(4-3i)/(4+3i)+(3+4i)/(3-4i)等于
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(4-3i)/(4+3i)+(3+4i)/(3-4i)等于
![(4-3i)/(4+3i)+(3+4i)/(3-4i)等于](/uploads/image/z/19190874-66-4.jpg?t=%284-3i%29%2F%284%2B3i%29%2B%283%2B4i%29%2F%283-4i%29%E7%AD%89%E4%BA%8E)
(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
= (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
复数除法 计算1+i/1-i,1/i,7+i/3+4i (-1+i)(2+i)/-i
复数(-1+根号3i)^4/(1-i)^8等于
1+i-2i^2+3i^3-4i^4+5i^5
计算(1+2i)+(2-3i)+(3+4i)+(4-5i)+...+(2008-2009i)
已知i为虚数单位 则复数4-3i/(2+i)^2等于
复数5+3i/4-i=1+i
几道高二数学题(5-3i)+(7-5i)-4i
复数(4+3i)(2+i)=
复数 5+3i/4-i
(1)i/1+i(2)2/(1+i)^2(3)(3-i)/(3+4i)(4)(3-4i)(1+2i)/2i
关于虚数、复数的问题:i+i²+i的3次方+i的4次方+.+i的2011次方 等于多少呢?
像3+4i