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FP1 2024 Jan Q9

A Level / Edexcel / FP1

IAL 2024 Jan Paper · Question 9

题目

Problem

Given that

3z−12=λ+5iλ−4i\frac{3z - 1}{2} = \frac{\lambda + 5\mathrm{i}}{\lambda - 4\mathrm{i}}

where λ\lambda is a real constant,

(a) determine zz, giving your answer in the form x+yix + y\mathrm{i}, where xx and yy are real and in terms of λ\lambda.

(4)

Given also that arg⁡z=π4\arg z = \dfrac{\pi}{4}

(b) find the possible values of λ\lambda.

(2)
题目中文翻译

已知

3z−12=λ+5iλ−4i\frac{3z - 1}{2} = \frac{\lambda + 5\mathrm{i}}{\lambda - 4\mathrm{i}}

其中 λ\lambda 为实常数，

(a) 确定 zz，答案以 x+yix + y\mathrm{i} 的形式表示，其中 xx 和 yy 为实数，且用含 λ\lambda 的式子表示。

(4)

已知 arg⁡z=π4\arg z = \dfrac{\pi}{4}，

(b) 求 λ\lambda 的可能值。

(2)

解答

(a)

解法一

思路

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先将右侧分式的分子、分母同乘分母的共轭，使它化为 Cartesian form；再由原等式解出 zz，并分别整理实部与虚部。

答题过程

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The numerator and denominator products are

 (λ+5i)(λ+4i)= λ2+4λi +5λi+20i2= λ2−20+9λi\begin{align*} &\,(\lambda+5\mathrm{i})(\lambda+4\mathrm{i})\\ =&\,\lambda^2+4\lambda\mathrm{i}\\ &\,\hspace{2pt}+5\lambda\mathrm{i}+20\mathrm{i}^2\\ =&\,\lambda^2-20+9\lambda\mathrm{i} \end{align*}

and

 (λ−4i)(λ+4i)= λ2−(4i)2= λ2+16.\begin{align*} &\,(\lambda-4\mathrm{i})(\lambda+4\mathrm{i})\\ =&\,\lambda^2-(4\mathrm{i})^2\\ =&\,\lambda^2+16. \end{align*}

Rationalising the right-hand side,

 λ+5iλ−4i= (λ+5i)(λ+4i)(λ−4i)(λ+4i)= λ2−20+9λiλ2+16.\begin{align*} &\,\frac{\lambda+5\mathrm{i}}{\lambda-4\mathrm{i}}\\ =&\,\frac{(\lambda+5\mathrm{i})(\lambda+4\mathrm{i})} {(\lambda-4\mathrm{i})(\lambda+4\mathrm{i})}\\ =&\,\frac{\lambda^2-20+9\lambda\mathrm{i}} {\lambda^2+16}. \end{align*}

Therefore,

 3z−1= 2(λ2−20+9λi)λ2+16.\begin{align*} &\,3z-1\\ =&\,\frac{2(\lambda^2-20+9\lambda\mathrm{i})}{\lambda^2+16}. \end{align*}

Adding 1 with the same denominator gives

 3z= 2(λ2−20+9λi)λ2+16+1= 2λ2−40+18λiλ2+16 +λ2+16λ2+16= 3λ2−24+18λiλ2+16.\begin{align*} &\,3z\\ =&\,\frac{2(\lambda^2-20+9\lambda\mathrm{i})}{\lambda^2+16}+1\\ =&\,\frac{2\lambda^2-40+18\lambda\mathrm{i}}{\lambda^2+16}\\ &\,\hspace{2pt}+\frac{\lambda^2+16}{\lambda^2+16}\\ =&\,\frac{3\lambda^2-24+18\lambda\mathrm{i}}{\lambda^2+16}. \end{align*}

Hence

z= λ2−8λ2+16 +6λλ2+16i.\boxed{ \begin{aligned} z=&\,\frac{\lambda^2-8}{\lambda^2+16}\\ &\,\hspace{2pt}+\frac{6\lambda}{\lambda^2+16}\mathrm{i} \end{aligned} }.

解法二

思路

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先由原等式直接解出 zz 的复分式形式，再进行有理化。这样会先约去公因数 3，运算更短。

答题过程

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When the right-hand side is put over a common denominator, its numerator is

 A= 2(λ+5i)+(λ−4i)= 2λ+10i+λ−4i= 3λ+6i.\begin{align*} &\,A\\ =&\,2(\lambda+5\mathrm{i})+(\lambda-4\mathrm{i})\\ =&\,2\lambda+10\mathrm{i}+\lambda-4\mathrm{i}\\ =&\,3\lambda+6\mathrm{i}. \end{align*}

From the given equation,

 3z= 2(λ+5i)λ−4i+1= Aλ−4i= 3λ+6iλ−4i.\begin{align*} &\,3z\\ =&\,\frac{2(\lambda+5\mathrm{i})}{\lambda-4\mathrm{i}}+1\\ =&\,\frac{A}{\lambda-4\mathrm{i}}\\ =&\,\frac{3\lambda+6\mathrm{i}}{\lambda-4\mathrm{i}}. \end{align*}

Thus

z=λ+2iλ−4i.z=\frac{\lambda+2\mathrm{i}} {\lambda-4\mathrm{i}}.

