高二数学复数的四则运算计算(1i)^7/(1-i)(1-i)^7/(1i)-(3-4i)(计算(1i)^7/(1-i)(1-i)^7/(1i)-(3-4i)(22i)^3/(43i)

Question

高二数学复数的四则运算计算(1 i)^7/(1-i) (1-i)^7/(1 i)-(3-4i)(
计算(1 i)^7/(1-i) (1-i)^7/(1 i)-(3-4i)(2 2i)^3/(4 3i)

Answer 1

先约分
原式=(1+i)^6(1-i)^6-(3-4i)[2^3(1+i)^2(1+i)](4-3i)/(4+3i)(4-3i)
=[(1+i)(1-i)]^6-[(3-4i)(4-3i)[8(1+2i-1)(1+i)]/(16+9)
=(1+1)^6-[(12-9i-16i-12)[16i(1+i)]/25
=64-(-25i)[16i(1+i)]/25
=64-16(1+i)
=48-16i