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FP1 2022 June Q4

A Level / Edexcel / FP1

IAL 2022 June Paper · Question 4

题目

Problem

4.

 f(z)= 2z4−19z3 +Az2+Bz−156\begin{align*} &\,f(z)\\ =&\,2z^4-19z^3\\ &\,\hspace{2pt}+Az^2+Bz-156 \end{align*}

where AA and BB are constants.

The complex number 5−i5 - \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.

(5)

(c) Determine the value of AA and the value of BB.

(2)
题目中文翻译
 f(z)= 2z4−19z3 +Az2+Bz−156\begin{align*} &\,f(z)\\ =&\,2z^4-19z^3\\ &\,\hspace{2pt}+Az^2+Bz-156 \end{align*}

其中 AA 和 BB 是常数。

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

(a) 写出该方程的另一个复根。

(1)

(b) 完全解方程 f(z)=0f(z) = 0。

(5)

(c) 求 AA 和 BB 的值。

(2)

解答

(a)

解法一

思路

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多项式系数均为实数，因此非实复根必定成共轭对出现。

答题过程

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Since f(z)f(z) has real coefficients, non-real roots occur in conjugate pairs. Therefore, another root is

z=5+i.\boxed{z=5+\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。再设剩余二次因式的首项为 2z22z^2；利用常数项与 z3z^3 系数即可确定其余两项，最后解这个二次因式。

答题过程

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The quadratic factor formed from the conjugate pair is

 [z−(5−i)][z−(5+i)]= (z−5)2+1= z2−10z+26.\begin{align*} &\,[z-(5-\mathrm{i})][z-(5+\mathrm{i})]\\ =&\,(z-5)^2+1\\ =&\,z^2-10z+26. \end{align*}

Write

 f(z)= (z2−10z+26) ×(2z2+cz+d).\begin{align*} &\,f(z)\\ =&\,(z^2-10z+26)\\ &\,\hspace{2pt}\times(2z^2+cz+d). \end{align*}

Comparing constant terms gives

26d=−156,26d=-156,

so d=−6d=-6. Comparing the coefficients of z3z^3 gives

c−20=−19,c-20=-19,

so c=1c=1. Hence

 f(z)= (z2−10z+26) ×(2z2+z−6).\begin{align*} &\,f(z)\\ =&\,(z^2-10z+26)\\ &\,\hspace{2pt}\times(2z^2+z-6). \end{align*}

The second factor can be factorised as

 2z2+z−6= (2z−3)(z+2).\begin{align*} &\,2z^2+z-6\\ =&\,(2z-3)(z+2). \end{align*}

Therefore, the complete set of solutions is

z= 5−i,z= 5+i,z= 32,z= −2.\begin{align*} z=&\,\boxed{5-\mathrm{i}},\\ z=&\,\boxed{5+\mathrm{i}},\\ z=&\,\boxed{\frac32},\\ z=&\,\boxed{-2}. \end{align*}

解法二

思路

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也可以先求 A,BA,B：将共轭根 5+i5+\mathrm{i} 代入 f(z)=0f(z)=0，分别比较实部和虚部，解出两个系数。随后代回原多项式并用已知二次因式完成分解。

答题过程

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Using the root z=5+iz=5+\mathrm{i},

z2= 24+10i,z3= 110+74i,z4= 476+480i.\begin{align*} z^2=&\,24+10\mathrm{i},\\ z^3=&\,110+74\mathrm{i},\\ z^4=&\,476+480\mathrm{i}. \end{align*}

Substituting into f(z)=0f(z)=0 gives

0= 2(476+480i) −19(110+74i) +A(24+10i) +B(5+i)−156.\begin{align*} 0=&\,2(476+480\mathrm{i})\\ &\,-19(110+74\mathrm{i})\\ &\,+A(24+10\mathrm{i})\\ &\,+B(5+\mathrm{i})-156. \end{align*}

Equating real and imaginary parts,

the real terms give

 952−2090+24A +5B−156= −1294+24A+5B,\begin{align*} &\,952-2090+24A\\ &\,\hspace{2pt}+5B-156\\ =&\,-1294+24A+5B, \end{align*}

and the coefficient of i\mathrm{i} is

 960−1406+10A+B= −446+10A+B.\begin{align*} &\,960-1406+10A+B\\ =&\,-446+10A+B. \end{align*}

Both must be zero, so

24A+5B= 1294,10A+B= 446.\begin{align*} 24A+5B=&\,1294,\\ 10A+B=&\,446. \end{align*}

