differentiability of complex function in a domain D Let $w=f(z)$ be a function of $\mathrm z$ defined in a domain $\mathrm D$
Then $f(z)$ is said to be differentiable at $$z=a$$ if the increment ratio  $$\frac{\Delta w}{\Delta z}=\frac{f(a+\Delta z)-f(a)}{\Delta z}$$
tends to  unique limit  as $\Delta z\rightarrow$0   as i.e. z $\rightarrow$0 along any path of domain $\mathrm D$
my question is what is the meaning of "along any path" of the domain $\mathrm D$ in the given defination of differentiablity of complex function
 A: You may know from ordinary calculus that a real function is differentiable at a point $a$ if and only if the limit
$$ \lim_{\Delta x \to 0} \frac{f(a+\Delta x) - f(a)}{\Delta x} $$
exists. In 1D, there are only two directions you can approach $0$ from, and that is from the left and from the right. You may recall that to prove a limit exists, you have to show that the left- and right-hand limits exist and are equal.
In 2D, the idea is similar, but instead of two directions you have to check, you now have to check infinitely many (i.e. you have to check that it works if you approach from the north, south, east, or west, or any other direction). This is obviously hard to do! Although you can often disprove a limit exists by showing that the limit along one path does not equal the limit along another path.
$\newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}}
\renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}}
\newcommand{\eps}{\varepsilon}$
For example: consider the function $f(z) = z/|z|$. We can take the limit along a path from the positive real line like so
\begin{align}
\lim_{z \in \R^+ \to 0} \frac{z}{|z|} &= \lim_{x \to 0^+} \frac{x}{|x|} = \lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} 1 = 1
\end{align}
and along the positive imaginary line like so
\begin{align}
\underset{\Im(z) > 0}{\lim_{z \to 0}} \frac{z}{|z|} &= \lim_{y \to 0^+} \frac{yi}{|yi|} = \lim_{y \to 0^+} \frac{yi}{|y|} = i \lim_{y \to 0^+} \frac{y}{|y|} = i \cdot 1 = i
\end{align}
The two limits are different, hence the general limit $\lim_{z \to 0} z/|z|$ does not exist.
If you want to prove a limit does exist, the most general way to do so is to get a guess as to what you think the limit should be, and then attempt to prove it using the definition of a limit.
i.e. If you think $\lim_{z \to z_0} f(z) = L$ you need to show that for any positive number $\eps$ (no matter how small): you can find another positive number $\delta$ such that whenever $0 < |z-z_0| < \delta$ (that is, whenever $z$ is within a $\delta$-radius of $z_0$), we have that $|f(z)-L| < \eps$ (that is, the image of $z$ is within an $\eps$-radius of $L$).
That's the definition, but often times when computing the derivative limit, you don't have to go thru all that trouble. If you have a continuous function, then you know that
$$ \lim_{z \to z_0} f(z) = f(z_0) $$
So if you wanted to compute the complex derivative of $f(z) = z^2$ at $z=a$ the easy(-er) way, then you could manipulate the increment ratio like so
\begin{align}
\frac{(a+\Delta z)^2 - a^2}{\Delta z} &= \frac{a^2 + 2a\Delta z + \Delta z^2 - a^2}{\Delta z} \\
&= \frac{2a\Delta z + \Delta z^2}{\Delta z} \\
&= 2a + \Delta z
\end{align}
This is a continuous function! Hence
$$ \lim_{\Delta z \to a} (2a + \Delta z) = 2a + 0 = 2a $$
