1.1.18 就以下各种情况,分别求arg z:-|||-(a) =dfrac (-2)(1+sqrt {3)i} ;-|||-(b) =dfrac (i)(-2-2i);-|||-(c) =((sqrt {3)-i)}^6.

步骤 1：计算 (a) $z=\dfrac {-2}{1+\sqrt {3}i}$ 的 argz
首先，将分母实部化，即乘以共轭复数：
$$
z=\dfrac {-2}{1+\sqrt {3}i} \cdot \dfrac {1-\sqrt {3}i}{1-\sqrt {3}i} = \dfrac {-2(1-\sqrt {3}i)}{1+3} = \dfrac {-2+2\sqrt {3}i}{4} = \dfrac {-1}{2} + \dfrac {\sqrt {3}}{2}i
$$
然后，计算 argz，即复数的主幅角：
$$
argz = \arctan\left(\dfrac {\sqrt {3}/2}{-1/2}\right) = \arctan(-\sqrt {3}) = \dfrac {2\pi}{3}
$$
步骤 2：计算 (b) $z=\dfrac {i}{-2-2i}$ 的 argz
首先，将分母实部化，即乘以共轭复数：
$$
z=\dfrac {i}{-2-2i} \cdot \dfrac {-2+2i}{-2+2i} = \dfrac {i(-2+2i)}{4+4} = \dfrac {-2i-2}{8} = -\dfrac {1}{4} - \dfrac {1}{4}i
$$
然后，计算 argz，即复数的主幅角：
$$
argz = \arctan\left(\dfrac {-1/4}{-1/4}\right) = \arctan(1) = \dfrac {-3\pi}{4}
$$
步骤 3：计算 (c) $z={(\sqrt {3}-i)}^{6}$ 的 argz
首先，计算复数的主幅角：
$$
arg(\sqrt {3}-i) = \arctan\left(\dfrac {-1}{\sqrt {3}}\right) = -\dfrac {\pi}{6}
$$
然后，计算 $z={(\sqrt {3}-i)}^{6}$ 的 argz：
$$
argz = 6 \cdot arg(\sqrt {3}-i) = 6 \cdot -\dfrac {\pi}{6} = -\pi
$$