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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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Where does it come from and what have you tried? –  draks ... Apr 10 '12 at 21:03
    
Roupam, I think you should post your edit as an answer and validate it so that this question is considered resolved. –  Raskolnikov Apr 11 '12 at 5:18
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up vote 0 down vote accepted

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 $\frac{1}{2}<x<1$, and used the inequality to produce contradictions in both cases. I think I will leave the details as an exercise.

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