Holomorphic function is continuous?

jaci55555
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Prove that any f: D -> C(complex) which is holomorphic in D subset of C is continuous in D



f is holomorphic in D if it is differentiable at every c element of D.
A function is differentiable at c if lim(h->0) (f(c+h) - f(c))/h exists.



I know from reals that a function is only differential if it is continuous... but not for complex numbers.

I have no idea how to prove a function continuous without drawing a graph or using the epsilon-delta method. Please help!
 
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So you know that a real differentiable function is continuous. How did you prove that? Doesn't the exact same proof carry over to this case?
 
I didn't prove it, it was given. But I will definitely go try find a proof for it. Any other help will be greatly appreciated.
 
The following observation will be useful:

f(y) - f(x) = \frac{f(y) - f(x)}{y - x} (y - x)

valid for y \neq x.
 
What can you say about the partial derivatives of a complex function in the neighborhood of a point where that function is not continuous?
 
holomorphic said:
What can you say about the partial derivatives of a complex function in the neighborhood of a point where that function is not continuous?
Nothing yet... what should i be able to say?

I tried it like this:

THere is an open D subset of C with c element of D and f holomorphic in D. F is holomorphic at every c element of D

lim(z-> c) = [f(z) - f(c)] = lim(z->c)[(f(x) - f(z))/(z-c) * (z-c)] = lim(z->c) [(f(z) - f(c))/(z-c)] * lim(z->c)[z-c] = f'(c) * 0 = 0

Therefore f is continuous in D
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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