Question

Given $z=f\left(x\right)+ig\left(x\right)$ where f, $g:(0,1)\to (0,1)$ are real valued functions. Then which of the following does not hold good?

• A. $z=\frac{1}{1-ix}+i\left(\frac{1}{1+ix}\right)$
• B. $z=\frac{1}{1+ix}+i\left(\frac{1}{1-ix}\right)$
• C. $z=\frac{1}{1+ix}+i\left(\frac{1}{1+ix}\right)$
• D. $z=\frac{1}{1-ix}+i\left(\frac{1}{1-ix}\right)$

Answer 1

Answer: (A), (C), (D)

$z=\frac{1}{1-ix}+i\left(\frac{1}{1+ix}\right)$
$z=\frac{1}{1+ix}+i\left(\frac{1}{1+ix}\right)$
$z=\frac{1}{1-ix}+i\left(\frac{1}{1-ix}\right)$

Choice (a) on simplification gives
$z=\frac{1+x}{1+x^2}+i\frac{1+x}{1+x^2}$
For $x=0.5,f\left(0.5\right)>1$ which is out of range, Hence, (a) is not correct.

From choice (b),
$z=\frac{1-x}{1+x^2}+i\frac{1-x}{1+x^2}$
f(x) and g(x) $ϵ(0,1)$ if $xϵ(0,1)$. Hence, (b) is correct.

From choice (c).
$z=\frac{1+x}{1+x^2}+i\frac{1-x}{1+x^2}$
Hence, (c) is not correct. From choice (d),
$z=\frac{1-x}{1+x^2}+i\frac{1+x}{1+x^2}$
Hence, (d) is not correct.