解答
化简 (1−3i)16
解答
−32768+327683i
求解步骤
(1−3i)16
使用二项式定理: (a+b)n=i=0∑n(in)a(n−i)bia=1,b=−3i
=i=0∑16(i16)⋅1(16−i)(−3i)i
展开求和
=0!(16−0)!16!⋅116(−3i)0+1!(16−1)!16!⋅115(−3i)1+2!(16−2)!16!⋅114(−3i)2+3!(16−3)!16!⋅113(−3i)3+4!(16−4)!16!⋅112(−3i)4+5!(16−5)!16!⋅111(−3i)5+6!(16−6)!16!⋅110(−3i)6+7!(16−7)!16!⋅19(−3i)7+8!(16−8)!16!⋅18(−3i)8+9!(16−9)!16!⋅17(−3i)9+10!(16−10)!16!⋅16(−3i)10+11!(16−11)!16!⋅15(−3i)11+12!(16−12)!16!⋅14(−3i)12+13!(16−13)!16!⋅13(−3i)13+14!(16−14)!16!⋅12(−3i)14+15!(16−15)!16!⋅11(−3i)15+16!(16−16)!16!⋅10(−3i)16
化简 0!(16−0)!16!⋅116(−3i)0:1
化简 1!(16−1)!16!⋅115(−3i)1:−163i
化简 2!(16−2)!16!⋅114(−3i)2:360i2
化简 3!(16−3)!16!⋅113(−3i)3:−16803i3
化简 4!(16−4)!16!⋅112(−3i)4:16380i4
化简 5!(16−5)!16!⋅111(−3i)5:−393123i5
化简 6!(16−6)!16!⋅110(−3i)6:216216i6
化简 7!(16−7)!16!⋅19(−3i)7:−3088803i7
化简 8!(16−8)!16!⋅18(−3i)8:1042470i8
化简 9!(16−9)!16!⋅17(−3i)9:−9266403i9
化简 10!(16−10)!16!⋅16(−3i)10:1945944i10
化简 11!(16−11)!16!⋅15(−3i)11:−10614243i11
化简 12!(16−12)!16!⋅14(−3i)12:1326780i12
化简 13!(16−13)!16!⋅13(−3i)13:−4082403i13
化简 14!(16−14)!16!⋅12(−3i)14:262440i14
化简 15!(16−15)!16!⋅11(−3i)15:−349923i15
化简 16!(16−16)!16!⋅10(−3i)16:6561i16
=1−163i+360i2−16803i3+16380i4−393123i5+216216i6−3088803i7+1042470i8−9266403i9+1945944i10−10614243i11+1326780i12−4082403i13+262440i14−349923i15+6561i16
化简 1−163i+360i2−16803i3+16380i4−393123i5+216216i6−3088803i7+1042470i8−9266403i9+1945944i10−10614243i11+1326780i12−4082403i13+262440i14−349923i15+6561i16:327683i−32768
=327683i−32768
Rewrite in standard complex form: −32768+327683i
=−32768+327683i