Eq. 19 of Gershgorin et. al. (2010) states that given a complex Gaussian variable $Z$ and a real Gaussian variable $X$, the mean of $Z\,e^{b\,X}$ is given by
$$\left\langle Z\,e^{b\,X} \right\rangle = \left( \left\langle Z \right\rangle + b\,\text{Cov}\left( Z,\ X \right) \right)\,e^{b\,\left\langle X \right\rangle + \frac{b^2}{2}\,\text{Var}\left( X \right)}.$$
The paper mentions that this quantity can be computed using the characteristic function of a Gaussian distribution, but I'm not sure how one would go about this.
I understand that the characteristic function is the inverse Fourier transform of the probability distribution function, and that the characteristic function of a linear combination of probability distribution functions is a product of the individual characteristic functions, but I don't know how the characteristic functions of $Z$ and $X$ would relate to the characteristic function of $Z\,e^{b\,X}$.
EDIT: I've made some progress on this problem. If we assume that $X$ and $Z$ are independent, with probability density functions $f_{X}\left( x \right)$ and $f_{Z}\left( z \right)$, then we may use the Law of the Unconscious Statistician
$$E\left[ g\left( X \right) \right] = \int_{\mathbb{R}} g\left( x \right)\,f_{X}\left( x \right)\,dx$$
to evaluate
\begin{align*} E\left[ Z\,e^{b\,X} \right] &= \int_{\mathbb{C}} \int_{\mathbb{R}} z\,e^{b\,x}\,f_{X}\left( x \right)\,f_{Z}\left( z \right)\,dx\,dz \\ &= \int_{\mathbb{C}} z\,f_{Z}\left( z \right)\,dz\,\int_{\mathbb{R}} e^{b\,x}\,f_{X}\left( x \right)\,dx \\ &= \left\langle Z \right\rangle\,M_{X}\left( b \right) \end{align*}
where $M_{X}\left( t \right)$ is the moment-generating function of $X$
$$M_{X}\left( t \right) = E\left[ e^{t\,X} \right].$$
Since $X$ is Gaussian, we have
$$M_{X}\left( b \right) = e^{b\,\left\langle X \right\rangle + \frac{b^2}{2}\,\text{Var}\left( X \right)},$$
which means for independent $X$ and $Z$ we have
$$\left\langle Z\,e^{b\,X} \right\rangle = \left\langle Z \right\rangle\,e^{b\,\left\langle X \right\rangle + \frac{b^2}{2}\,\text{Var}\left( X \right)}$$
which is almost there! I am guessing that once we drop the assumption that the desired identity falls out.
Alright, I figured it out based on some of the references ([15, 16]) in that Gershgorin et. al. 2010 paper.
To begin, let's write $Z = U + i\,V$ so that we may discuss its real and complex parts, and define a vector $\overrightarrow{V} = \left[ U\ V\ X \right]^T$. Note that $\overrightarrow{V}$ is normally distributed since each of $U$, $V$, and $X$ are as well.
The quantity we would like to find $\left\langle Z\,e^{b\,X} \right\rangle$ will come from finding its real and complex parts $\left\langle U\,e^{b\,X} \right\rangle$ and $\left\langle V\,e^{b\,X} \right\rangle$.
