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IAL 2024 Jan FP1 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 + yi, 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 + yi 的形式表示，其中 xx 和 yy 为实数，且用含 λ\lambda 的式子表示。

已知 arg⁡z=π4\arg z = \dfrac{\pi}{4}，

(b) 求 λ\lambda 的可能值。

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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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+16,3z= 3λ2−24+18λiλ2+16.\begin{align*} 3z-1 =&\,\frac{2(\lambda^2-20+9\lambda\mathrm{i})} {\lambda^2+16},\\ 3z =&\,\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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From the given equation,

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

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.(3x+3y\mathrm{i}-1)(\lambda-4\mathrm{i}) =2\lambda+10\mathrm{i}.

Equating real and imaginary parts,

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

Eliminating yy 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}.

Similarly,

(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*}

Therefore,

λ=3±17.\lambda=3\pm\sqrt{17}.

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}}.