ij˵110000Ԫi=4%ӵ3ĩ7ĩÿȶȡʽÿӦȡ٣

I130II7II01-1IX=I2I求X.大哥大姐,用矩阵的初等变换写~I215II4II130II7II01-1IX=I2I求X.大哥大姐,用矩阵的初等变换写~I215II4II130II7II0

几道高二数学题(5-3i)+(7-5i)-4i(5-3i)+(7-5i)-4i(-2-4i)-(-2+i)+(1+7i)(-2-3i)(-5+i)(1+i)(2+i)(3+i)(3-i)z=4+2i最

复数除法计算1+i/1-i,1/i,7+i/3+4i（-1+i)(2+i)/-i复数除法计算1+i/1-i,1/i,7+i/3+4i（-1+i)(2+i)/-i复数除法计算1+i/1-i,1/i,7+

i为虚数单位,1/i+1/i^3+1/i^5+1/i^7=A.iB.-iC.1+iD.1-i,i为虚数单位,1/i+1/i^3+1/i^5+1/i^7=A.iB.-iC.1+iD.1-i,i为虚数单位

复数5+3i/4-i=1+i复数5+3i/4-i=1+i复数5+3i/4-i=1+i5+3i/4-i=(5+3i)(4+i)/(4-i)(4+i)=(20+12i+5i+3i^2)/(16-i^2)=

计算；（1）,（-8-7i)(-3i)(2),(4-3i)(-5-4i)(3),2i/2-i(4),2+(4).2+i/7+4i计算；（1）,（-8-7i)(-3i)(2),(4-3i)(-5-4i)

复数(4+3i)(2+i)=复数(4+3i)(2+i)=复数(4+3i)(2+i)=解(4+3i)(2+i)=8+6i+4i+3i^2=8+10i-3=5+10i希望对你有帮助学习进步O(∩_∩)O

#defineN20fun(inta[],intn,intm){inti;for(i=m;i>n;i--)a[i+1]=a[i];returnm;}voidmain()#defineN20fun(in

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数学复数的运算(2+i)^3(4

(1+7i)^2=?(3+2i)*(3-2i)*(-4+i)=?根号3i分之1=?8+3i分之5-2i=?1-2i分之1+2i-1+2i分之1-2i=?(1+7i)^2=?(3+2i)*(3-2i)*

Dima,i%a=array(1,2,3,4,5,6,7)Fori=Lbound(a)toUbound(a)a(i)=a(i)*a(i)nextiprinta(i)下标越界了,为什么啊Dima,i%a

i/3+4i=1+i/3-3i=求化简过程,i/3+4i=1+i/3-3i=求化简过程,i/3+4i=1+i/3-3i=求化简过程,前面那个等式应该改是i在分母上吧?在分母上乘以i然后化成一元二次方程

复数方程(1+i)^7(1-i)+(1+4i)/(3-2i)复数方程(1+i)^7(1-i)+(1+4i)/(3-2i)复数方程(1+i)^7(1-i)+(1+4i)/(3-2i)(1+i)^7(1-

[(1-2i)^2/(3-4i)]-[(2+i)^2/(4-3i)]=?[(1-2i)^2/(3-4i)]-[(2+i)^2/(4-3i)]=?[(1-2i)^2/(3-4i)]-[(2+i)^2/(

计算(1-2i)-(2-3i)+(3-4i)-(4-5i)=计算(1-2i)-(2-3i)+(3-4i)-(4-5i)=计算(1-2i)-(2-3i)+(3-4i)-(4-5i)=(1-2i)-(2-

(1+i)^7/1-i+(1-i)^7/1+i-(3-4i)(2+2i)^3/4+3i（要过程,麻烦你们帮帮忙!）(1+i)^7/1-i+(1-i)^7/1+i-(3-4i)(2+2i)^3/4+3i

#includemain(){inti=1;while(!((i%2=#includemain(){inti=1;while(!((i%2==1)&&(i%3==2)&&(i%5==4)&&(i%6=

已知i为虚数单位,则(4+2i)/(1-i)=（）A.1+3iB.1-3iC.3-iD.3+i已知i为虚数单位,则(4+2i)/(1-i)=（）A.1+3iB.1-3iC.3-iD.3+i已知i为虚数

试求i^1,i^2,i^3,i^4,i^5,i^6,i^7,i^8i是虚数由题目推测出i^n的值的规律,用式子表达试求i^1,i^2,i^3,i^4,i^5,i^6,i^7,i^8i是虚数由题目推测出

i+2i^2+3i^3+4i^4+…+2010i^2010=?要有步骤``i+2i^2+3i^3+4i^4+…+2010i^2010=?要有步骤``i+2i^2+3i^3+4i^4+…+2010i^2