已知复数=dfrac (2-2i)(1+sqrt {3)i},则=dfrac (2-2i)(1+sqrt {3)i}=dfrac (2-2i)(1+sqrt {
复数=dfrac ((sqrt {3)+i)(2-2i)}((sqrt {3)-i)(2+2i)}的三角形式为( )=dfrac ((sqrt {3)+i)(2
A . 0 B . =-dfrac (1-5i)(2+3i) C . =-dfrac (1-5i)(2+3i)D . =-dfrac (1-5i)(2+3i)设
1.设z=(1-sqrt(3)i)/(2),求|z|及Arg z.1.设$z=\frac{1-\sqrt{3}i}{2}$,求|z|及Arg z.
_(1)=5-5i, () _(2)=-3+4i 罒 (dfrac ({z)_(1)}({z)_(2)}) (整体取共-|||-A -dfrac (7)(5)-
1.2.1 求下面各题的所有的根、单根,并说明几何意义:-|||-(a) ((2i))^dfrac (1{2)};-|||-(b) ((1-sqrt {3)i)
1.求下列复数z的实部与虚部、共轭复数、模与辐角:-|||-(1) 1/(3+2i);-|||-(2) dfrac (1)(i)-dfrac (3i)(1-i)
dfrac ({mu )_(0)I}(pi R)(dfrac (1)(2)+dfrac (pi )(6))-|||-dfrac ({mu )_(0)I}(pi
求复数=dfrac (2i)(1-i)+dfrac (2(1-i))(i)的模与主辐角。求复数的模与主辐角。
7.计算下列复数的值:-|||-(1) _(1)=((dfrac {-1+sqrt {3)i}(2))}^6;