I know that $$E[e^{i \lambda B_T}] = e^{- \lambda}, \lambda >0$$ holds for a Complex Brownian motion $B(t)= \mathbf{i} + X_t + \mathbf{i} Y_t$, where $T= \inf\{ t \geq 0: Y_t = -1 \}$.
I need to use "symmetry" to show that actually $$E[e^{i \lambda B_T}] = e^{- | \lambda|}, \lambda \in \mathbb{R}$$
How can I move?
Edit
So, let $\lambda <0$.
Then $$E[e^{i (-\lambda)B_T}] = e^{-(-\lambda)} = e^{\lambda}$$ since $-\lambda >0$.
But $E[e^{i (- \lambda) B_T}]=E[e^{i \lambda(-B_T)}] = E[e^{i \lambda B_T}]$
where in the last equality I used the fact that $-(B_t)_t$ has the same distribution as $(B_t)_t$.
All in all $$E[e^{i \lambda B_T}]= e^{\lambda}, \qquad \lambda <0$$
So we can conclude $E[e^{i \lambda B_t}] = e^{- |\lambda|}, \quad \lambda \in \mathbb{R}$