# Computing pole order of $f(z)=\frac{e^z}{(z+1)^3(z-2)}$ at $z_0=2$

Determine the singularity type of $$f(z)=\frac{e^z}{(z+1)^3(z-2)}$$ at $$z_0=2$$.

$$\lim_{z\to -2}\frac{e^z}{(z+1)^3(z-2)}=\infty$$

so the singularity is pole.I know the solution where it states it is a pole of order 1. However I cannot see how I can find that out. If I take $$w=z-2$$ then the Laurent series:

$$\frac{1+(w-1)+\frac{(w-2)^2}{2!}\frac{(w-2)^3}{3!}+\frac{(w-2)^4}{4!}...}{(w-1)^3 w}$$

But I cannot see how the pole order is going to be 1.

Question:

Can someone help me computing the order of the pole?

If $$f$$ has a pole at $$z_0$$ then the order of the pole is the smallest positive integer $$k$$ with the property that $$(z-z_0)^k f(z)$$ has a removable singularity at $$z=z_0$$. Or equivalently, the unique positive integer $$k$$ with the property that $$\lim_{z \to z_0}(z-z_0)^k f(z) \text{ exists and is not zero.}$$

In your case, $$g(z) = (z-2)^1f(z) = \frac{e^z}{(z+1)^3}$$ is holomorphic in a neighborhood of $$z_0=2$$ with $$g(2) \ne 0$$, which shows that $$f$$ has a pole of order $$1$$ at $$z_0=2$$.

Or in terms of Laurent series: $$g$$ has a Taylor series $$g(z) = b_0 + b_1(z-2) + b_2 (z-2)^2 + \ldots$$ at $$z_0 = 2$$, with $$b_0 = g(2) \ne 0$$, so that $$f(z) =\frac{b_0}{z-2 } + b_1 + b_1 (z-2) + \ldots$$ is the Laurent series of $$f$$ at $$z_0=2$$.

You can also argue that $$\frac{1}{f(z)} = (z-2) h(z)$$ where $$h(z) = \frac{(z+1)^3}{e^z}$$ is holomorphic and not zero in a neighborhood of $$z_0=2$$, so that $$1/f$$ has a simple zero at $$z_0=2$$.

At $$z= 2$$ , only the term $$z-2$$ makes the function indefinite, whereas, $$e^z \ and \ (z+1)^3$$ give finite values at $$z=2$$ viz., $$e^2 \ and \ 27$$. So we need to consider only $$z-2$$ which is of order 1. Thus the pole is of order 1.

Similarly if we see at $$z= -1$$, only the term $$(z+1)^3$$ makes the function indefinite as $$e^{-1}$$ and $$-1-2 = -3$$ are finite. Here the root(pole) occurs thrice. So in this case, the pole is of order 3.

As another example, consider

$$g(z) = \frac{1}{z(e^z - 1)}$$

At $$z_0 = 0$$, both the terms $$z = 0$$ and $$e^0 - 1 = 1-1 = 0$$ make the function indefinite (as the denominator becomes zero) . So the pole is of order 2.

• Perhaps it would be helpful to mention that at $z=-1$ the pole is of order $3$. It isn't clear to me exactly where OP is confused. – Ryan Goulden May 4 at 16:30
• @RyanGoulden Thanks, I'll add it. – Ak19 May 4 at 16:31