解答
展开 (1+i1−i−1+i,−1−i)⋅(3−4i6−2i1+2i4+3i)
解答
1⋅3−1⋅4⋅6i−1⋅2⋅1i+1⋅2⋅4i+1⋅3i+1⋅3i−1⋅4⋅6ii−1⋅2⋅1ii+1⋅2⋅4ii+1⋅3ii−3i+4⋅6ii+2⋅1ii−2⋅4ii−3ii−1⋅3+1⋅4⋅6i+1⋅2⋅1i−1⋅2⋅4i−1⋅3i+i,3−i,4⋅6i−i,2⋅1i+i,2⋅4i+i,3i−1⋅3+1⋅4⋅6i+1⋅2⋅1i−1⋅2⋅4i−1⋅3i−3i+4⋅6ii+2⋅1ii−2⋅4ii−3ii
求解步骤
(1+i1−i−1+i,−1−i)(3−4i6−2i1+2i4+3i)
=(1+1i−i−1+i,−1−i)(3−4⋅6i−2⋅1i+2⋅4i+3i)
打开括号=1⋅3+1⋅(−4i6)+1⋅(−2i1)+1⋅2i4+1⋅3i+i1⋅3+i1⋅(−4i6)+i1⋅(−2i1)+i1⋅2i4+i1⋅3i+(−i)⋅3+(−i)(−4i6)+(−i)(−2i1)+(−i)⋅2i4+(−i)⋅3i+(−1)⋅3+(−1)(−4i6)+(−1)(−2i1)+(−1)⋅2i4+(−1)⋅3i+i,3+i,(−4i6)+i,(−2i1)+i,2i4+i,3i+(−1)⋅3+(−1)(−4i6)+(−1)(−2i1)+(−1)⋅2i4+(−1)⋅3i+(−i)⋅3+(−i)(−4i6)+(−i)(−2i1)+(−i)⋅2i4+(−i)⋅3i
使用加减运算法则+(−a)=−a,(−a)(−b)=ab=1⋅3−1⋅4⋅6i−1⋅2⋅1i+1⋅2⋅4i+1⋅3i+1⋅3i−1⋅4⋅6ii−1⋅2⋅1ii+1⋅2⋅4ii+1⋅3ii−3i+4⋅6ii+2⋅1ii−2⋅4ii−3ii−1⋅3+1⋅4⋅6i+1⋅2⋅1i−1⋅2⋅4i−1⋅3i+i,3−i,4⋅6i−i,2⋅1i+i,2⋅4i+i,3i−1⋅3+1⋅4⋅6i+1⋅2⋅1i−1⋅2⋅4i−1⋅3i−3i+4⋅6ii+2⋅1ii−2⋅4ii−3ii