are these two terms equivalent: analytic and differentiable?
No.
Counter example: $\displaystyle f(z) = \left | z \right|^2$.
A function f of the complex variable z is analytic at a point $\displaystyle z_{0}$ if its derivative exists not only at $\displaystyle z_{0}$ but at each point z in some neighbourhood of $\displaystyle z_{0}$.
The function $\displaystyle f(z) = \left | z \right|^2$ is not analytic at any point since its derivative exists only at $\displaystyle z = 0$ and not throughout any neighbourhood (proof available upon request). So it's differentiable at $\displaystyle z = 0$ but not analytic at $\displaystyle z = 0$.
now, after reading this, i got confused..
Complex Differentiable -- from Wolfram MathWorld
Analytic Function -- from Wolfram MathWorld
you don't have to prove.. i know how to do it.. thanks anyways..
A function that is analytic is also complex differentiable. But a function that is complex differentiable may not always be analytic.
In order for a complex differentiable function to be analytic it must be complex differentiable at every point of the region in question.
thanks for that but it doesn't help much since you just stated the definition..
to give you idea of what i am talking about, for a given function, you are to show that the Cauchy-Riemann equations is satisfied at a point, say z0 but also show that it is not differentiable at that point.
consider this thread.. http://www.mathhelpforum.com/math-he...-analysis.html
basically, the question is similar except at the last. instead of showing that it is not holomorphic, you have to show that it is not differentiable..
A function is differenciable at a point $\displaystyle z_0$ iff it is defined in the neighborhood of this point and $\displaystyle \lim_{z\to z_0} \tfrac{f(z)-f(z_0)}{z-z_0}$ exists. A function is analytic at a point $\displaystyle z_0$ iff there is a neighborhood around $\displaystyle z_0$ such that the function is differenciable at all those point in the neighborhood. The notion of a function being differenciable at only one point and nowhere else around the neighborhood really does not appear in complex analysis. Most functions that are considered are also differenciable in the neighorhood as well. It turns out that if a function is analytic then it must be infinitely differenciable and the Taylor series must converge to the function. One of the many shocking results in complex analysis. Therefore, the term "analytic" is consistent with the other meaning of the word "analytic" i.e. having a power series expansion.
where? this one?
he showed that the function is not differentiable at a neighborhood and then concluded not analytic..No.
Counter example: .
A function f of the complex variable z is analytic at a point if its derivative exists not only at but at each point z in some neighbourhood of .
The function is not analytic at any point since its derivative exists only at and not throughout any neighbourhood (proof available upon request). So it's differentiable at but not analytic at .
what if, i found it not analytic at the point? i can conclude that there exists a neighborhood of the point that is not differentiable, right? but, can i conclude that it is not differentiable at the point?
(waah, i am tempted to put the original question i have, but due to intellectual honesty, i can't! )
EDIT: this is the original question but i did not include the function and the point..
Given this function f(z) = ????, show that it is continuous and satisfies Cauchy-Riemann equations at z=???? but is not differentiable there.
i am having trouble with the last part, that is to show that it is not differentiable there..
what i did, i showed that it is not analytic at that point. can i conclude that it is not differentiable there? if not, what can i do?
thanks a lot!!