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)

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

Given: $z=f\left(x\right)+ig\left(x\right)$ where

$f,g:(0,1)\to (0,1)$ are real valued functions.

From option (A),

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

$\Rightarrow z=\frac{(1+ix)+i(1-ix)}{(1-ix)(1+ix)}=\frac{1+ix+i+x}{1+x^2}$

$\Rightarrow z=\frac{1+x}{1+x^2}+i\frac{1+x}{1+x^2}$

For $x=0.5,f\left(0.5\right)>1$ which is out of range.

Hence, (A) is not correct.

From option (B),

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

$\Rightarrow z=\frac{1-x}{1+x^2}+i\frac{1-x}{1+x^2}$

$f\left(x\right)$ and $g\left(x\right)ϵ(0,1)$ if $xϵ(0,1)$.Hence, (B) is correct.

From option (C),

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

$\Rightarrow z=\frac{1-ix+i+x}{1+x^2}=\frac{1+x}{1+x^2}+i\frac{1-x}{1+x^2}$

For $x=0.5,f\left(0.5\right)>1$ which is out of range.

Hence, (C) is not correct.

From option (D),

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

$\Rightarrow z=\frac{1+ix+i-x}{1+x^2}=\frac{1-x}{1+x^2}+i\frac{1+x}{1+x^2}$

For $x=0.5,g\left(0.5\right)>1$ which is out of range.

Hence, $\left(D\right)$ is not correct.

Hence, answer is options $\left(A\right),\left(C\right),\left(D\right)$