The real and imaginary parts of $X_t = \exp[iuN_t - \lambda t(e^{iu}-1)]; \; 0 \le t < \infty$ are martingales

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This is Problem 3.21 from Chapter 1 of Karatzas and Shreve.

Let $\{N_t, \mathscr{F}_t : 0 \le t < \infty\}$ be a Poisson process with parameter $\lambda>0.$ For $u \in \mathbb{C}$ and $i=\sqrt{-1}$, define the process $$X_t = \exp[iuN_t - \lambda t(e^{iu}-1)]; \; 0 \le t < \infty.$$

Show that $\{Re(X_t), \mathscr{F}_t; 0 \le t < \infty\}$, and $\{Im(X_t), \mathscr{F}_t: 0 \le t < \infty\}$ are martingales.

I am lost from start to this problem. How can we get the real and imaginary parts from $X_t$?

My attempt: Let $u := v + iw$ for $v,w$ real. Then $iuN_t - \lambda t (e^{iu}-1)=ivN_t - wN_t - \lambda te^{iv-w} + \lambda t = \lambda t - wN_t + ivN_t -\lambda t e^{-w}e^{iv} = \lambda t - wN_t + ivN_t - \lambda e^{-w}t( \cos v + i\sin v) \lambda t - wN_t - \lambda e^{-w} \cos v t + i(vN_t + \sin v) .$

Hence the Real part of $X$ becomes $\exp[\lambda t - wN_t - \lambda e^{-w} \cos v t]\cdot \cos (vN_t + \sin v)$ and the Imaginary part of X is $\exp[\lambda t - wN_t - \lambda e^{-w} \cos v t]\cdot \sin (vN_t + \sin v)$.

I know that $N_t - \lambda t$ is a martingale, but I don't know how to apply the conditional expectation to this complicated function.