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I require some help with identifying and classifying the isolated singularities of complex-valued functions.

  • For example: $f(z)=\frac{(z^3+1)}{z^2(z+1)}$, here I understand that there is a pole of order 2 @ $z=0$, and a removable singularity @$z=1$, but this is really only from intuition, and would appreciate some clarification on how to formally demonstrate this.
  • Another example: $f(z)=\frac{cos(z)}{z^2+1}+4z$, here the singularities are less obvious to me, and further I am particularly confused by $cos(z)$.

So far I am inclined to use a Laurent series, however defining an $R$ is unclear to me, as is how to simply for intricate functions as mentioned above. Thanks.

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Note that we obtain these functions by dividing entire functions. The quotient rule tells us that we will get differentiability everywhere the functions are defined: i.e. where the denominators are $0$. That is, our functions will be analytic everywhere except where the denominators are $0$.

So, in the first case, this is everywhere except $z = 0$ and $z = -1$ (not $z = +1$ - careful!). In the second case, this is everywhere except $z = \pm i$.

To show that the singularity $z_0$ of a function $f$ is a pole and compute its order, you multiply $f(z)$ by some power of $z - z_0$ until the singularity becomes removable (note that this isn't always possible). So, in the case of the first function with $z = 0$, we multiply $$z f(z) = \frac{z^3 + 1}{z(z + 1)},$$ which is not removable at $0$ since the numerator tends to a non-zero number as $z \to 0$. If we try a higher power, $$z^2 f(z) = \frac{z^3 + 1}{z + 1},$$ on its domain, which is $z \neq 0, -1$. Note that the resulting function can be extended to an analytic function at $z = 0$. Thus, the singularity is removable at $z = 0$, and the pole of $f$ at $z = 0$ was order $2$.

For the second function, note that the singularities at $z = \pm i$ are not removable, since $$\cos(z) \to \frac{e^{-1} - e}{2} \neq 0$$ as $z \to i$, and similarly for $z \to -i$. However, multiplying $f$ by $z - i$ or by $z + i$ will make the singularities removable, hence the singularities are poles of order $1$.

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