ans

018553
$25$的平方根是$5$与$-5$, $-25$的平方根是$5\mathrm{i}$与$-5\mathrm{i}$

024821
两根为$1\pm \dfrac{\sqrt{6}}{2}\mathrm{i}$, 因式分解的结果为$2(x-1-\dfrac{\sqrt{6}}{2}\mathrm{i})(x-1+\dfrac{\sqrt{6}}{2}\mathrm{i})$

018555
$p=8$, $q=26$

018556
有错误, 错在$|x_1|^2$不能整理成$x_1^2$; 正确的解答可以通过分判别式$\ge 0$与$<0$进行, $m=-15$或$16$

009657
$(-1,0)$

009659
$-\dfrac{1}{3}$

018557
(1) $r=2$, $\theta=\dfrac{\pi}{6}$, 三角形式为$2(\cos\dfrac{\pi}{6}+\mathrm{i}\sin\dfrac{\pi}{6})$;\\
(2) $r=\sqrt{2}$, $\theta=\dfrac{3\pi}{4}$, 三角形式为$\sqrt{2}(\cos\dfrac{3\pi}{4}+\mathrm{i}\sin\dfrac{3\pi}{4})$;\\
(3) $r=1$, $\theta=\pi$, 三角形式为$2(\cos\pi+\mathrm{i}\sin\pi)$;\\
(4) $r=5$, $\theta=\pi+\arctan\dfrac{4}{3}$, 三角形式为$5(\cos(\pi+\arctan\dfrac{4}{3})+\mathrm{i}\sin(\pi+\arctan\dfrac{4}{3}))$.

018558
(1) $\cos(-\theta)+\mathrm{i}\sin(-\theta)$; (2) $2(\cos(\pi+\alpha)+\mathrm{i}\sin(\pi+\alpha))$

018559
$-3-\sqrt{3}\mathrm{i}$

018560
B

009660
(1) 是, 理由略; (2) 不是, 理由略; (3) 是, 理由略; (4) 不是, 理由略; (5) 不是, 理由略; (6) 不是, 理由略

009661
(1) $3(\cos 0+\mathrm{i}\sin 0)$; (2) $2(\cos\dfrac{3\pi}{2}+\mathrm{i}\sin\dfrac{3\pi}{2})$; (3) $\sqrt{2}(\cos\dfrac{\pi}{4}+\mathrm{i}\sin\dfrac{\pi}{4})$; (4) $2(\cos\dfrac{2\pi}{3}+\mathrm{i}\sin\dfrac{2\pi}{3})$.

018561
$\cos(\dfrac{\pi}{2}-\alpha)+\mathrm{i}\sin(\dfrac{\pi}{2}-\alpha)$

018562
$\begin{cases}\dfrac{1}{\cos\theta}(\cos\theta+\mathrm{i}\sin\theta), & \theta \in (0,\dfrac{\pi}{2}),\\ -\dfrac{1}{\cos\theta}(\cos(\pi+\theta)+\mathrm{i}\sin(\pi+\theta)), & \theta \in (\dfrac{\pi}{2},\pi)\end{cases}$

018563
$\dfrac{3}{2}+\dfrac{3\sqrt{3}}{2}\mathrm{i}$

018564
$(1-\sqrt{3})-(1+\sqrt{3})\mathrm{i}$

018565
$2\mathrm{i}$

018566
$-1024$

018568
$x=\pm(\sqrt{3}+\mathrm{i})$或$\pm(1-\sqrt{3}\mathrm{i})$

018569
$n$的最小值为$6$, 相应的$z$的值为$-\dfrac{27}{64}$

009662
(1) $16\mathrm{i}$; (2) $3\mathrm{i}$; (3) $-2\sqrt{6}+2\sqrt{2}\mathrm{i}$

009663
$\pm 1$与$\pm \mathrm{i}$