已知 $\dfrac{2+3a\mathrm{i}}{a+2\mathrm{i}}=\lambda(2-\mathrm{i})$，其中 $a$ 和 $\lambda$ 为实常数。

(a) 证明 $3a^2+4a-4=0$。

(b) 由此求 $a$ 的可能值及对应的 $\lambda$ 的值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

先清去复分母，展开后比较实部、虚部。消去 $\lambda$ 时可交叉相乘，不必除以可能为零的含参式。

Multiplying by $a+2\mathrm{i}$,

Equating real and imaginary parts,

Multiply the first equation by $4-a$ and the second by $2a+2$. Their right sides are equal, giving

Thus

也可以先用共轭化实分母，再与 $\lambda(2-\mathrm{i})$ 比较实部、虚部，消去 $\lambda$。

Multiplying by the conjugate,

Equating this to $2\lambda-\lambda\mathrm{i}$ gives

The first equation plus twice the second gives

Hence $\boxed{3a^2+4a-4=0}$.

分解(a)的二次式，再把每个 $a$ 代回同一个实部方程求相应的 $\lambda$。

From part (a),

Thus $a=2/3$ or $a=-2$.

Using $2=\lambda(2a+2)$:

• If $a=2/3$, then $2=(10/3)\lambda$, giving $\lambda=3/5$.
• If $a=-2$, then $2=-2\lambda$, giving $\lambda=-1$.

Therefore the pairs are