Question

已知z=1+i,w=(2-i)z-2

Answer 1

w=(2-i)z-2 = (2-i)*(1+i) - 2 = 2-i+2i+1-2=1+i

1. |w| = √2

2.

w的共厄复数／w = (1-i) / (1+i) = (1-i)(1-i) / (1+1) = (1-1-2i) / 2 = -i

az+b=a(1+i)+b=a+b+ai=-i

得 a=-1 a+b=0,b=1

因此 a=-1 b=1