[(4-3i)^2(3-i)^4/(1-2i)^4]的模=
复数除法 计算1+i/1-i,1/i,7+i/3+4i (-1+i)(2+i)/-i
[(4-3i)^2(3-i)^4/(1-2i)^4]的模=
1+i-2i^2+3i^3-4i^4+5i^5
计算(1+2i)+(2-3i)+(3+4i)+(4-5i)+...+(2008-2009i)
[(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
复数(4+3i)(2+i)=
(1)i/1+i(2)2/(1+i)^2(3)(3-i)/(3+4i)(4)(3-4i)(1+2i)/2i
[(1-√3i)^5-(1+√3i)^4]/[i*(-1+i^8)*(1/2+1/2i)]的答案是2√3-4i,
int i=4;则表达式(i==2)?i:i-1的值是多少,
数学 复数的运算(2+i)^3(4-2i) / 5i(1+i)=1-7i 怎么算的 求详细过程