Question

已知复数Z满足IZI=1那么I X-1+i I=多少?

Answer 1

z=a+bi

|z|=√(a^2+b^2)=1

a^2+b^2=1

|z-1+i|=|a+bi-1+i|=|a-1+(b+1)i|=√[(a-1)^2+(b+1)^2

=√(a^2-2a+1+b^2+2b+1)

=√(a^2+b^2-2a+2b+2)

=√(1-2a+2b+2)

=√(-2a+2b+3)