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Given $0 \leq x < 1$, $0 < Re(\rho) < 1$.

Then does this equation contain no solutions for x other than $x = \frac{1}{2}$? $2^\rho + \frac{1}{2} = \frac{1}{(1-x)^\rho} + x$

I am unable to find any answer.

EDIT: I have solved it. The answer is $x = \frac{1}{2}$ is the only solution for $0 \leq x < 1$. Basically I separated the real and imaginary parts and checked for conditions $0 \leq x < \frac{1}{2}$ and $\frac{1}{2} < x < 1$, and used the inequality to produce contradictions in both cases. I think I will leave the details as an exercise.

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    Roupam, I think you $s$hould po$s$t your edit a$s$ an an$s$wer and validate it $s$o that this question is considered resolved.2012-04-11

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As per @Raskolnikov's suggestion I am adding my edit as an answer.

The answer is $x=\frac{1}{2}$ is the only solution for 0≤x<1. Basically I separated the real and imaginary parts and checked for conditions $0≤x<\frac{1}{2}$ and $\frac{1}{2}, and used the inequality to produce contradictions in both cases. I think I will leave the details as an exercise.