已知 $\frac{2+a\mathrm{i}}{a-3\mathrm{i}}=\lambda(1-5\mathrm{i})$，其中 $a$ 和 $\lambda$ 为实常数。

(a) 证明 $a^2-5a+6=0$。

(b) 由此求 $a$ 的可能值以及对应的 $\lambda$ 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。

先乘去原分母，在右边展开复数乘积。实部与虚部分别相等，然后消去 $\lambda$。

Multiplying by $a-3\mathrm{i}$ gives

Equating parts,

The first equation implies $a\ne15$ and $\lambda=2/(a-15)$. Substitution gives

也可将原分式的分母实数化，再比较两个实部与虚部。

Multiplying numerator and denominator by $a+3\mathrm{i}$,

Thus $\lambda=-a/(a^2+9)$ and $-5\lambda=(a^2+6)/(a^2+9)$. Substituting the first into the second gives $5a=a^2+6$, or $a^2-5a+6=0$.

还可直接把 $\lambda$ 表示成一个复分式。由于 $\lambda$ 是实数，实数化后分子的虚部必须为零。

Dividing by $1-5\mathrm{i}$ gives

The squared modulus of the denominator is $(a-15)^2+(5a+3)^2>0$. Multiplying by its conjugate, the imaginary part of the numerator is

Since $\lambda$ is real, this must vanish, giving the required equation.

分解 (a) 的二次式。对每一个根分别用实部方程求对应的 $\lambda$。

The equation factors as $(a-2)(a-3)=0$. Using $\lambda=-a/(a^2+9)$,