Given `z=f(x)+ig(x)` where `f,g:(0,1) to (0,1)` are real valued functions. Then which of the following does not hold good?
a.`z=1/(1-i x)+i1/(1+i x)`
b. `z=1/(1+i x)+i1/(1-i x)`
c. `z=1/(1+i x)+i1/(1+i x)`
d. `z=1/(1-i x)+i1/(1-i x)`

Options

A. `z = (1)/(1-ix) + i((1)/(1+ix))`
B. `z =(1)/(1+ix) +i((1)/(1-ix))`
C. `z= (1)/(1+ix)+i((1)/(1+ix))`
D. `z = (1)/(1-ix)+i((1)/(1-ix))`

Answer

A, C, D

Explanation

Choice (a) on simplification given
`z= (1+x)/(1+x^(2)) + i(1+x)/(1+x^(2))`
For x = 0.5. `( 0.5) gt 1` which is out of range , Hence ,(1) is not
correct . Form choice (20,
`z = (1-x)/(1+x^(2)) + i (1-x)/(1+x^(2))`
f(x) and g (x) `in (0,1) if x in (0,1)`. Hence (2) is correct. From choice (3),
`z = (1+ x)/(1+x^(2)) + (1-x)/(1+x^(2))i`
Hence, (3) is not correct. From choice (4),
`z= (1-x)/(1+x^(2)) + (1+x)/(1+x^(2))i`
Hence, (4) is not correct.