共50条信息
在复平面内，复数\(z=\dfrac{1}{2+i}+{{i}^{2018}}\)对应的点位于\((\) \()\)
设\(f(n)={\left( \dfrac{1+i}{1-i}\right)}^{n}+{\left( \dfrac{1-i}{1+i}\right)}^{n}\left(n∈N\right) \)，则集合\(\left\{x \left|x=f(n) \right.\right\} \)的子集个数是 ．
已知\(i\)为虚数单位，复数\(z\)满足\(z(1-i)=1+i\)，则\({{z}^{2017}}=\) \((\) \()\)
\(\left( \left. \dfrac{1+i}{1-i} \right. \right)^{6} + \dfrac{ \sqrt{2}+ \sqrt{3}i}{ \sqrt{3}- \sqrt{2}i}=\)_______．
\({{{i}}^{{2018}}}\)的值为\((\) \()\)
设复数\(x=\dfrac{2i}{1-i}(i\)是虚数单位\()\)，则\(C_{2018}^{1}x+C_{2018}^{2}{x}^{2}+C_{2018}^{3}{x}^{3}+…+C_{2018}^{2018}{x}^{2018}= \)
\(i\)为虚数单位，则\({{(\dfrac{1+i}{1-i})}^{2018}}=\)_______
复数\(x < 0\)等于 \((\) \()\)
计算\((1)\dfrac{2+2i}{1-i}+{{\left( \dfrac{\sqrt{2}}{1+i} \right)}^{2016}}\) \((2)\)计算 \(\int_{-2}^{0}{\sqrt{4-{{x}^{2}}}}dx\)
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