共50条信息
设\(i\)为虚数单位，则\((\dfrac{1{+}i}{i})^{2014}\)等于\(({ })\)
若\(n\)是大于\(2000\)的奇数，则复数\({{\left( \dfrac{{1}+{i}}{{1}-{i}} \right)}^{2n}}+{{\left( \dfrac{{1}-{i}}{{1}+{i}} \right)}^{2n}}\)的值是 \((\) \()\)
\({{{i}}^{{2018}}}\)的值为\((\) \()\)
设复数\(x=\dfrac{2i}{1-i}(i\)是虚数单位\()\)，则\(C_{2018}^{1}x+C_{2018}^{2}{x}^{2}+C_{2018}^{3}{x}^{3}+…+C_{2018}^{2018}{x}^{2018}= \)
设\(z= \dfrac{1}{2}+ \dfrac{ \sqrt{3}}{2}i(i\)是数单位\()\)，则\(z+2z^{2}+3z^{3}+4z^{4}+5z^{5}+6z^{6}=(\) \()\)
复数\(x < 0\)等于 \((\) \()\)
复数 \( \dfrac{2- \sqrt{3i}}{i} \) \((i\)为虚数单位\()\)的虚部是\((\) \()\)
\((1)\)计算\({{\left[ (1+2i)\cdot {{i}^{100}}+{{(\dfrac{1-i}{1+i})}^{5}} \right]}^{2}}-{{(\dfrac{1+i}{\sqrt{2}})}^{20}}\)
\((2)\)已知\(z\)，\(ω \)为复数，\((1+3i)·z\)为纯虚数，\(ω= \dfrac{z}{2+i} \)，且\(|ω|=5 \sqrt{2} \)，求复数\(z\)．
进入组卷