For the numerator product,

 (λ+2i)(λ+4i)= λ2+4λi +2λi+8i2= λ2−8+6λi.\begin{align*} &\,(\lambda+2\mathrm{i})(\lambda+4\mathrm{i})\\ =&\,\lambda^2+4\lambda\mathrm{i}\\ &\,\hspace{2pt}+2\lambda\mathrm{i}+8\mathrm{i}^2\\ =&\,\lambda^2-8+6\lambda\mathrm{i}. \end{align*}

Also,

 (λ−4i)(λ+4i)= λ2−(4i)2= λ2+16.\begin{align*} &\,(\lambda-4\mathrm{i})(\lambda+4\mathrm{i})\\ =&\,\lambda^2-(4\mathrm{i})^2\\ =&\,\lambda^2+16. \end{align*}

Rationalising,

 z= (λ+2i)(λ+4i)(λ−4i)(λ+4i)= λ2−8+6λiλ2+16.\begin{align*} &\,z\\ =&\,\frac{(\lambda+2\mathrm{i})(\lambda+4\mathrm{i})} {(\lambda-4\mathrm{i})(\lambda+4\mathrm{i})}\\ =&\,\frac{\lambda^2-8+6\lambda\mathrm{i}} {\lambda^2+16}. \end{align*}

Therefore,

z= λ2−8λ2+16 +6λλ2+16i.\boxed{ \begin{aligned} z=&\,\frac{\lambda^2-8}{\lambda^2+16}\\ &\,\hspace{2pt}+\frac{6\lambda}{\lambda^2+16}\mathrm{i} \end{aligned} }.

解法三

思路

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设 z=x+yiz=x+y\mathrm{i}，交叉相乘后比较实部与虚部，得到关于 x,yx,y 的二元一次方程组。联立求解便能直接得到 Cartesian form。

答题过程

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Let

z=x+yi.z=x+y\mathrm{i}.

Cross-multiplying the given equation gives

 (3x+3yi−1) ×(λ−4i)= 2λ+10i.\begin{align*} &\,(3x+3y\mathrm{i}-1)\\ &\,\hspace{2pt}\times(\lambda-4\mathrm{i})\\ =&\,2\lambda+10\mathrm{i}. \end{align*}

Expanding the product on the left gives

 (3x+3yi−1) ×(λ−4i)= 3λx−λ+12y +(−12x+4+3λy)i.\begin{align*} &\,(3x+3y\mathrm{i}-1)\\ &\,\hspace{2pt}\times(\lambda-4\mathrm{i})\\ =&\,3\lambda x-\lambda+12y\\ &\,\hspace{2pt}+(-12x+4+3\lambda y)\mathrm{i}. \end{align*}

Equating real and imaginary parts gives

3λx+12y= 3λ,−12x+3λy= 6.\begin{align*} 3\lambda x+12y=&\,3\lambda,\\ -12x+3\lambda y=&\,6. \end{align*}

Dividing both equations by 3,

λx+4y= λ,−4x+λy= 2.\begin{align*} \lambda x+4y =&\,\lambda,\\ -4x+\lambda y =&\,2. \end{align*}

Multiplying the first equation by λ\lambda and the second by 4 gives

λ2x+4λy= λ2,−16x+4λy= 8.\begin{align*} \lambda^2x+4\lambda y=&\,\lambda^2,\\ -16x+4\lambda y=&\,8. \end{align*}

Subtracting eliminates yy and gives

(λ2+16)x=λ2−8,(\lambda^2+16)x=\lambda^2-8,

so

x=λ2−8λ2+16.x=\frac{\lambda^2-8}{\lambda^2+16}.

To eliminate xx, multiply the first equation by 4 and the second by λ\lambda:

4λx+16y= 4λ,−4λx+λ2y= 2λ.\begin{align*} 4\lambda x+16y=&\,4\lambda,\\ -4\lambda x+\lambda^2y=&\,2\lambda. \end{align*}

Adding gives

(λ2+16)y=6λ,(\lambda^2+16)y=6\lambda,

so

y=6λλ2+16.y=\frac{6\lambda}{\lambda^2+16}.

Hence

z= λ2−8λ2+16 +6λλ2+16i.\boxed{ \begin{aligned} z=&\,\frac{\lambda^2-8}{\lambda^2+16}\\ &\,\hspace{2pt}+\frac{6\lambda}{\lambda^2+16}\mathrm{i} \end{aligned} }.

(b)

解法一

思路

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辐角为 π/4\pi/4 表示复数位于第一象限且实部等于虚部。先令 (a) 的实部与虚部相等，解出两个候选值，再利用第一象限要求排除负值。

答题过程

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Since arg⁡z=π4\arg z=\frac\pi4, the real and imaginary parts of zz are equal and positive. Hence

λ2−8λ2+16=6λλ2+16.\frac{\lambda^2-8}{\lambda^2+16} =\frac{6\lambda}{\lambda^2+16}.

Since λ2+16>0\lambda^2+16>0,

λ2−8= 6λλ2−6λ−8= 0.\begin{align*} \lambda^2-8 =&\,6\lambda\\ \lambda^2-6\lambda-8 =&\,0. \end{align*}

The discriminant is

Δ= (−6)2−4(1)(−8)= 36+32= 68.\begin{align*} \Delta=&\,(-6)^2-4(1)(-8)\\ =&\,36+32\\ =&\,68. \end{align*}

Therefore,

λ= 6±Δ2= 6±2172= 3±17.\begin{align*} \lambda=&\,\frac{6\pm\sqrt\Delta}{2}\\ =&\,\frac{6\pm2\sqrt{17}}2\\ =&\,3\pm\sqrt{17}. \end{align*}

For λ=3−17<0\lambda=3-\sqrt{17}<0, the imaginary part

6λλ2+16\frac{6\lambda}{\lambda^2+16}

is negative, so zz is not in the first quadrant. This value must be rejected.

Hence

λ=3+17.\boxed{\lambda=3+\sqrt{17}}.