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.