If `|z-1|=1`, then

Options

A. `arg((z-1-i)//z)` can be equal to `-pi//4`
B. `(z-2)//z` is purely imaaginary number
C. `(z-2)//z` is purely real number
D. if `arg(z) = theta,` where `z ne 0` and `theta` is acute, then `1-2//z=itan theta`

Answer

A, B, D

Explanation

Since `arg ((z-1-i)//z)` is the angle substended by the chord joining the point O and `1+i` at the cuircumference of the circle `|z-1|=1` , so it is equal to`-pi//4`. The line joining the points z=0 and z=2 +0i is the diameter.
`arg(z-2)/(z) = pm (pi)/(2)`
`rArr (z-2)/(z-0)`is purely imaginary
We have
`/_OPA = (pi)/(2)`
`rArr arg ((2-z)/(0-z)) = (pi)/(2) rArr (z-2)/(z) = (AP)/(OP)i`
Now in `DeltaOAP`
`tan theta = (AP)/(OP)`
Thus , `(z-2)/(z) = i tan theta`