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IAL 2022 June FP1 Q4

A Level / Edexcel / FP1

IAL 2022 June Paper · Question 4

题目

Problem

4. f(z)=2z4−19z3+Az2+Bz−156f(z) = 2z^4 - 19z^3 + Az^2 + Bz - 156

where AA and BB are constants.

The complex number 5−i5 - 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
  1. f(z)=2z4−19z3+Az2+Bz−156f(z) = 2z^4 - 19z^3 + Az^2 + Bz - 156

其中 AA 和 BB 是常数。

复数 5−i5 - 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。

(b) 完全解方程 f(z)=0f(z) = 0。

(c) 求 AA 和 BB 的值。

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).f(z)=(z^2-10z+26)(2z^2+cz+d).

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).f(z)=(z^2-10z+26)(2z^2+z-6).

The second factor can be factorised as

2z2+z−6=(2z−3)(z+2).2z^2+z-6=(2z-3)(z+2).

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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官方评分资料也允许先求 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,

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

Solving simultaneously gives

A= 36,B= 86.\begin{align*} A=&\,36,\\ B=&\,86. \end{align*}

Thus

f(z)=2z4−19z3+36z2+86z−156.f(z)=2z^4-19z^3+36z^2+86z-156.

Dividing by the known factor z2−10z+26z^2-10z+26 gives

f(z)=(z2−10z+26)(2z2+z−6).f(z)=(z^2-10z+26)(2z^2+z-6).

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).f(z)=(z^2-10z+26)(2z^2+z-6).

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