Multiplying the second equation by 55 gives

50A+5B= 2230,24A+5B= 1294.\begin{align*} 50A+5B=&\,2230,\\ 24A+5B=&\,1294. \end{align*}

Subtracting gives

26A= 936,A= 36.\begin{align*} 26A=&\,936,\\ A=&\,36. \end{align*}

Then

B= 446−10(36)= 86.\begin{align*} B=&\,446-10(36)\\ =&\,86. \end{align*}

Thus

 f(z)= 2z4−19z3+36z2 +86z−156.\begin{align*} &\,f(z)\\ =&\,2z^4-19z^3+36z^2\\ &\,\hspace{2pt}+86z-156. \end{align*}

To divide by the known factor, write

g(z)=z2−10z+26.g(z)=z^2-10z+26.

The first quotient term is 2z22z^2. Subtracting its product with g(z)g(z) leaves

 f(z)−2z2g(z)= (2z4−19z3+36z2 +86z−156) −(2z4−20z3 +52z2)= z3−16z2 +86z−156.\begin{align*} &\,f(z)-2z^2g(z)\\ =&\,\big(2z^4-19z^3+36z^2\\ &\,\hspace{2pt}+86z-156\big)\\ &\,\hspace{4pt}-\big(2z^4-20z^3\\ &\,\hspace{6pt}+52z^2\big)\\ =&\,z^3-16z^2\\ &\,\hspace{2pt}+86z-156. \end{align*}

The next quotient term is zz. Subtracting zg(z)zg(z) gives

 (z3−16z2 +86z−156) −(z3−10z2+26z)= −6z2+60z−156= −6g(z).\begin{align*} &\,\big(z^3-16z^2\\ &\,\hspace{2pt}+86z-156\big)\\ &\,\hspace{2pt}-\big(z^3-10z^2+26z\big)\\ =&\,-6z^2+60z-156\\ =&\,-6g(z). \end{align*}

Thus the final quotient term is −6-6, with zero remainder, and

 f(z)= (z2−10z+26) ×(2z2+z−6).\begin{align*} &\,f(z)\\ =&\,(z^2-10z+26)\\ &\,\hspace{2pt}\times(2z^2+z-6). \end{align*}

As in the first method,

 2z2+z−6= (2z−3)(z+2).\begin{align*} &\,2z^2+z-6\\ =&\,(2z-3)(z+2). \end{align*}

Therefore,

z= 5−i,z= 5+i,z= 32,z= −2.\begin{align*} z=&\,\boxed{5-\mathrm{i}},\\ z=&\,\boxed{5+\mathrm{i}},\\ z=&\,\boxed{\frac32},\\ z=&\,\boxed{-2}. \end{align*}

(c)

解法一

思路

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承接 (b) 解法一，把两个二次因式相乘，并将 z2z^2 项与 zz 项的系数和原多项式比较。

答题过程

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From part (b),

 f(z)= (z2−10z+26) ×(2z2+z−6).\begin{align*} &\,f(z)\\ =&\,(z^2-10z+26)\\ &\,\hspace{2pt}\times(2z^2+z-6). \end{align*}

Expanding,

 f(z)= 2z4+z3−6z2 −20z3−10z2+60z +52z2+26z−156= 2z4−19z3+36z2 +86z−156.\begin{align*} &\,f(z)\\ =&\,2z^4+z^3-6z^2\\ &\,-20z^3-10z^2+60z\\ &\,+52z^2+26z-156\\ =&\,2z^4-19z^3+36z^2\\ &\,\hspace{2pt}+86z-156. \end{align*}

Comparing coefficients gives

A=36,B=86.\boxed{A=36,\qquad B=86}.

解法二

思路

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沿用 (b) 解法二中代入复根、比较实部和虚部所得的联立方程，直接读取已经求出的系数。

答题过程

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As shown in the second method of part (b), substituting z=5+iz=5+\mathrm{i} gives

24A+5B= 1294,10A+B= 446.\begin{align*} 24A+5B=&\,1294,\\ 10A+B=&\,446. \end{align*}

From the second equation,

B=446−10A.B=446-10A.

Substituting into the first equation,

 24A+5(446−10A)= 1294, 24A+2230−50A= 1294, −26A=−936, A=36.\begin{align*} &\,24A+5(446-10A)\\ =&\,1294,\\ &\,24A+2230-50A\\ =&\,1294,\\ &\,-26A=-936,\\ &\,A=36. \end{align*}

Hence

B=446−10(36)=86.B=446-10(36)=86.

Therefore,

A=36,B=86.\boxed{A=36,\qquad B=86}.