共50条信息
若复数\(z\)满足\(\left( 1+i \right)z=1-2i\),则复数\(z\)的虚部为( )
已知复数\(z=\dfrac{2}{-1+i}\),则
已知\(\dfrac{x}{1-i}=1-yi\),其中\(x\),\(y\)是实数,\(i\)是虚数单位,则\(x+yi\)的共轭复数为\((\) \()\)
设\(z_{1}\),\(z_{2}\)是复数,则下列命题中的假命题是\((\) \()\)
当实数\(m\)为何值时,复数\(z=(m\)\({\,\!}^{2}\)\(+m-2)+(m\)\({\,\!}^{2}\)\(-1)i\)是:
\((\)Ⅰ\()\)实数; \((\)Ⅱ\()\)虚数; \((\)Ⅲ\()\) 纯虚数.
复数\( \dfrac{5i}{1+2i} \)的虚部是( )
复数\(z= \dfrac{1-3i}{1+2i} \),则( )
若复数 \(z \) 满足\(\left(3-4i\right)z=\left|4+3i\right| \),则\(z \)的虚部为\((\) \()\)
设\(i\)为虚数单位,则复数\(z=\dfrac{1{+}2i}{i}\)的虚部为
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