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FP1 2019 May Q7

A Level / Edexcel / FP1

IAL 2019 May Paper · Question 7

题目

Problem

7.

f(z)= z4−6z3 +az2−44z +b\begin{align*} f(z)=&\,z^4-6z^3\\ &\,\hspace{2pt}+az^2-44z\\ &\,\hspace{4pt}+b \end{align*}

where aa and bb are real constants.

Given that −1−3i-1 - 3\mathrm{i} is a root of the equation f(z)=0f(z) = 0

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

(1)

(b) Hence find the other roots of the equation f(z)=0f(z) = 0.

(6)
题目中文翻译

7.

f(z)= z4−6z3 +az2−44z +b\begin{align*} f(z)=&\,z^4-6z^3\\ &\,\hspace{2pt}+az^2-44z\\ &\,\hspace{4pt}+b \end{align*}

其中 aa 和 bb 是实常数。

已知 −1−3i-1 - 3\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数根。

(1)

(b) 由此求方程 f(z)=0f(z) = 0 的其他根。

(6)

解答

(a)

解法一

思路

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多项式的系数都是实数，因此非实复根必成共轭对出现。

答题过程

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Since f(z)f(z) has real coefficients, the complex conjugate of −1−3i-1-3\mathrm{i} is also a root. Hence another root is

−1+3i.\boxed{-1+3\mathrm{i}}.

(b)

解法一

思路

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承接 (a)，先由一对共轭根组成实系数二次因式。再把剩余因式设为首一二次式，通过比较 z3z^3 项与 zz 项的系数求出它，最后解该二次方程。

答题过程

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Using the conjugate roots from part (a), their quadratic factor is

D(z)= (z+1)2+9= z2+2z+10.\begin{align*} D(z)=&\,(z+1)^2+9 \\ =&\,z^2+2z+10. \end{align*}

Equivalently, this follows from multiplying the factors z+1+3iz+1+3\mathrm{i} and z+1−3iz+1-3\mathrm{i}. For some real constants cc and dd,

f(z)=D(z)(z2+cz+d).f(z)=D(z)(z^2+cz+d).

Comparing the coefficients of z3z^3 gives

c+2=−6,c+2=-6,

so c=−8c=-8. Comparing the coefficients of zz then gives

2d+10c=−44.2d+10c=-44.

Hence

2d−80=−44,2d-80=-44,

so d=18d=18. Thus

f(z)=D(z)(z2−8z+18).f(z)=D(z)(z^2-8z+18).

The remaining roots satisfy

z2−8z+18=0.z^2-8z+18=0.

Therefore,

z= 8±64−722= 8±−82= 4±2 i.\begin{align*} z=&\,\frac{8\pm\sqrt{64-72}}{2} \\ =&\,\frac{8\pm\sqrt{-8}}{2} \\ =&\,4\pm\sqrt{2}\,\mathrm{i}. \end{align*}

Hence the other roots are

z1= 4+2 i,z2= 4−2 i.\begin{align*} z_1=&\,\boxed{4+\sqrt2\,\mathrm{i}},\\ z_2=&\,\boxed{4-\sqrt2\,\mathrm{i}}. \end{align*}

解法二

思路

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也可以使用多项式长除法。用 (a) 得到的二次因式直接除原四次式；余式必须为零，这会同时确定未知系数，并给出剩余二次因式。

答题过程

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From part (a), let D(z)=z2+2z+10D(z)=z^2+2z+10. The first quotient term is z2z^2. Subtracting z2D(z)z^2D(z) leaves

R1(z)= −8z3 +(a−10)z2 −44z+b.\begin{align*} R_1(z)=&\,-8z^3\\ &\,\hspace{2pt}+(a-10)z^2\\ &\,\hspace{4pt}-44z+b. \end{align*}

The next quotient term is −8z-8z. Subtracting −8zD(z)-8zD(z) from R1(z)R_1(z) leaves

R2(z)= (a+6)z2 +36z+b.\begin{align*} R_2(z)=&\,(a+6)z^2\\ &\,\hspace{2pt}+36z+b. \end{align*}

The final quotient term is a+6a+6. Subtracting (a+6)D(z)(a+6)D(z) gives the remainder

R(z)= (24−2a)z +b−10a−60.\begin{align*} R(z)=&\,(24-2a)z\\ &\,\hspace{2pt}+b-10a-60. \end{align*}

Since D(z)D(z) is a factor, R(z)R(z) is identically zero. Thus

24−2a= 0,b−10a−60= 0,\begin{align*} 24-2a=&\,0,\\ b-10a-60=&\,0, \end{align*}

so a=12a=12 and b=180b=180. The quotient is therefore

 z2−8z+(a+6)= z2−8z+18.\begin{align*} &\,z^2-8z+(a+6)\\ =&\,z^2-8z+18. \end{align*}

Hence the remaining quadratic factor is

z2−8z+18.z^2-8z+18.

Solving it,

z2−8z+18= 0(z−4)2+2= 0(z−4)2= −2.\begin{align*} z^2-8z+18=&\,0 \\ (z-4)^2+2=&\,0 \\ (z-4)^2=&\,-2. \end{align*}

Therefore, the other roots are

z=4±2 i.\boxed{z=4\pm\sqrt{2}\,\mathrm{i}}.