In the complex world there is a fundamental involution, namely the map $\gamma:\ z\mapsto\bar z$. An involution of some set $X$ is a map $\iota:\ X\to X$ which is not the identity map ${\rm id}_X$, but whose square $\iota\circ\iota$ is the identity.
Given any region $\Omega\subset{\mathbb C}$ and a function $f:\ \Omega\to{\mathbb C}$, one can compose $f$ with $\gamma$ in various ways. Your cases 2. and 3. produce for a given $f$ the functions
$$f\circ \gamma:\quad z\mapsto f\bigl(\bar z)$$
and
$$\gamma\circ f:\quad z\mapsto\overline{f(z)}\ .$$
When $f$ is a holomorphic function of $z$ then both $f\circ\gamma$ and $\gamma\circ f$ are antiholomorphic, which means, e.g., that nonoriented angles between tangent vectors at points $z_0\in\Omega$ are preserved, but the orientation of small circles around $z_0$ is reversed.
Most interesting is your case 1. Here the function $f$ is transformed (also termed conjugated) into the new function
$$\bar f:=\gamma\circ f\circ\gamma:\quad z\mapsto\overline{ f(\bar z)}\ .$$
When $f$ is a holomorphic function on $\Omega$ then it easy to check by means of the CR-equations and the chain rule that the function $\bar f$ is a holomorphic function on $\bar\Omega:=\{\bar z|z\in\Omega\}$.
The case when $\Omega$ and $\bar\Omega$ share an interval on the real axis is of special interest, because then we can ask the question: Could it be that in fact $\bar f(z)\equiv f(z)$ for all $z\in \Omega\cap\bar\Omega\ $? After all, in the definition of $\bar f$ there are two conjugations involved.
To investigate this case, assume that $0\in\Omega\cap\bar\Omega$ and that
$$f(z)=\sum_{k=0}^\infty a_kz^k\qquad(|z|<\rho)$$
with certain complex coefficients $a_k$. Then
$$\bar f(z)=\overline{\sum_{k=0}^\infty a_k(\bar z)^k}=\sum_{k=0}^\infty \bar a_k z^k\ ,$$
and this is $\ \equiv f(z)$ iff all $\bar a_k=a_k$, i.e., if all $a_k$ are in fact real. When the latter is the case then automatically $f(z)$ is real for real $z$, and it is not difficult to show the converse: If a holomorphic $f(z)$ is real for real $z$ (as in the case $f:=\sin$) then $\bar f=f$. This is the so-called reflection principle.
Couple of years late, but I came across this issue too and did some digging.
The key point is that TensorFlow defines the "gradient" of a complex-valued function $f(z)$ of a complex variable as "the gradient of the real map $F: (x,y)\mapsto Re({f(x+iy)})$, expressed as a complex number" (the gradient of that real map is a vector in $\mathbb R^2$, so we can express it as a complex number in the obvious way).
Presumably the reason for that definition is that in TF one is usually concerned with gradients for the purpose of running gradient descent on a loss function, and in particular for identifying the direction of maximum increase/decrease of that loss function. Using the above definition of gradient means that a complex-valued function of complex variables can be used as a loss function in a standard gradient descent algorithm, and the result will be that the real part of the function gets minimised (which seems to me a somewhat reasonable interpretation of "optimise this complex-valued function").
Note that an equivalent way to write that definition of gradient is
$$ gradient(f) := \frac{dF}{dx} + i\frac{dF}{dy}=\overline{\frac{df}{dz} + \frac{d\overline{f}}{dz}},$$
which for a holomorphic function simply reduces to $\overline{\frac{df}{dz}}$.
I wrote up a longer explanation on the GitHub issue you linked: https://github.com/tensorflow/tensorflow/issues/3348#issuecomment-512101921
Best Answer
For simplicity, I assume that $A = a = 1$. In general, for real $t$, $$ \overline{e^{it}} = \overline{\cos t + i\sin t} = \cos t - i\sin t = \cos(-t) + i\sin(-t) = e^{-it}. $$ Hence, for your expression you get $$ \overline{e^{mx+it}} = \overline{e^{mx}e^{it}} = e^{mx}e^{-it} = e^{mx-it}. $$ Hope this helps for understanding. If something is unclear, do not hesitate to ask in the comments.