Q:

two sided brownian motion hit time +1 -1

Given standard brownian motion a common way to calculate expectation of hitting time $t$ of either +1 or -1 is considering martingale $W_t^2-t$.
Thus we get to $\frac{1}{2}((+1)^2-t) +\frac{1}{2}((-1)^2-t)=0$ and we get $t=1$ . 
But if we consider another martingale $ e^{W_t-\frac{t}{2}}$ then the $t$ we get is not 1 anymore.  
i.e. the equation becomes $ \frac{1}{2}e^{+1-\frac{t}{2}}+\frac{1}{2}e^{-1-\frac{t}{2}}=1$ and 1 is not a solution anymore.  where is the flaw? 

A:

You are computing $\mathbb E[e^{T/2}]$ in your second try, and this is notoriously distinct from $e^{\mathbb E[T]/2}$, unless your random variable $T$ is constant, by Jensen (which it is not, thus).