Question

已知复数z=(-1+3i)*(1-i)/i-(1+3i)/i,w=z+ai(a属于R),当|w/z|≤根号2时,求a的取

Answer 1

z=(i^2+3i)*(1-i)/i-(3i-i^2)/i

=(i+3)*(1-i)-(3-i)

=1-i

==>w=z+ai=1+(a-1)i

==>|w|^2=a^2-2a+2,|z|^2=2

|w/z|^2≤1/2

==>a^2-2a+2≤1

==>(a-1)^2≤0

==>a=1