Why does the Laurent series require negative powers in order to represent complex functions? I briefly studied Laurent Series during a Complex Analysis course and was told that they are a generalisation of the Taylor Series. The intuition behind why the Taylor Series can represent any smooth function is clear to me, though I do not understand why the Laurent Series requires negative powers to represent complex functions and I don't believe this was ever explained to us. Any assistance would be great!
 A: If $f$ is meromorphic and have a pole at $a$ it means that $f(z)=(z-a)^{-n} g(z)$ for some $n\in \mathbb{N}$ in a neighborhood of $a$. There $g$ is holomorphic and $g(a)\neq 0$.
As $g$ is holomorphic it have a Taylor expansion in this neighborhood, say $g(z)=\sum_{k\geqslant 0}c_k (z-a)^k$, then we will have
$$
f(z)=\sum_{k\geqslant 0}c_k (z-a)^{k-n}
$$
what is known as the Laurent series of $f$ at $a$.
EDIT: I will sketch the case when $f$ is neither holomorphic nor meromorphic but it is differentiable in a neighborhood of a point and at this point the function is undefined.
Let $f$ holomorphic at $\mathbb{C}\setminus \{a\}$, and suppose that there is no $n\in \mathbb{N}$ such that $(z-a)^n f(z)$ is holomorphic in a neighborhood of $a$, then we says that $f$ have an essential singularity at $a$. In this case it can be shown that we can expand $f$ as a pointwise convergent power series centered at $a$ with infinitely many negative powers, this will be called the Laurent series of $f$ at $a$, it will be of the form
$$
f(z)=\sum_{k\in \mathbb{Z}}c_k (z-a)^k
$$
where infinitely many $c_k$ with $k<0$ will be nonzero.
However this case is far from intuitive, not as the previous case of a meromorphic function, and I will not expand on this matter more here.
A: Taylor series only describe the behavior of functions in a neighborhood of points where the function is actually defined. For instance, $\sum_{k=0}^\infty\frac{z^k}{k!}$ is a Taylor expansion of $\exp(z)$ around $z_0=0$. The point $z_0$ is a point is inside the domain of $\exp$. But functions like $f(z)=\frac1z$ or $g(z)=\exp(\frac1z)$ don't have Taylor expansions at $0$, since they have singularities there: points where the functions aren't defined.
However, we often want to understand the behavior of functions close to such singularities. For instance, singularities play an essential part in the theory of contour integrals. In this context, it turns out that any function $f$ that is holomorphic in a neighborhood of a point $z_0$, but has a singularity at $z_0$, can be split into an analytic and a singular part $f=g+h$, where the analytic part $g$ is a function which can be holomorphically extended to $z_0$, and the singular part $h$ has a singularity at $z_0$, but vanishes far away from $z_0$. For instance, the function $f(z)=\frac{1+z^2}{z}$ splits into $z+\frac1z$, where $z$ is holomorphically extendable to $0$, while $\frac1z$ has a singularity there, but vanishes at infinity.
Now the analytic part's behavior around the singularity can be described by a plain old Taylor series. But the singular part cannot. However, it turns out that the singular part can be described by something similar to a Taylor series, but where all the smooth powers of $z-z_0$ are replaced by the singular powers (that is, those with negative exponents). And then we can explain the behavior of the singularity using this new kind of series, which is itself singular at $z_0$.
