Question

已知z=[1+i+√3(1-i)]/(4+4i),求z^2+1/z的值

Answer 1

z=[(1+√3)+(1-√3)i]/(4+4i)

z^2=[(4+2√3)+2(1-3)i-(4-2√3)]/(16+32i-16)

=(4√3-4i)/(32i)

=(√3-i)/(8i)

=-(1+√3i)/8

1/z=(4+4i)/[(1+√3)+(1-√3)i]

=(4+4i)[(1+√3)-(1-√3)i]/[(1+√3)+(1-√3)i][(1+√3)-(1-√3)i]

=[(4+4√3)+(4-4√3)+(4+4√3)i-(4-4√3)i]/[(4+2√3)+(4-2√3)]

=(8+8√3i)/8

所以原式=(7+7√3i)/8