1.解方程z^2-4iz-(4-9i)=0.

将方程 $z^2 - 4iz - (4 - 9i) = 0$ 整理为标准形式，其中 $a = 1$，$b = -4i$，$c = -4 + 9i$。使用二次公式： \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4i \pm \sqrt{(-4i)^2 - 4 \cdot 1 \cdot (-4 + 9i)}}{2} \] 计算判别式： \[ \Delta = (-4i)^2 - 4(-4 + 9i) = -16 + 16 - 36i = -36i \] 求 $\sqrt{-36i}$： \[ -36i = 36 \left( \cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right) \right) \] \[ \sqrt{-36i} = 6 \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right) = 3\sqrt{2} - 3\sqrt{2}i \quad \text{或} \quad -3\sqrt{2} + 3\sqrt{2}i \] 代入二次公式得解： \[ z_1 = \frac{4i + 3\sqrt{2} - 3\sqrt{2}i}{2} = \frac{3\sqrt{2}}{2} + \left(2 - \frac{3\sqrt{2}}{2}\right)i \] \[ z_2 = \frac{4i - 3\sqrt{2} + 3\sqrt{2}i}{2} = -\frac{3\sqrt{2}}{2} + \left(2 + \frac{3\sqrt{2}}{2}\right)i \] **答案：** \[ \boxed{\frac{3\sqrt{2}}{2} + \left(2 - \frac{3\sqrt{2}}{2}\right)i, -\frac{3\sqrt{2}}{2} + \left(2 + \frac{3\sqrt{2}}{2}\right)i} \]