共50条信息
从集合\(\left\{0,1,2,3,4,5\right\} \)中任取两个互不相等的数\(a,b\)组成\(a+bi\),其中虚数有( )个
复数\(\dfrac{1-i}{1+i}+1\)的虚部是
已知\(i\)为虚数单位,若复数\(i·z= \sqrt{2}-i \),则\({复数}Z{的虚部是}(\) \()\)
已知\({i}\)为虚数单位,复数\(z={{\left( a-{i} \right)}^{2}}\),\(a\in \mathrm{R}\),若复数\(z\)是纯虚数,则\(\left| z \right|=\)
已知复数\({z}_{1}=\left(2x+1\right)+2i,{z}_{2}=-x-yi\left(x,y∈R\right) \)
\((2)\)若复数\(\left(1+i\right)·{z}_{1} \)为纯虚数,求复数\({{z}_{1}}\)的模\(\left| {{z}_{1}} \right|\)。
已知\(z\)是复数,\(z+2i\),\(\dfrac{z}{2-i}\)均为实数\((i\)为虚数单位\()\),且复数\((z+mi)^{2}\)在复平面上对应的点在第一象限.
\((\)Ⅰ\()\)求复数\(z\);
\((\)Ⅱ\()\)求实数\(m\)的取值范围.
复数\(z\)满足\(\overline{z}\left( 1-i \right)=\left| 1+i \right|\),则复数\(z\)的实部与虚部之和为\((\) \()\)
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