(8-3i)-(4-5i)=
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(8-3i)-(4-5i)=
(8-3i)-(4-5i)=
(8-3i)-(4-5i)=
(8-3i)-(4-5i)=4-2i
4+2i。。。
4+8i
4 2i
(8-3i)-(4-5i)=8-3i-4+5i=4+2i
(8-3i)-(4-5i)=
计算;(1),(-8-7i)(-3i) (2),(4-3i)(-5-4i) (3),2i/2-i (4),2+(4).2+i/7+4i
复数计算:(1)i+i^2+i^3+.+i^100(2)i^10+i^20+i^30+.+i^80(3)i*i^2*i^3*.*i^100(4)i*i^3*i^5*.*i^99(5)[(1+i)/(1-i)])[(1+i)/(1-i)]^2)[(1+i)/(1-i)]^3.)[(1+i)/(1-i)]^100
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
1+i-2i^2+3i^3-4i^4+5i^5
计算(1+2i)+(2-3i)+(3+4i)+(4-5i)+...+(2008-2009i)
[(1-√3i)^5-(1+√3i)^4]/[i*(-1+i^8)*(1/2+1/2i)]的答案是2√3-4i,
复数5+3i/4-i=1+i
(1-3i)-(2+5i)+(-4+9i)等于?
I 1 3 0 I I 7 I I 0 1 -1 I X= I 2 I求X.大哥大姐,用矩阵的初等变换写~I 2 1 5 I I 4 I
(2-5i)(5-2i)-(10-17i);;;;;;(4+3i)^2-(2-i)(2+i^5)
计算(4-3i)(-5-4i)
求解(-3+2i)-(4-5i)
#define N 20 fun(int a[],int n,int m) {int i; for(i=m;i>n;i--)a[i+1]=a[i]; return m; } void main() #define N 20fun(int a[],int n,int m){int i;for(i=m;i>n;i--)a[i+1]=a[i];return m;}void main(){ int i,a[N]={1,2,3,4,5,6,7,8,9,10};fun(a,0,N/2);for(i=0;i
计算((根号2+根号2 i)^3(4+5i))/((5-4i)(1-i))
计算(1-2i)-(2-3i)+(3-4i)-(4-5i)=
计算(3-4i)/(1+2i)+iˆ15-(1+i)ˆ8
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 最好 有比较清楚的过程