If 'z, lies on the circle `|z-2i|=2sqrt2`, then the value of `arg((z-2)/(z+2))` is the equal to

Options

A. `(pi)/(3)`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(pi)/(2)`

Answer

B

Explanation

`CA = CB = 2sqrt(2), OC = 2`
` rArr OA = OB = 2`
` rArr A -= 2+ )I, B = - 2 + 0i, B = - 2 + 0i`

Clearly,
`/_BCP= pi//2`
` rArr /_ BPA = pi//4`
`rArr arg ((z-2)/(z+ 2)) = (pi)/(4)`