Question

设复数z=(a-i)/(1+i),a属于R,当复数u=z(z+2i)的虚部与实部之差取到最大值时,求|z+1|的值.

Answer 1

z=(a-i)/(1+i)=（a-i)(1-i)/2=[a-1-(a+1)i]/2,

z+2i=[a-1+(3-a)i]/2,

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

=[(a-1)^+(a+1)(3-a)+(a-1)(2-2a)i]/4

=[4-2(a-1)^i]/4,

依题意[-2(a-1)^-4]/4取到最大值时a=1,z=-i,

∴|z+1|=|1-i|=√2.