If z lies on the circle |z−1|=1, then z−2z equals
None of these
Note that |z−1|=1 represents a circle with the segment joining z=0 and z=2+0i as a diameter.
See the figure. If z lies on the cirlce, z−2z−0 is purley imaginary.
Alternatively, assume z=x+iy
Then (x−1)2+y2=1
⇒x2+y2−2x=0.
The complex number z−2z
=(x−2)+iyx+iy=((x−2)+iy)(x−iy)x2−y2=x2−2x−(x−2)iy−i2y2x2−y2=x2+y2−2x−(x−2)iyx2−y2=−(x−2)iyx2−y2
This is a purely imaginery complex number.