Mean of exponential Ito integral Suppose I have the quantity
$$M_\omega(0,L) = \exp{\bigg[\int_{0}^{L}\tanh{(\theta(z)/2)}(dW_1(z) + idW_2(z))\bigg]}$$
where $\theta(z)$ takes values $\in[0,\infty),$ can I say then that the mean
$\mathbb{E}[M_\omega(0,L)] = 1$ since $\mathbb{E}[M_\omega] = \exp{\bigg[\mathbb{E}[\int_{0}^{L}\tanh{(\theta(z)/2)}(dW_1(z) + idW_2(z))]\bigg]} = \exp(0) = 1$ by considering the Brownian increment $\Delta W_{t_{i+1}} = W_{t_{i+1}} - W_{t_i}$. By the definition of a Wiener process, if $\Delta t = t_{i+1}-t_i$, then (1) $\Delta W_{t_{i+1}} \sim \mathcal{N}(0,\Delta t) $ and (2) $W_{t_i}$ is independent from $\Delta W_{t_{i+1}}$.
Thus, expanded as a sort of Ito-Riemann sum:
\begin{align}
\mathbb{E}\left[ \int_0^t f(W_s)\;dW_s \right]
&\approx \mathbb{E}\left[ \sum_i f(W_{t_i})[W_{t_{i+1}} - W_{t_i}] \right] \\
&= \sum_i \mathbb{E}\left[ f(W_{t_i})\Delta W_{t_{i+1}} \right] \\
&= \sum_i \mathbb{E}\left[ f(W_{t_i})\right] \mathbb{E}\left[\Delta W_{t_{i+1}} \right]\;\;\;\;\;\;\;\;\;\;\; \\
&= \sum_i \mathbb{E}\left[ f(W_{t_i})\right] 0 \;\;\;\;\;\;\;\;\;\;\;\;\,\;\;\;\;\;\;\;\;\;\;\; \\
&=0?
\end{align}
Also, how can I show that this quantity $M_\omega$ is a martingale?
 A: The logic in your question is incorrect, since it would apply equally well to real valued random variables. For an arbitrary real valued random variable $X$ the equality
$$\mathbb{E}[e^X] = e^{\mathbb{E}[X]}$$
holds if and only if X is degenerate (i.e. $\mathbb{P}(X=a) = 1$ for some $a \in \mathbb{R}$). This is a consequence of Jensen's inequality. Now, your random variable is complex valued, but the lesson is the same. As a general rule, exponentials typically do not commute with expectations.
In this case, you can compute the mean of the random variable you are interested in directly because the components of standard complex Brownian motion are independent and so the resulting stochastic integrals in this case are independent (I am assuming here that $\theta$ is a deterministic function here).
$$
\int_0^t \tanh(\theta(z)/2)dW_1(z) \sim N\left(0, \int_0^t \tanh^2(\theta(z)/2)dz\right)
$$
and
$$
\int_0^t \tanh(\theta(z)/2)dW_2(z) \sim N\left(0, \int_0^t \tanh^2(\theta(z)/2)dz\right)
$$
If $X$ and $Y$ are independent real normal random variables with a common variance $\sigma^2$ and mean zero,
$$
\mathbb{E}[e^{X+iY}] = \mathbb{E}[e^X] \mathbb{E}[e^{iY}] = e^{\sigma^2/2}e^{-\sigma^2/2} = 1.
$$
Therefore
$$
\mathbb{E}\left[e^{\int_0^t \tanh(\theta(z)/2)dW_1(z) + i \int_0^t \tanh(\theta(z)/2)dW_2(z)}\right] = 1. 
$$
Now, complex Brownian motion $W(t) = W_1(t) + i W_2(t)$ is a conformal martingale. The stochastic integral
$$
Z_t = \int_0^t \tanh(\theta(z)/2)dW(z)
$$
is usually written in differential form as
$$
dZ_t = \tanh(\theta(t)/2)dW(t)
$$
is also a conformal local martingale. We can see this by computing its bracket:
$$
d\langle Z,Z\rangle_t = \tanh^2(\theta(t)/2)d\langle W,W \rangle_t = 0
$$
Analytic functions of conformal local martingales are local martingales (see Corollary 37 in that note). To see that it is an actual martingale, just compute as above.
