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.


Can someone help me computing the order of the pole?

Thanks in advance!


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.

  • $\begingroup$ 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. $\endgroup$ – Ryan Goulden May 4 at 16:30
  • $\begingroup$ @RyanGoulden Thanks, I'll add it. $\endgroup$ – Ak19 May 4 at 16:31

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