If `z` is a complex number having least absolute value and `|z-2+2i=1,t h e nz=` `(2-1//sqrt(2))(1-i)` b. `(2-1//sqrt(2))(1+i)` c. `(2+1//sqrt(2))(1-i)""` d. `(2+1//sqrt(2))(1+i)`

Options

A. `(2-1//sqrt(2))(1-i)`
B. `(2-1//sqrt(2))(1+i)`
C. `(2+1//sqrt(2)(1-i)`
D. `(2+1//sqrt(2))(1+i)`

Answer

A

Explanation

We have,
`|z-2+ 2i|= 1`
`rArr |z- (2-2i)|=1`

Hence, z lies on a circle having centre at (2,-2) and radius 1.
It is evident form the figure that the required complex number z is given by the point P.
We find the OP makes an angle `pi//4` with OX and
`OP = OC- CP`
` = sqrt(2^(2) + 2^(2)) - 1 = 2sqrt(2)-1`
So, coordinates of P are `[(2sqrt(2)-1)cos (pi//4),-(2sqrt(2)-1)sin(pi//4)]`i.e.,`((2-1//sqrt(2)), -(2-1//sqrt(2)))`
Hence,
`z = (2-(1)/(sqrt(2)))+{(2-(1)/(sqrt(2)))}i = (2-(1)/(sqrt(2)))(1-i)`