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IAL 2019 May FP1 Q7

A Level / Edexcel / FP1

IAL 2019 May Paper · Question 7

Question

Problem

f(z)=z4−6z3+az2−44z+bf(z) = z^4 - 6z^3 + az^2 - 44z + b

where aa and bb are real constants.

Given that −1−3i-1 - 3i 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)

中文翻译

f(z)=z4−6z3+az2−44z+bf(z) = z^4 - 6z^3 + az^2 - 44z + b

其中 aa 和 bb 是实常数。

已知 −1−3i-1 - 3i 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 two roots from part (a),

 (z−(−1−3i))(z−(−1+3i))= (z+1+3i)(z+1−3i)= (z+1)2+9= z2+2z+10.\begin{align*} &\,\big(z-(-1-3\mathrm{i})\big) \big(z-(-1+3\mathrm{i})\big) \\ =&\,(z+1+3\mathrm{i})(z+1-3\mathrm{i}) \\ =&\,(z+1)^2+9 \\ =&\,z^2+2z+10. \end{align*}

Therefore, for some real constants cc and dd,

f(z)=(z2+2z+10)(z2+cz+d).f(z)=(z^2+2z+10)(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)=(z2+2z+10)(z2−8z+18).f(z)=(z^2+2z+10)(z^2-8z+18).

The remaining roots satisfy

z2−8z+18=0.z^2-8z+18=0.

Therefore,

z= 8±(−8)2−4(1)(18)2= 8±−82= 4±2 i.\begin{align*} z=&\,\frac{8\pm\sqrt{(-8)^2-4(1)(18)}}{2} \\ =&\,\frac{8\pm\sqrt{-8}}{2} \\ =&\,4\pm\sqrt{2}\,\mathrm{i}. \end{align*}

Hence the other roots are

4+2 iand4−2 i.\boxed{4+\sqrt{2}\,\mathrm{i}\quad\text{and}\quad 4-\sqrt{2}\,\mathrm{i}}.

解法二

思路

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官方评分资料也接受多项式长除法。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), z2+2z+10z^2+2z+10 is a factor of f(z)f(z). Dividing gives

f(z)= (z2+2z+10)(z2−8z+18) +(a−12)z2+(b−180).\begin{align*} f(z)=&\,(z^2+2z+10)(z^2-8z+18) \\ &\,+(a-12)z^2+(b-180). \end{align*}

Since the division must have zero remainder,

a=12andb=180.a=12\qquad\text{and}\qquad b=180.

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}}.