\,6z^3+z^2+2z+21. \end{align*}

Comparing coefficients with

 f(z)= 6z3+Az2+Bz+C,\begin{align*} &\,f(z)\\ =&\,6z^3+Az^2+Bz+C, \end{align*}

we obtain

A=1,B=2,C=21.\begin{gathered} \boxed{A=1},\qquad\boxed{B=2},\\ \boxed{C=21}. \end{gathered}

解法二

思路

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也可以使用三次方程的根与系数关系。分别计算三根之和、两两乘积之和与三根之积，再与 −A6-\frac A6、B6\frac B6、−C6-\frac C6 比较。

答题过程

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Let the roots be

α=23+173i,β=23−173i,γ=−32.\begin{gathered} \alpha=\frac23+\frac{\sqrt{17}}3\mathrm{i},\\ \beta=\frac23-\frac{\sqrt{17}}3\mathrm{i},\\ \gamma=-\frac32. \end{gathered}

Their sum is

α+β+γ=43−32=−16.\alpha+\beta+\gamma =\frac43-\frac32 =-\frac16.

Since the sum of the roots is −A6-\frac A6,

−A6=−16⟹A=1.-\frac A6=-\frac16 \quad\Longrightarrow\quad A=1.

Also,

αβ= (23)2+(173)2= 49+179= 73.\begin{align*} \alpha\beta =&\,\big(\tfrac23\big)^2+\big(\tfrac{\sqrt{17}}3\big)^2\\ =&\,\frac49+\frac{17}{9}\\ =&\,\frac73. \end{align*}

Hence the pairwise product sum is

 αβ+αγ+βγ= αβ+γ(α+β)= 73−32(43)= 13.\begin{align*} &\,\alpha\beta+\alpha\gamma+\beta\gamma\\ =&\,\alpha\beta+\gamma(\alpha+\beta)\\ =&\,\frac73-\frac32\big(\frac43\big)\\ =&\,\frac13. \end{align*}

Since this equals B6\frac B6,

B6=13⟹B=2.\frac B6=\frac13 \quad\Longrightarrow\quad B=2.

Finally,

αβγ=73(−32)=−72.\alpha\beta\gamma =\frac73\big(-\frac32\big) =-\frac72.

Since the product of the roots is −C6-\frac C6,

−C6= −72C= 21.\begin{align*} -\frac C6=&\,-\frac72\\ C=&\,21. \end{align*}

Therefore,

A=1,B=2,C=21.\begin{gathered} \boxed{A=1},\qquad\boxed{B=2},\\ \boxed{C=21}. \end{gathered}

解法三

思路

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把实根代入多项式，得到一个实系数方程。再计算已知复根的平方与立方，代入后分别令实部和虚部为零，得到另外两个方程。最后联立消元，求出三个系数。

答题过程

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Since −32-\frac32 is a root,

 0= 6(−32)3+A(−32)2 −32B+C= −814+94A−32B+C.\begin{align*} &\,0\\ =&\,6\big(-\tfrac32\big)^3+A\big(-\tfrac32\big)^2\\ &\,\hspace{2pt}-\tfrac32B+C\\ =&\,-\frac{81}{4}+\frac94A-\frac32B+C. \end{align*}

Multiplying by 4 gives

9A−6B+4C=81.(1)9A-6B+4C=81.\tag{1}

Write the given non-real root as

w=23+173i.w=\frac23+\frac{\sqrt{17}}3\mathrm{i}.

Using i2=−1\mathrm{i}^2=-1,

w2= 4−179+4179i= −139+4179i.\begin{align*} w^2 =&\,\frac{4-17}{9}+\frac{4\sqrt{17}}9\mathrm{i}\\ =&\,-\frac{13}{9}+\frac{4\sqrt{17}}9\mathrm{i}. \end{align*}

Multiplying this by ww gives

w3= −26−6827 +(8−13)1727i= −9427−51727i.\begin{align*} w^3 =&\,\frac{-26-68}{27}\\ &\,\hspace{2pt}+\frac{(8-13)\sqrt{17}}{27}\mathrm{i}\\ =&\,-\frac{94}{27}-\frac{5\sqrt{17}}{27}\mathrm{i}. \end{align*}

Now f(w)=0f(w)=0. Its real part gives

 −1889−13A9 +2B3+C=0,\begin{align*} &\,-\frac{188}{9}-\frac{13A}{9}\\ &\,\hspace{2pt}+\frac{2B}{3}+C=0, \end{align*}

so

13A−6B−9C=−188.(2)13A-6B-9C=-188.\tag{2}

Its imaginary part gives

179(−10+4A+3B)=0,\frac{\sqrt{17}}9(-10+4A+3B)=0,

so

4A+3B=10.(3)4A+3B=10.\tag{3}

From (3), 6B=20−8A6B=20-8A. Substituting into (1) and (2), respectively, gives

 9A−(20−8A)+4C=81 17A+4C=101\begin{align*} &\,9A-(20-8A)+4C=81\\ &\,17A+4C=101 \end{align*}

and

 13A−(20−8A) −9C=−188 21A−9C=−168 7A−3C=−56.\begin{align*} &\,13A-(20-8A)\\ &\,\hspace{2pt}-9C=-188\\ &\,21A-9C=-168\\ &\,7A-3C=-56. \end{align*}

Multiplying these last two equations by 3 and 4 gives

51A+12C=303,28A−12C=−224.\begin{gathered} 51A+12C=303,\\ 28A-12C=-224. \end{gathered}

Adding eliminates CC:

79A= 79A= 1.\begin{align*} 79A=&\,79\\ A=&\,1. \end{align*}

Then

17+4C= 1014C= 84C= 21.\begin{align*} 17+4C=&\,101\\ 4C=&\,84\\ C=&\,21. \end{align*}

Finally, substituting A=1A=1 into (3),

4+3B= 103B= 6B= 2.\begin{align*} 4+3B=&\,10\\ 3B=&\,6\\ B=&\,2. \end{align*}

Thus

A=1,B=2,C=21.\begin{gathered} \boxed{A=1},\qquad\boxed{B=2},\\ \boxed{C=21}. \end{gathered}

(c)

解法一

思路

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在同一 Argand 图上标出三个根：实根 −32-\frac32 位于负实轴；共轭根的实部同为 23\frac23，虚部分别为正、负，因此它们在第一、第四象限并关于实轴对称。

答题过程

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The roots are represented by the points

(−32,0),(23,173),(23,−173).\begin{gathered} \big(-\tfrac32,0\big),\\ \big(\tfrac23,\tfrac{\sqrt{17}}3\big),\\ \big(\tfrac23,-\tfrac{\sqrt{17}}3\big). \end{gathered}

The conjugate pair must be vertically aligned and symmetric about the real axis.