Changing order of limit and complex derivative Assume $f(z,t)$ is complex differentiable at $z=z_0$, $t$ is a real parameter. Under which condition is $\lim_{t \rightarrow t_0} f(z,t)$ also complex differentiable at $z = z_0$?
I know some theorems and their proofs from real analysis, but I was curious whether I could prove the corresponding theorems in the complex domain but I am unsure what the condition of limit switching theorem might be!
EDIT
How does the result extends to interchangeability of derivatives? I.e. assume $f(z,t)$ is both $z \in \mathbb{C}$ and $t \in \mathbb{R}$ differentiable in some neighborhood of $z_0$ and $t_0$. Under which conditions we have that $\frac{\partial f}{\partial t}$ is also $z$ differentiable?
ADDENDUM
(Comment to @Bobby Ocean answer)
Consider a following: Somehow you have proven that $z^t$ is $z$ complex differentiable for each real $t$ (except branches). Then you define a function $g(z) = \lim_{t \rightarrow 0} (z^t - 1)/t$, how do you prove this is also complex differentiable? I would like to define the logaritmic function this way, and proving $g'(z)=1/z$ would require knowing whether I can interchange the derivatives.
 A: My answer is in response to your second question: "Under which conditions we have that $\frac{\partial f}{\partial t}$ is also $z$ differentiable?"
Suppose $U$ is open in $\mathbb C$ and $g:U\to \mathbb C.$ Assume $g\in C^1(U).$ Define the operator
$$D=\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}.$$
Then $g$ is analytic in $U$ iff $Dg\equiv 0$ in $U.$ This follows easily from the Cauchy-Riemann equations.
Now suppose $f:U\times I\to \mathbb C$ with $f\in C^2(U\times I).$ (Here $U$ is open in $\mathbb C$ and $I$ is open in $\mathbb R.$) This is your $f(z,t)$ here. Think of $z=x+iy.$
Assume that for fixed $t,$ $f$ is complex differentiable in $z.$ The $C^2$ hypothesis implies switching the order of derivatives in $x,y,t$ leaves the result unchanged. Thus
$$D\frac{\partial}{\partial t} f(x+iy,t) =  \frac{\partial}{\partial t} Df(x+iy,t).$$
By hypothesis, the inner partial derivative on the right is $0$ for all $(z,t)\in U\times I.$ Hence the left side vanishes identically. This proves $\dfrac{\partial}{\partial t} f(z,t)$ is analytic in $U$ for each fixed $t.$
Thus $C^2$ smoothness (in the real sense) is a sufficient condition for $\partial f/\partial t $ to be complex differentiable with respect to $z.$
A: The existence of a pointwise limit at the point $t_0$ is insufficient to guarantee a differentiable function. In general, one needs uniform convergence to preserve various properties like differentiation, integration, continuity, etc.
A simple example would be $f(z,t):=z^t$ on $t>0$ and $z\in\mathbb{C}-\text{B}$ (where $\text{B}$ is some branch cut). Note, $f(z,t)$ is real-differentiable in $t\in\mathbb{R}$ and complex-differentiable in $z\in\mathbb{C}-\text{B}$. Notice, for each fixed $z$ we have, $$h(z):=\lim_{t\to 0} f(z,t) = \left\{\matrix{1 & z\neq 0 \\ 0 & z = 0}\right.$$ a function that is clearly not differentiable at $z=0$.

The the ability to interchange derivatives is not the same thing as the existence of the derivatives. Not only must you assume that the function is $C^2$ (example where $f_{xy}(0,0)\neq f_{yx}(0,0)$ because we don't have $C^2$), but you must also assume that the mixed partials already exist. You are not guaranteed the mixed partial derivatives will even exist, simply because the function is $C^2$. Let me say it another way, $C^2$ isn't good enough to allow you to take mixed partial derivatives.
Explicitly, $f(z,t)$ could be infinitely differentiable with respect to variables $t$ and $z$, and yet $\frac{\partial}{\partial z}\frac{\partial}{\partial t} f(z,t)$ could fail to exist at some point in $\mathbb{C}\times\mathbb{R}$. This is simply another way of saying that $\frac{\partial}{\partial t} f(z,t)$ could fail to be differentiable with respect to $z$ at some point $z_0$. Or as I will show by way of example, $\frac{\partial}{\partial t} f(z,t)$ could still be a discontinuous function with respect to the variable $z$.
Example.
To be clear, this is a non-obvious example that took me awhile to construct. First we define for each $\alpha\in\mathbb{C}$, $$g_\alpha(z) := \left\{\matrix{(e^{\alpha z}-1)/z & z\neq 0 \\ \alpha & z=0}\right\} = \sum_{k=1}^\infty \frac{\alpha^k}{k!}z^{k-1}$$ Note that this is a family of entire functions that are infinitely differentiable in $z$ everywhere in $\mathbb{C}$ (this is easily seen, since the power series representation has an infinite radius of convergence). Now define, $$f(z,t) := \left\{\matrix{t\cdot g_{-1/t^2}(z) & t\neq 0 \\ 0 & t=0}\right.$$ Astonishingly $f(z,t)$ is infinitely differentiable with respect to $t\in\mathbb{R}$ (including at the point $0$), likewise, $f(z,t)$ is infinitely differentiable with respect to $z\in\mathbb{C}$ (again, including the point at $0$) (literally, for each fixed $t$, $f(z,t)$ is entire). However, when we attempt to mix partial derivatives we will get a problem, take $\frac{\partial}{\partial t}$, $$\frac{\partial}{\partial t} f(z,t) = \left\{\matrix{2e^{-z/t^2}/t^2 + g_{-1/t^2}(z) & t\neq 0 \\ -1/z & t=0}\right.$$ Hence, $$\lim_{t\to 0}\frac{\partial}{\partial t} f(z,t) = -\frac{1}{z}$$ a clearly discontinuous function in the variable $z$, with no derivative at zero. Thus, even though $f(z,t)$ was infinitely differentiable at $t=0$ with respect to the variable $t\in\mathbb{R}$ and is infinitely differentiable at $z=0$ with respect to the variable $z\in\mathbb{C}$, that meant nothing in determining if $\frac{\partial}{\partial t} f(z,t)$ would be differentiable with respect $z$ at $z=0$ when $t=0$. We even point out that $\frac{\partial}{\partial t} f(z,t)$ is differentiable with respect to $z$ at $z=0$ for any $t\neq 0$.
Please note that the reason this example works is because we can utilize the "infinitely real-differentiable flat function" $e^{-1/t^2}$. Such function is not complex-differentiable at zero, even though it is real-differentiable at zero. The above example would not apply to multi-variate complex-differentiable functions, $f(z,w)$, that would be a different discussion.
