The set `{R e((2i z)/(1-z^2)): zi sacom p l e xnu m b e r ,|z|=1,z=+-1}` is________.

Options

A. `(-infty, -1)(1,infty)`
B. `(-infty,0) uu (1,infty)`
C. `[2,infty)`
D. `(-infty,-1) uu [1,infty)`

Answer

D

Explanation

Let `z=x+iy`. Then,
`(2iz)/(1-z^(2))=(2i(x+iy))/(1-(x+iy)^(2))=(-2y+2ix)/(1-(x^(2)-y^(2)+2ixy))`
`rArr (2iz)/(1-z^(2))=-1/y`
`therefore {"Re"(2iz)/(1-z^(2)):z` is a complex number, `|z|=1, z ne +-1}`
`={-1/y:x^(2)+y^(2)=1, x ne +-1}={+-1/sqrt(1-x^(2)): x ne +-1}`
`=(-infty, -1) uu (1,infty)`