Why don't we approach the point angularly to find the derivative of Complex functions? In case of Real functions, we approach the point from left and right and see if both derivatives are equal. But in case of Complex functions, the point can be approached from infinitely many directions. Then how can we check that the derivatives from all the directions are equal?
I think in case of Complex functions, we should approach the point angularly. For example, to find a derivative at a point $z$, we can rotate through an angle $d\theta$ about the origin clockwise or anti-clockwise and then find the difference in the value of the function, then divide by the angle change and apply the limit $d\theta \rightarrow0$. In this case, we'll have clock-wise and anti-clockwise derivatives and we can see if they're equal. So, if the argument of $z$ is $\arg{(z)}$, then the derivative would be:
$$\lim_{d\theta \rightarrow0}\frac{f(|z|(\cos{(\arg{(z)}+d\theta})+i\sin{(\arg{(z)}+d\theta})))-f(z)}{d\theta}$$
I think approaching the point angularly should be more convenient than approaching the point from all directions. So, why do we approach the point linearly even in case of Complex functions?
EDIT: I've one more idea, we can even approach the point from outward and inward directions along the line joining the point with the origin. So, we have outward derivative and inward derivative.
In this case the angle remains constant but the modulus changes by an infinitesimal amount. It will be:
$$\lim_{dz\rightarrow 0}\frac{f((|z|+dz)(\cos{(\arg{(z)})}+i\sin{(\arg{(z)})})-f(z)}{dz}$$
So, if a Complex function is given as a function of modulus ($|z|$) and argument ($\theta$ )instead as a function of the co-ordinates, then my 'angluar' derivative is the partial derivative of the function w.r.t $\theta$ and my 'modular' derivative is the partial derivative of the function w.r.t. $|z|$.
Together, both my angular and modular derivatives can get from value of the function to another.
For example, let $f(|z|,\theta)$ be a complex function. Let it's partial derivative w.r.t $\theta$ be $a(|z|,\theta)$ and the partial derivative w.r.t $|z|$ be $m(|z|,\theta)$. Let $z_1$ and $z_2$ be two points. Then,
$$f(z_2)=f(z_1)+\int_{\arg{(z_1)}}^{\arg{(z_2)}}a(|z_1|, \theta)d\theta+\int_{|z_1|}^{|z_2|}m({|z|, \arg{(z_2)}})d|z|$$
Also, what if the definition of the complex derivative is $$f'(z) = \lim_{w\to z} \frac{f(w)-f(z)}{w-z}$$
That doesn't make a difference. Problem is, how do we evaluate that limit? If we substitute $w=z+h$, then that's the same thing as approaching $z$ from the right along the line parallel to the the real axis passing through $z$. If we substitute $w=z+ih$, then that's the same as approaching $z$ from upward along the line parallel to the imaginary axis passing through $z$. If we substitute $w=z+h(cos45+isin45)$, then that's the same as approaching $z$ from North-East along the line passing through $z$ and inclined 45 degrees to the real axis. But how do we know that all those approaches are equal? If they are, only then the derivative exists.
 A: Note that a multivariable function may have a derivative in every linear direction but yet have no tangent plane at the point! So even the idea of approaching the point at every angle is logically incorrect because it is insufficient to guarantee differentiability. One example of such a function is:
$
\def\lfrac#1#2{{\Large\frac{#1}{#2}}}
$

Let $f(x,y) = \cases{ \lfrac{x}{y} \sqrt{x^2+y^2} & if $y \ne 0$ \\ 0 & if $y = 0$}$.
Then $\lfrac{d(f(rt,st))}{dt}$ exists for every pair of reals $(r,s)$ such that $r^2+s^2 = 1$.
For instance if $s \ne 0$ then $\lfrac{d\Big(\lfrac{rt}{st}\sqrt{(rt)^2+(st)^2}\Big)}{dt} = \lfrac{r}{s}$, which in particular holds at $t = 0$.
Thus $f$ has all directional derivatives (and is in fact linear in every direction).
But $f$ has no tangent plane at the origin and is hence not differentiable there.

I'll emphasize that complex differentiability is much stronger than multivariable differentiability, but at least you should realize that it's wrong to think of differentiation as simply approaching in all directions.
A: We do indeed approach points "angularly". The definition is
$$
f'(z) = \lim_{w\to z} \frac{f(w)-f(z)}{w-z}.
$$
That means $w$ can approach $z$ from any direction at all, or can follow a spiral toward $z$ so that you can't identify a direction from which it approaches, and no matter how it approaches, the limit must be the same; otherwise the derivative does not exist at all.
Consider, for example, this:
$$
f(z) = \begin{cases} e^{-1/z^2} & \text{if  }z\ne0, \\ 0 & \text{if } z=0. \end{cases}
$$
If we restrict all attention to real numbers, then this is differentiable at $0$. In fact, it is posisble to show it has derivatives of all orders at $0$ (but that takes a lot more work). However, consider what happens as $w$ approaches $0$ from the imaginary direction. We then conclude that $f$ is not differentiable at $0$ as a function of a complex variable.
The exclusion of functions like this from among those that are differentiable as functions of a complex variable is the reason why the only functions differentiable as functions or a complex variable are quite well behaved, so that some very strong theorems apply to them that do not apply to functions of a real variable.
A: Limits of complex functions approach from "all directions" because we want continuity to mean that $z \approx w$ implies $f(z) \approx f(w)$. You don't get that property if you fix a particular direction.
To wit, let $g$ be any discontinuous function of the nonegative real numbers. Then the function $f$ defined by $f(z) = g(|z|)$ would be 'angularly' continuous,  but it would not be continuous.

That said, it only looks like infinitely many directions when you pretend the complexes are a real plane; in some sense there is only one "complex" direction, and many basic limits are still obvious to compute; e.g.
$$ \lim_{h \to 0} \frac{(z+h)^2 - z^2}{h} = \lim_{h \to 0} \frac{2zh + h^2}{h} = \lim_{h \to 0} (2z + h) = 2z $$
A: If there is $dx$ in denominator, you are taking $x$ in two directions.If there is $d\theta$ in denominator, you are taking $\theta$ in all directions.OK? Then why are you switching back along a line in polar coordinates still?
A: There is one obvious problem with this idea: it doesn't give a sensible definition of "a derivative" at the point $z = 0$. No doubt that could be fixed up in some arbitrary fashion.
A bigger, but less obvious problem, is that this definition won't have many useful properties. The important point about the standard definition is that the limit is the same for any approach to the point $z$, and because of that, you can form a power series approximation to the function (analogous to a Taylor series for real variables) which is valid for all points in a region around $z$, not just for points lying on some particular line in the complex plane.
This leads to the concept of an "analytic function," which is one of the most important theoretical and practical ideas in Complex Analysis. The standard definition of a complex derivative may seem a bit peculiar, but it turns out that it divides complex functions into those that are "well behaved" (and in fact much better behaved than one might initially have guessed) and those that are not well behaved at all.
Of course, for analytic functions, the OP's definition will give the same derivative as the standard definition, but it won't answer the important question of whether the function is analytic or not.
