I know that counter examples exist, so this must be wrong. I just want to know what's wrong with this reasoning:
If we consider the limit:
$$L=\lim_{h\rightarrow 0} \frac{f(z+h(\cos{p}+i\sin{p})) - f(z)}{h(\cos{p}+i\sin{p})}$$
If this limit is independent of $p$, then clearly, the complex function $f$ has the same derivative for all straight-line approaches to the point $z$.
Now, I want to show that, if the straight line approaches give the same derivative, then approaching the point $z$ along any other curve must also give the same derivative (this has to be wrong because counter examples exist).
Let $z=a+ib$. Suppose this point lies on the curve $y=g(x)$. If we change $a$ by a small amount $h$ along the curve, then we arrive at the point $a+h+i(g(a+h))$. Then the derivative by approaching the point along this curve should be:
$$L'=\lim_{h\rightarrow 0} \frac{f(z+h+i(g(a+h)-g(a)))-f(z)}{h+i(g(a+h)-g(a))}$$
Consider $h+i(g(a+h)-g(a))$. This equals $h'(\cos{d}+i\sin{d})$, where $h'=\sqrt{h^2+(g(a+h)-g(a))^2}$, and $d=\arctan{\frac{g(a+h)-g(a)}{h}}=\arctan{g'(a)}$. As $h$ tends to zero, $h'$ also tends to zero. So, limit $L'$ becomes:
$$\lim_{h'\rightarrow 0} \frac{f(z+h'(\cos{d}+i\sin{d})) - f(z)}{h'(\cos{d}+i\sin{d})}$$
This is the same limit as limit $L$.