To find $\left\langle U\,e^{b\,X} \right\rangle$, we begin by considering the partial derivatives of the moment-generating function $M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)$ for $\overrightarrow{V}$
$$M_{\overrightarrow{V}} \left( \overrightarrow{t} \right) = \int_{\mathbb{R}^{3}} e^{\overrightarrow{t}^{T}\,\overrightarrow{v}}\,f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)\,d\overrightarrow{v},$$
where $f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)$ is the probability density funciton for $\overrightarrow{V}$. We will differentiate this function with respect to the first and second entries of $\overrightarrow{t}$, denoted $t_1$ and $t_2$ respectively
\begin{align*} \frac{\partial M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)}{\partial t_1} &= \int_{\mathbb{R}^{3}} u\,e^{\overrightarrow{t}^{T}\,\overrightarrow{v}}\,f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)\,d\overrightarrow{v},\\ \frac{\partial M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)}{\partial t_2} &= \int_{\mathbb{R}^{3}} v\,e^{\overrightarrow{t}^{T}\,\overrightarrow{v}}\,f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)\,d\overrightarrow{v}. \end{align*}
Evaluating these partial derivatives at $\overrightarrow{t} = \left[ 0\ 0\ b \right]^{T}$, we obtain
\begin{align*} \left.\frac{\partial M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)}{\partial t_1}\middle|_{\overrightarrow{t} = \left[ 0\ 0\ b \right]^{T}}\right. &= \int_{\mathbb{R}^{3}} u\,e^{b\,x}\,f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)\,d\overrightarrow{v} = \left\langle U\,e^{b\,X} \right\rangle,\\ \left. \frac{\partial M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)}{\partial t_2}\middle|_{\overrightarrow{t} = \left[ 0\ 0\ b \right]^{T}} \right. &= \int_{\mathbb{R}^{3}} v\,e^{b\,x}\,f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)\,d\overrightarrow{v} = \left\langle V\,e^{b\,X} \right\rangle. \end{align*}
Now, since $\overrightarrow{V}$ is a normally distributed, its moment-generating function is
$$M_{\overrightarrow{V}} \left( \overrightarrow{t} \right) = e^{\overrightarrow{t}^{T}\,\left(\overrightarrow{\mu}_{\overrightarrow{V}} + \frac{1}{2}\,\Sigma\,\overrightarrow{t}\right)}$$
where $\Sigma$ is the covariance matrix of $U$, $V$, and $X$. Hence,
\begin{align*} \left\langle U\,e^{b\,X} \right\rangle &= \left.\frac{\partial M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)}{\partial t_1}\middle|_{\overrightarrow{t} = \left[ 0\ 0\ b \right]^{T}}\right. \\ &= \left. \left( \left\langle U \right\rangle + t_1\,\text{Cov}\left( U,\ U \right) + t_2\,\text{Cov}\left( U,\ V \right) + t_3\,\text{Cov}\left( U,\ X \right) \right)\,e^{\overrightarrow{t}^{T}\,\left(\overrightarrow{\mu}_{\overrightarrow{V}} + \frac{1}{2}\,\Sigma\,\overrightarrow{t}\right)} \middle|_{\overrightarrow{t} = \left[ 0\ 0\ b \right]^{T}}\right. \\ &= \left( \left\langle U \right\rangle + b\,\text{Cov}\left( U,\ X \right) \right)\,e^{b\,\left\langle X \right\rangle + \frac{1}{2}\,b^2\,\text{Var}\left( X \right)}, \end{align*}
and similarly
\begin{align*} \left\langle V\,e^{b\,X} \right\rangle &= \left( \left\langle V \right\rangle + b\,\text{Cov}\left( V,\ X \right) \right)\,e^{b\,\left\langle X \right\rangle + \frac{1}{2}\,b^2\,\text{Var}\left( X \right)}, \end{align*}
Therefore,
\begin{align*} \left\langle Z\,e^{b\,X} \right\rangle &= \left\langle U\,e^{b\,X} \right\rangle + i\,\left\langle V\,e^{b\,X} \right\rangle \\ &= \left( \left( \left\langle U \right\rangle + i\,\left\langle V \right\rangle \right) + b\,\left( \text{Cov}\left( U,\ X \right) + i\,\text{Cov}\left( V,\ X \right) \right) \right)\,e^{b\,\left\langle X \right\rangle + \frac{1}{2}\,b^2\,\text{Var}\left( X \right)} \\ &= \left( \left\langle Z \right\rangle + b\,\text{Cov}\left( Z,\ X \right) \right)\,e^{b\,\left\langle X \right\rangle + \frac{1}{2}\,b^2\,\text{Var}\left( X \right)} \end{align*}
as desired.