By the way, if f has a pole at z0, then you could look at 1 over f,

and then that function will have a removal singularity at z0 and vice versa.

Why is that the case?

Well, if f is a pole,

we know that means that the absolute value of f goes to infinity, as z approaches z0.

Our Riemann's theorem, we learned about removable singularities.

If I look at 1 over f(z), because f(z) goes to infinity,

1 over f (z) must go to 0, and that's clearly a dominant function here, is z0,

in Riemann's theorem that means I'm dealing with a removable singularity.

Finally, let's look at essential singularities.

Remember that an essential singularity means that the Laurent series of f

has infinitely many negative powers of z minus z0.

Here's an example, f (z) equals e to the 1 over z.

How do I find the Laurent series?

We remember the Taylor series for the exponential function.

e to the w = 1 + w + w squared over 2 factorial

+ w cubed over 3 factorial, and this is actually true for any w.

So this is the Taylor series centered 0.

If you plug in w =1 over z,

then we find e to the 1 over z = 1 + 1 over z.

We simply replace every w with a 1 over z.

So 1 over 2 factorial z squared and so forth.

And so that's the series that you see right here.

Fully there are infinite negative powers of z in that series.

Again, because the Laurent series is unique and because it just found

one series this must be the Laurent series for e to the 1 over z.

So we see that this series has infinitely many negative powers of z,

therefore, e to the 1 over z has an essential singularity at the origin.

Let's look at what happens when z approaches 0.

What happens to e to the 1 over z?

Let's start with zs approaching the origin along the x- axis.

And in fact, why don't we start by approaching from the right?