For checking differentiability of real functions, we simply use the limit f(x+h)-f(x) divided by h. As h tends to 0 the sequence f(x+h) approaches f(x). But there are other sequences too which approach f(x). For example, f(x)h as h tends to 1, or f(x+h^{2} ) as h tends to 0. But we don’t bother to check if those sequences of derivatives approach the same value.

In the complex case, we can have this limit: f(z+h(cosp+isinp))-f(z) divided by h(cosp+isinp). As h tends to 0, this limit represents approaching the point z along a straight line at an angle p. If this limit is independent of p, then the limit is the same in all directions, so the function should be differentiable at the point.

But, I came to know that that limit is not enough to show complex differentiability. In the complex case, we also have to consider approaching the point along a parabola, spiral, and, in fact, along every curve approaching z. All those limits must be the same for the derivative to exist.

Why is that the case? Doesn’t the limit: f(z+h(cosp+isinp))-f(z) divided by h(cosp+isinp) being independent of p already show that the derivative is same in all directions of approach?

submitted by /u/fbi767

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