Power Series Expansion about Point

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Expanding a function into a power series about a point, such as z = 4 + i, involves using the formula for power series expansion for complex variables. The general form is f(z) = ∑(n=0 to ∞) [f^(n)(a)/n!](z-a)^n, where 'a' is the point of expansion and 'f(z)' is the function being expanded. In the given example, the function is f(z) = 1/(e^z - 1) with a = 4 + i. This method applies to analytic functions, ensuring the series converges in a neighborhood around the point of expansion. Understanding this concept is essential for working with complex analysis and power series.
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So this is a REALLY elementary question but I can't seem to find the answer on the net, or maybe I did but just keep skipping over it some how. (by the way, this is with respect to complex numbers z \in C which is used in Complex Analysis, thus why I chose this forum). I know what it means when someone gives a function and says "Expand into a power series about z", but what does it mean when they say to expand about, say, z = 4 + i for a function like (e^{z} - 1)^{-1}.

I don't need to know the answer, just the general formula for something like that. Again, elementary I know. Thanks! Any help appreciated.
 
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In a regular power series expansion, when you expand the function f(x) about a point a, you get the expansion

f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n.
(when it exists and all that jazz).

If f is a function of a complex variable z, then for analytic functions you get the same form of the series expansion,

f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(z-a)^n.
where both z and a may be complex numbers. In your example, a = 4-i and f(z) = 1/(exp(z)-1).
 

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