共50条信息
已知\(x\)是纯虚数,\(y\)是实数,且\(2x-1+i=y-(3-y)i\),则\(x+y\)等于
若\(m+i=\left( 1+2i \right)\cdot ni(m,n\in R,i\)是虚数单位\()\),则\(n-m\)等于\((\) \()\)
实数\(m\)为何值时,复数\(z=({{m}^{2}}-8m+15)+({{m}^{2}}-5m)i\)是:
\((1)\)纯虚数; \((2)\)等于\(3+6i\); \((3)\)所对应的点在第四象限.
复数\(4-3a-{a}^{2}i \)与复数\({a}^{2}+4ai \)相等,则实数\(a\)的值为__________.
设复数\(z= \dfrac{{\left(1+i\right)}^{2}+3\left(1-i\right)}{2+i} \),若\(z^{2}+az+b=1+i\),求实数\(a\),\(b\)的值
\((1)\)已知复数\(z\)满足\(z+\left| z \right|=3+i,\)求复数\(z\);
\((2)\)计算:\({{\left( \dfrac{1+{i}}{1-{i}} \right)}^{4}}+\dfrac{\sqrt{2}+\sqrt{3}{i}}{\sqrt{3}-\sqrt{2}{i}}\).
\({z}_{1}={m}^{2}-3m+{m}^{2}i,{z}_{2}=4+\left(5m+6\right)i \),其中\(m\)为实数,\(i\)为虚数单位,若\({{z}_{1}}-{{z}_{2}}=0\),则\(m\)的值为\((\) \()\)
对于复数\(z=a+bi(a,b\in R)\),若\(z+i=\dfrac{2-i}{1+2i}\),则\(b=\)_________.
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