第1个回答 2012-06-25
10、(1)√2﹙cosπ/4+isinπ/4﹚×2﹙cosπ/3+isinπ/3﹚÷﹙cosπ/6+isinπ/6﹚
=√2﹙√2/2+√2i/2﹚×2﹙1/2+√3i/2﹚÷﹙√3/2﹢i/2﹚
=2﹙1+i﹚﹙1+√3i﹚/﹙√3+i﹚
=2﹙1+√3i+i+√3i²﹚/﹙√3+i﹚
=2[1-√3+﹙√3+1﹚i]/﹙√3+i﹚
=√3-1+﹙√3+1﹚i
题目太长我就不抄一遍了,下面五次方六次方你将就看吧
﹙2﹚=[2﹙√2/2+√2i/2﹚﹚]^5×[√3﹙√3/2+i/2﹚]^6÷[√6﹙0+i﹚]^2
=﹙√2+√2i﹚^5×﹙3/2+√3i/2﹚^6÷6i^2
=4√2﹙1+i﹚^5×﹙3+√3i﹚^6/64÷﹙﹣6﹚
=﹣√2﹙1+i﹚^5×﹙3+√3i﹚^6/96
=﹣√2﹙1+2i+i^2﹚^3×﹙9+6√3i+3i^2﹚^4/96
=﹣√2×8i^3×﹙6+6√3i﹚^4/96
=√2i﹙36+72i+108i^2﹚^2/12
=√2i﹙﹣72+72i﹚^2/12
=√2i﹙5184-10368i+5184i^2﹚/12
=-10368√2i^2/12
=864√2
(3)=[﹙√3+i﹚^6+﹙√3-i﹚^6]/64
=[﹙3+2√3i+i^2﹚^4+﹙3-2√3i+i^2﹚^4]/64
=[﹙4+8√3i+4i^2﹚^2+﹙4-8√3i+4i^2﹚^2]/64
=﹙192i^2+192i^2﹚/64
=-6
11、
(1)x²+7
=x²-7i²
=﹙x+√7i﹚﹙x-√7i﹚
(2)2x²-6x+5
=﹙x²-4x+4﹚+﹙x²-2x+1﹚
=﹙x-2﹚²+﹙x-1﹚²
=﹙x-2﹚²-﹙x-1﹚²i²
=[x-2+﹙x-1﹚i][﹙x-2-﹙x-1﹚i]
12、ω=-1/2+√3i/2
(1)1+ω+ω²
=1+﹙-1/2+√3i/2﹚+﹙-1/2+√3i/2﹚²
=1-1/2+√3i/2+1/4-√3i/2+3i²/4
=1-1/2+√3i/2+1/4-√3i/2-3/4
=0
﹙2﹚﹙1+ω-ω²﹚³-﹙1-ω+ω²﹚³
=﹙1+ω+ω²-2ω²﹚³-﹙1+ω+ω²-2ω﹚³
=﹙-2ω²﹚³-﹙-2ω﹚³
=-8[﹙-1/2+√3i/2 ﹚²]³+8﹙-1/2+√3i/2﹚³
=-8﹙1/4-√3i/2+3i²/4﹚³+8﹙-1/2+√3i/2﹚³
=8﹙1/2+√3i/2﹚³+8﹙-1/2+√3i/2﹚³
=8[﹙1/2+√3i/2﹚³+﹙-1/2+√3i/2﹚³]
=8﹙1/2+√3i/2-1/2+√3i/2﹚[﹙1/2+√3i/2﹚²-﹙1/2+√3i/2﹚﹙-1/2+√3i/2﹚+﹙-1/2+√3i/2﹚²
=8√3i[﹙1/4+√3i/2+3i²/4﹚-﹙-1/4+3i²/4﹚+﹙1/4-√3i/2+3i²/4﹚]
=8√3i×0
=0
补充说明:^后面加数字表示次方,^4四次方,^5五次方,^6六次方。
i²=﹣1
立方和公式:
a³﹢b³=﹙a﹢b﹚﹙a²-ab﹢b²﹚