A conventional Hamiltonian is Hermitian. Hence, if it contains a non-Hermitian term, it must either also contain its Hermitian conjuagte as another term, or have 0 weight. In this particular case, since $Z\otimes X\otimes Y$ is Hermitian itself, the coefficient would have to be 0. So, if you're talking about conventional Hamiltonians, you've probably made a mistake in your calculation. Note that if the Hermitian conjugate of the term is not present, you cannot simply fix things by adding it in; it will give you a completely different result.

On the other hand, you might be wanting to implement a non-Hermitian Hamiltonian. These things do exist, often for the description of noise processes, but are not nearly so widespread. You need to explicitly include the "non-Hermitian" terminology, otherwise everyone will just think that what you're doing is wrong because it's not Hermitian, and a Hamiltonian should be Hermitian. I'm not overly familiar with what capabilities the various simulators provide, but I'd be surprised if they have non-Hermiticity built in.

However, you can simulate it, at the cost of non-deterministic implementation. There will be more sophisticated methods than this (see the links in this answer), but let me describe a particularly simply one: I'm going to assume there's only one non-Hermitian component, which is a $i\times$(a tensor product of Paulis). I'll call this tensor product of Paulis $K$. The rest of the Hamiltonian is $H$. You want to create the evolution $$ e^{-iHt+Kt} $$ We start by Trotterising the evolution, $$ e^{-iHt+Kt}\approx \prod_{i=1}^Ne^{-iH\delta t+K\delta t} $$ where $N\delta t=t$. Now we work on simulating an individual term $e^{-iH\delta t+K\delta t}\approx e^{-iH\delta t}e^{K\delta t}$. You already know how to deal with the Hermitian part so, focus on $$e^{K\delta t}=\cosh(\delta t)\mathbb{I}+\sinh(\delta t)K.$$

We introduce an ancilla qubit in the state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, and we use this as the control qubit in a controlled-$K$ gate. Then we measure the ancilla in the $\{|\psi\rangle,|\psi^\perp\rangle\}$ basis (where $\langle\psi|\psi^\perp\rangle=0$). If the outcome is $|\psi\rangle$, then on the target qubits we have implemented the operation $|\alpha|^2\mathbb{I}+|\beta|^2K$, up to normalisation. So, if you fix $(1-|\alpha|^2)/|\alpha|^2=\tanh(\delta t)$, you have perfectly implemented that operation. If the measurement fails, then it's up to you whether you want to try to recover (this may well not be possible) or start again.

A conventional Hamiltonian is Hermitian. Hence, if it contains a non-Hermitian term, it must either also contain its Hermitian conjuagte as another term, or have 0 weight. In this particular case, since $Z\otimes X\otimes Y$ is Hermitian itself, the coefficient would have to be 0. So, if you're talking about conventional Hamiltonians, you've probably made a mistake in your calculation. Note that if the Hermitian conjugate of the term is not present, you cannot simply fix things by adding it in; it will give you a completely different result.

On the other hand, you might be wanting to implement a non-Hermitian Hamiltonian. These things do exist, often for the description of noise processes, but are not nearly so widespread. You need to explicitly include the "non-Hermitian" terminology, otherwise everyone will just think that what you're doing is wrong because it's not Hermitian, and a Hamiltonian should be Hermitian. I'm not overly familiar with what capabilities the various simulators provide, but I'd be surprised if they have non-Hermiticity built in.

However, you can simulate it, at the cost of non-deterministic implementation. There will be more sophisticated methods than this (see the links in this answer), but let me describe a particularly simply one: I'm going to assume there's only one non-Hermitian component, which is $i\times$(a tensor product of Paulis). I'll call this tensor product of Paulis $K$. The rest of the Hamiltonian is $H$. You want to create the evolution $$ e^{-iHt+Kt} $$ We start by Trotterising the evolution, $$ e^{-iHt+Kt}= \prod_{i=1}^Ne^{-iH\delta t+K\delta t} $$ where $N\delta t=t$. Now we work on simulating an individual term $e^{-iH\delta t+K\delta t}\approx e^{-iH\delta t}e^{K\delta t}$ (which becomes more accurate at large $N$). You already know how to deal with the Hermitian part so, focus on $$e^{K\delta t}=\cosh(\delta t)\mathbb{I}+\sinh(\delta t)K.$$

We introduce an ancilla qubit in the state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, and we use this as the control qubit in a controlled-$K$ gate. Then we measure the ancilla in the $\{|\psi\rangle,|\psi^\perp\rangle\}$ basis (where $\langle\psi|\psi^\perp\rangle=0$). If the outcome is $|\psi\rangle$, then on the target qubits we have implemented the operation $|\alpha|^2\mathbb{I}+|\beta|^2K$, up to normalisation. So, if you fix $(1-|\alpha|^2)/|\alpha|^2=\tanh(\delta t)$, you have perfectly implemented that operation. If the measurement fails, then it's up to you whether you want to try to recover (this may well not be possible) or start again.