作业帮(4-3i)/(4+3i)+(3+4i)/(3-4i)等于
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/15 09:39:55
x͏A EL(Sh҄"uc7װV.\ϼ<: ɵ+T y?m9)y&5b-i\ ;f CBh&)=Zy
s] H̍HS$
zup\=8G[@6!. $k7r
(4-3i)/(4+3i)+(3+4i)/(3-4i)等于
(4-3i)/(4+3i)+(3+4i)/(3-4i)等于
(4-3i)/(4+3i)+(3+4i)/(3-4i)等于
(4-3i)/(4+3i)+(3+4i)/(3-4i)
= (4-3i)(4-3i)/[(4+3i)(4-3i)] + (3+4i)(3+4i)/[(3+4i)(3-4i)]
= (4-3i)(4-3i)/25 + (3+4i)(3+4i)/25
= (25- 24i)/25 + (25+24i)/25
= 50/25
= 2
等于0
(4-3i)/(4+3i)+(3+4i)/(3-4i)
= (4-3i)(4-3i)/[(4+3i)(4-3i)] + (3+4i)(3+4i)/[(3+4i)(3-4i)]
= (4-3i)(4-3i)/25 + (3+4i)(3+4i)/25
= (16-9- 24i)/25 + (9-16+24i)/25
= 0
[(1-2i)^2/(3-4i)]-[(2+i)^2/(4-3i)] =[1-4-4i)/(3-4i)]-[(4-1+4i)/(4-3i)] =-(3+4i)/(3-4i)-(3+4i)/(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
计算（1+2i)+(2-3i)+(3+4i)+(4-5i)+...+(2008-2009i)
1+i-2i^2+3i^3-4i^4+5i^5
(i-2)(1-2i)(3+4i)计算过程
计算 (1-2i)(2+i)(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)
计算；（1）,（-8-7i)(-3i) (2),(4-3i)(-5-4i) (3),2i/2-i (4),2+(4).2+i/7+4i
计算(1-2i)(3+4i)
计算(1+2i)/(3-4i)
求解(-3+2i)-(4-5i)
|(3+2i)-(4-i)|等于
复数(4+3i)(2+i)=
2+3i分之4-i