Question

复数z=x+yi,(x属于R,y属于R）且|z|=1,复数w=2z-3+4i,|w|的取值范围是

Answer 1

w=2z-3+4i

=2(x+yi)-3+4i

=2x+2yi-3+4i

=(2x-3)+(2y+4)i

|w|^2=4x^2-12x+9+4y^2+16y+16

=-12x+16y+29

=4(-3x+4y)+29

x^2+y^2=1

求-3x+4y极限

设-3x+4y=t

y=0.75x+t/4

数形结合.

t=±5

|w|^2=>4*5+29=49

=>4*(-5)+29=9

所以,|w|的取值范围是.[3,7]

A