Find Harmonic Numbers for Imaginary and Complex Values
Let us consider the simplest case with an imaginary argument of the harmonic number, namely $H_i$. This can easily be generalized to the requested case $H_{ji}$.
The harmonic number can be defined for complex $z$ as
$$H_z = \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+z} \right)\tag{1}$$
Notice that you must not split the sum into two parts, because both sums
$$\sum_{k=1}^\infty \left( \frac{1}{k} \right),\sum_{k=1}^\infty \left( \frac{1}{k+z} \right) $$
are divergent.
Now for $z=i$ we have
$$H_i = \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+i} \right)\tag{2}$$
writing the summand as
$$ \frac{1}{k} - \frac{1}{k+i} = \frac{1}{k} - \frac{k-i}{(k+i)(k-i)} = \frac{1}{k} - \frac{k-i}{k^2+1}=\frac{1}{k} - \frac{k}{k^2+1}+i \frac{1}{k^2+1}\\=\frac{1}{k(k^2+1)} +i \frac{1}{k^2+1} $$
we get
$$H_i = \sum_{k=1}^\infty \left(\frac{1}{k(k^2+1)} +i \frac{1}{k^2+1}\right)\tag{3}$$
Now splitting is permitted because the two sums are convergent. In fact, the real and imaginary parts of $H_i$, $f$ and $g$, respectively, are
$$f = \sum_{k=1}^\infty \left(\frac{1}{k(k^2+1)}\right)\tag{4}$$
$$g = \sum_{k=1}^\infty \left( \frac{1}{k^2+1}\right)\tag{5}$$
I stop here for a while to let you calculate $f$ and $g$, i.e. find expressions which are not just real and imaginary part of $H_i$.
EDIT
Now, what can be said about $f$ and $g$?
$g$ has a closed form
$$g = \sum_{k=1}^\infty \left( \frac{1}{k^2+1}\right)= \frac{1}{2}\left(\sum_{k=-\infty}^\infty \left( \frac{1}{k^2+1}\right)-1 \right) \\= \frac{1}{2} (\pi \coth (\pi )-1) = \dfrac{\pi-1}{2}+\dfrac{\pi}{e^{2\pi}-1}\tag{6}$$
which has been given in several places in this forum, for instance here How to prove $\sum_{n=0}^{\infty} \frac{1}{1+n^2} = \frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}$ (notice that there the sum starts at $k=0$) and derived with complex contour integration here How to sum $\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}$?.
For $f$ I have not found a closed expression other than the information that $f$ is the real part of $H_i$. However, normally this would be considered a closed form as well.
Technically, the deeper reason for the different behaviour of $f$ and $g$ is that whereas $g$ can be written as a symmetric sum from $-\infty$ to $\infty$ which allows complex contour integration with a kernel $\pi \cot(\pi z)$, $f$ is a one-sided sum which has the kernel $H_{-z}$. The latter kernel then just brings us back to where we came from. Usage of contour integrals for infinite sums is described for example in chapter 2 of https://projecteuclid.org/download/pdf_1/euclid.em/1047674270