To be a bit more precise, we say a function f(z) is analytic at a point z0 if it can be represented by a power series in a neighborhood about that point:
f(z) = \sum_{n=0}^\infty a_n (z-z_0)^n
For example, take f(z)=1/(1-z). We can come up with a power series for f(z) expanded about z0=0, namely
\frac{1}{1-z} = \sum_{n=0}^\infty z^n
so f(z) is analytic at z=0. Similarly, we can also find a power series for f(z) about the point z=3,
\frac{1}{1-z} = \frac{1}{-2-(z-3)} = -\frac{1}{2(1+(z-3)/2)} = -\frac{1}{2} \sum_{n=0}^\infty (-1)^n\left(\frac{z-3}{2}\right)^n
so f(z) is analytic at z=3. In fact, you can find a power series about every point in the complex plane except for z=1, so this function is analytic everywhere except z=1.
Note that asking if a function is analytic at some point z is not the same as asking if a particular series for the function is valid at z. The fact that the first series above doesn't converge at z=3 says nothing about whether f(z) is analytic there.
So say you have a function f(z) and you expand it in a power series about the point z0. This expansion is good for some neighborhood of z0, so you'd like to know how big this neighborhood actually can be. The answer is that the series is only going to be valid up until you run into a singularity. The distance between z0 and the singularity is the radius of convergence of the series. This is the idea that your professor was trying to get across to you.
For example, the function f(z)=1/(1-z) is singular at z1=1. The first series was for f(z) expanded about z0=0, so its radius of convergence is |z1-z0|=1. The second series was for f(z) expanded about z0=3, so the radius of convergence for that series is 2, the distance from z0=3 to the singularity at z=1.
Unlike a power series, a Laurent series includes negative powers of (z-z0). This allows you to express f(z) as a series about z0 that converges in a region where the power series won't. Again, if we consider f(z)=1/(1-z) expanded about z0=0, we have
\frac{1}{1-z} = \sum_{n=0}^\infty z^n
for the region |z|<1, and
\frac{1}{1-z} = -\frac{1}{z}\sum_{n=0}^\infty \left(\frac{1}{z}\right)^n
for the region |z|>1. The different regions will have different Laurent series.
Determining the regions of convergence is similar to finding the radius of convergence of a power series: you look at where the singularities are. For example, take the function
g(z)=\frac{1}{(z-1)(z+4)}
It has singularities at z=1 and z=-4. Say we want to expand about the point z0=0. The closest singularity is at a distance of 1, and the other one is at a distance of 4. So you'll have three regions to worry about: |z|<1, 1<|z|<4, and |z|>4.
Suppose you want to find series about the point z0=2. The closest singularity is z=1, which is at a distance of 1 from z0=2, and the other singularity is at a distance of 6 from z0=2. So, in this case, the regions are: |z-2|<1, 1<|z-2|<6, and |z-2|>6.
