Obtaining martingales from Poisson process All processes here are continuous. Suppose we have a Poisson process $(N_t)_{t\geq 0}$ with parameter $\lambda > 0$ and adapted to the filtration $(\mathcal{F}_t)_{t\geq 0}$. Fix $u\in\mathbb{C}$, let $i=\sqrt{-1}$ and define the process 
$$X_t = e^{iuN_t - \lambda t(e^{iu}-1)}.$$
From this, I want to show that $(Re(X_t))_{t\geq 0}$ and $(Im(X_t))_{t\geq 0}$ are $\mathcal{F}_t$-martingales. 
Here is what I have done so far: We can assume $u\neq 0$. The first thing I did was write $$X_t = \frac{e^{iuN_t}}{E[e^{iuY}]},$$
where $Y$ is any Poisson random variable with parameter $\lambda t$. Now we have that
$$X_t = \frac{e^{iuN_t}}{E[e^{iuY}]} = $$
$$ = \frac{\cos(uN_t)E[\cos(uY)] + \sin(uN_t)E[\sin(uY)] + i\big(\sin(uN_t)E[\cos(uY)] + \cos(uN_t)E[\sin(uY)]\big)}{E[\sin(uY)]^2 + E[\cos(uY)]^2}.$$
Therefore 
$$E[Re(X_t)]=\frac{E[\cos(uN_t)]E[\cos(uY)] + E[\sin(uN_t)]E[\sin(uY)]}{E[\sin(uY)]^2 + E[\cos(uY)]^2} < \infty$$
for the numerator is finite and the denominator is positive. The same reasoning applies to the imaginary part of $X_t$.
The hard part is to show that $E[Re(X_t)|\mathcal{F}_s] = Re(X_s)$ for all $t>s$. I need a help here.
Thanks!
 A: Now I know how to solve this problem. Thank you Did and saz for the help (let me know if there is anything wrong with my solution, I'll try to fix as soon as I can). 
First of all, the Poisson process $(N_t)_{t\geq 0}$ is such that $N_0=0$ a.s., and $E[N_t-N_s] = \lambda (t-s)$ for $s<t$. In particular, $E[N_t-N_0] = E[N_t] = \lambda t$ for any $t > 0$. Therefore,
$$X_t = e^{iuN_t - \lambda t(e^{iu}-1)} = \frac{e^{iuN_t}}{E[e^{iuN_t}]}.$$
From this, we have that $E[X_t] = 1$, so we get $E[Re(X_t)]<\infty$ and $E[Im(X_t)]<\infty$ in a more direct way.
Finally, let $0\leq s<t$, then 
$$E[X_t|\mathcal{F}_s] = $$
$$= E[e^{iuN_t - \lambda t(e^{iu}-1)}|\mathcal{F}_s] = $$
$$= e^{- \lambda t(e^{iu}-1)}E[e^{iuN_t}|\mathcal{F}_s] =$$
$$= e^{- \lambda t(e^{iu}-1)}E[e^{iuN_t+iu(N_s-N_s)}|\mathcal{F}_s] =$$ 
$$= e^{- \lambda t(e^{iu}-1)}E[e^{iuN_t-iuN_s}e^{iuN_s}|\mathcal{F}_s] =$$
$$= e^{- \lambda t(e^{iu}-1)}e^{iuN_s}E[e^{iu(N_t-N_s)}|\mathcal{F}_s].$$
The last equality is valid because $e^{iuN_s}$ is $F_s$-measurable. The increments are stationary, so we have that $E[e^{iu(N_t-N_s)}|\mathcal{F}_s] = E[e^{iu(N_t-N_s)}] = e^{\lambda(t-s)(e^{iu}-1)}$. Therefore,
$$E[X_t|\mathcal{F}_s] = $$
$$= e^{- \lambda t(e^{iu}-1)}e^{iuN_s}e^{\lambda(t-s)(e^{iu}-1)} = $$
$$= e^{- \lambda t(e^{iu}-1)+iuN_s+\lambda(t-s)(e^{iu}-1)} = $$
$$= e^{iuN_s-\lambda s(e^{iu}-1)} = X_s.$$
From this, we conclude that $(X_t)_{t\geq 0}$ is a $\mathcal{F}_t$-martingal. To prove that $(Re(X_t))_{t\geq 0}$ is also a $\mathcal{F}_t$-martingal, consider $0\leq s<t$, $A\in\mathcal{F}_s$ arbitrary and note that. 
$$E[X_t\cdot\textbf{I}_A] = E[X_s\cdot\textbf{I}_A] \implies \int_A X_t\ dP = \int_A X_s\ dP \implies$$
$$\implies \int_A Re(X_t)\ dP +i\int_A Im(X_t)\ dP = \int_A Re(X_s)\ dP +i\int_A Im(X_s)\ dP \implies$$ 
$$\int_A Re(X_t)\ dP =\int_A Re(X_s)\ dP $$ and
$$\int_A Im(X_t)\ dP =\int_A Im(X_s)\ dP. $$
Therefore, $E[Re(X_t)\cdot\textbf{I}_A]=E[Re(X_s)\cdot\textbf{I}_A]$ and $E[Im(X_t)\cdot\textbf{I}_A]=E[Im(X_s)\cdot\textbf{I}_A]$ for all $A\in\mathcal{F}_s$. This implies $E[Re(X_t)|\mathcal{F}_s] = Re(X_s)$ and $E[Im(X_t)|\mathcal{F}_s] = Im(X_s)$, and we are done.
