解答
展开 (4+6i)5(8−2i)10
解答
37129313035060672−i37129324424909472
求解步骤
(4+6i)5(8−2i)10
(4+6i)5=3904−19104i
=3904−19104i(8−2i)10
(8−2i)10=−927506432i+810−2193409024
=3904−19104i(−927506432i+810−2193409024)
使用复数算数法则: c+dia+bi=(c−di)(c+di)(c−di)(a+bi)=c2+d2(ac+bd)+(bc−ad)ia=810−2193409024,b=−927506432,c=3904,d=−19104=39042+(−19104)2((810−2193409024)⋅3904+(−927506432)(−19104))+(−927506432⋅3904−(810−2193409024)(−19104))i
整理后得=380204032(810⋅3904+9156014047232)+(810⋅19104−4.55239E13)i
化简 380204032(810⋅3904+9156014047232)+(810⋅19104−4.55239E13)i:380204032810⋅3904+9156014047232+i(810⋅19104−4.55239E13)
=380204032810⋅3904+9156014047232+i(810⋅19104−4.55239E13)
乘开 810⋅3904+9156014047232+i(810⋅19104−4.55239E13):810⋅3904+9156014047232+810⋅19104i−4.55239E13i
=380204032810⋅3904+9156014047232+810⋅19104i−4.55239E13i
使用分式法则: ca±b=ca±cb380204032810⋅3904+9156014047232+810⋅19104i−4.55239E13i=380204032810⋅3904+3802040329156014047232+380204032810⋅19104i−3802040324.55239E13i=380204032810⋅3904+3802040329156014047232+380204032810⋅19104i−3802040324.55239E13i
380204032810⋅3904=3712934093640704
3802040329156014047232=28561687801536
380204032810⋅19104i=37129320031995904i
3802040324.55239E13i=285613419761952i
=3712934093640704+28561687801536+37129320031995904i−285613419761952i
将 3712934093640704+28561687801536+37129320031995904i−285613419761952i 改写成标准复数形式:37129313035060672−37129324424909472i
=37129313035060672−37129324424909472i