{(3-4i)(2+2i)^3}/(4+3i)=?
复数除法 计算1+i/1-i,1/i,7+i/3+4i (-1+i)(2+i)/-i
复数(4+3i)(2+i)=
1+i-2i^2+3i^3-4i^4+5i^5
计算(1+2i)+(2-3i)+(3+4i)+(4-5i)+...+(2008-2009i)
i^0!+i^1!+i^2!+i^3!+...+i^100!
[(1-2i)^2/(3-4i)]-[(2+i)^2/(4-3i)]=?
计算(1-2i)-(2-3i)+(3-4i)-(4-5i)=
已知i为虚数单位,则(4+2i)/(1-i)=( ) A.1+3i B.1-3i C.3-i D.3+i
i+2i^2+3i^3+4i^4+…+2010i^2010=?要有步骤``
(1)i/1+i(2)2/(1+i)^2(3)(3-i)/(3+4i)(4)(3-4i)(1+2i)/2i
设i为虚数单位,则1+i+i^2+i^3+.+i^10=
设i为虚数单位,则1+i+i^2+i^3+A+i^10=?