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FP1 2021 Oct Q4

A Level / Edexcel / FP1

IAL 2021 Oct Paper · Question 4

题目

Problem

4.

 f(z)= 2z3−z2 +az+b\begin{align*} &\,f(z)\\ =&\,2z^3-z^2\\ &\,+az+b \end{align*}

where aa and bb are integers.

The complex number −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) Determine the value of aa and the value of bb.

(4)

(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
 f(z)= 2z3−z2 +az+b\begin{align*} &\,f(z)\\ =&\,2z^3-z^2\\ &\,+az+b \end{align*}

其中 aa 和 bb 是整数。

复数 −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根。

(1)

(b) 求 aa 和 bb 的值。

(4)

(c) 在同一幅 Argand 图上画出方程 f(z)=0f(z) = 0 的所有根。

(2)

解答

(a)

解法一

思路

展开

多项式的系数都是实数，因此非实复根成共轭对出现。

答题过程

展开

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

−1+3i.\boxed{-1+3\mathrm{i}}.

(b)

解法一

思路

展开

先由一对共轭复根构造实系数二次因式。再结合原多项式的首项，设剩余的一次因式为 2z+c2z+c，比较 z2z^2 的系数求出 cc，展开后即可读出 aa 和 bb。

答题过程

展开

The quadratic factor corresponding to the roots −1±3i-1\pm3\mathrm{i} is

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

Since the coefficient of z3z^3 in f(z)f(z) is 22, write

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

In this product, the terms in z2z^2 are

cz2+4z2=(c+4)z2.cz^2+4z^2=(c+4)z^2.

Comparing with the coefficient −1-1 in f(z)f(z) gives

c+4= −1,c= −5.\begin{align*} c+4=&\,-1,\\ c=&\,-5. \end{align*}

Hence

 f(z)= (z2+2z+10) ×(2z−5)= 2z3−5z2+4z2 −10z+20z−50= 2z3−z2+10z−50.\begin{align*} &\,f(z)\\ =&\,(z^2+2z+10)\\ &\,\hspace{2pt}\times(2z-5)\\ =&\,2z^3-5z^2+4z^2\\ &\,-10z+20z-50\\ =&\,2z^3-z^2+10z-50. \end{align*}

Therefore,

a=10,b=−50.\boxed{a=10,\qquad b=-50}.

解法二

思路

展开

把共轭根 −1+3i-1+3\mathrm{i} 直接代入 f(z)=0f(z)=0，分别比较实部与虚部，得到关于 aa、bb 的两个实方程。

答题过程

展开

Let z=−1+3iz=-1+3\mathrm{i}. Then

z2= (−1+3i)2= 1−6i+9i2= 1−6i−9= −8−6i.\begin{align*} z^2=&\,(-1+3\mathrm{i})^2\\ =&\,1-6\mathrm{i}+9\mathrm{i}^2\\ =&\,1-6\mathrm{i}-9\\ =&\,-8-6\mathrm{i}. \end{align*}

Also,

 z3= (−8−6i) ×(−1+3i)= 8−24i+6i−18i2= 8−18i+18= 26−18i.\begin{align*} &\,z^3\\ =&\,(-8-6\mathrm{i})\\ &\,\hspace{2pt}\times(-1+3\mathrm{i})\\ =&\,8-24\mathrm{i}+6\mathrm{i}-18\mathrm{i}^2\\ =&\,8-18\mathrm{i}+18\\ =&\,26-18\mathrm{i}. \end{align*}

Since f(z)=0f(z)=0,

0= 2(26−18i) −(−8−6i) +a(−1+3i)+b= 52−36i+8+6i −a+3ai+b= (60−a+b) +(−30+3a)i.\begin{align*} 0=&\,2(26-18\mathrm{i})\\ &\,-(-8-6\mathrm{i})\\ &\,+a(-1+3\mathrm{i})+b\\ =&\,52-36\mathrm{i}+8+6\mathrm{i}\\ &\,-a+3a\mathrm{i}+b\\ =&\,(60-a+b)\\ &\,+(-30+3a)\mathrm{i}. \end{align*}

Equating imaginary parts gives

−30+3a= 0,3a= 30,a= 10.\begin{align*} -30+3a=&\,0,\\ 3a=&\,30,\\ a=&\,10. \end{align*}

Equating real parts then gives

60−a+b= 0,60−10+b= 0,b= −50.\begin{align*} 60-a+b=&\,0,\\ 60-10+b=&\,0,\\ b=&\,-50. \end{align*}

Therefore,

a=10,b=−50.\boxed{a=10,\qquad b=-50}.

(c)

解法一

思路

展开

由 (b) 的因式分解可得第三个根为 52\frac52。在 Argand 图上分别标出两个共轭复根 (−1,3)(-1,3)、(−1,−3)(-1,-3)，以及实根 (52,0)\big(\frac52,0\big)。

答题过程

展开

From part (b),

 f(z)= (z2+2z+10) ×(2z−5).\begin{align*} &\,f(z)\\ =&\,(z^2+2z+10)\\ &\,\hspace{2pt}\times(2z-5). \end{align*}

The linear factor gives

2z−5= 0,z= 52.\begin{align*} 2z-5=&\,0,\\ z=&\,\frac52. \end{align*}

Thus the three roots are

z1= −1+3i,z2= −1−3i,z3= 52.\begin{align*} z_1=&\,-1+3\mathrm{i},\\ z_2=&\,-1-3\mathrm{i},\\ z_3=&\,\frac52. \end{align*}

These are represented on the Argand diagram by the points

 (−1,3), (−1,−3), (52,0).\begin{align*} &\,(-1,3),\\ &\,(-1,-3),\\ &\,\big(\tfrac52,0\big). \end{align*}