Ambiguous function pole definition












1












$begingroup$


As I have seen on the Internet each definition of a complex function pole says, that a pole is a point such that $$lim_{zto z_0} f(z) = infty$$
But what if this limit doesn't exist? What if
$$
begin{cases}
limlimits_{z to z_0 + 0} f(z) = infty \
limlimits_{z to z_0 - 0} f(z) = -infty end{cases}
$$

Does this mean, that $z_0$ is the pole of $f(z)$?
For example when I investigate
$$
f(z)=frac{1}{(z^2-4)(z^4-1)}
$$

All $limlimits_{zto z_k} f(z)$ where $z_k in { pm i, pm1, pm2}$ do not exist (as they give $pminfty$ depending on the way we approach those points), but doing a query on WolframAlpha says that all those points are function's poles.
Wolfram's output










share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you meany by $ z to $z_0 -$ and $ z to z_0 +$?
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 23:36












  • $begingroup$
    @KaviRamaMurthy Sorry, I am not an expert but I mean that when we approach $z_0$ on $Re z$ line from positive or negative zero we get different signs for infinity after taking the limit.
    $endgroup$
    – Veetaha
    Dec 19 '18 at 23:40


















1












$begingroup$


As I have seen on the Internet each definition of a complex function pole says, that a pole is a point such that $$lim_{zto z_0} f(z) = infty$$
But what if this limit doesn't exist? What if
$$
begin{cases}
limlimits_{z to z_0 + 0} f(z) = infty \
limlimits_{z to z_0 - 0} f(z) = -infty end{cases}
$$

Does this mean, that $z_0$ is the pole of $f(z)$?
For example when I investigate
$$
f(z)=frac{1}{(z^2-4)(z^4-1)}
$$

All $limlimits_{zto z_k} f(z)$ where $z_k in { pm i, pm1, pm2}$ do not exist (as they give $pminfty$ depending on the way we approach those points), but doing a query on WolframAlpha says that all those points are function's poles.
Wolfram's output










share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you meany by $ z to $z_0 -$ and $ z to z_0 +$?
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 23:36












  • $begingroup$
    @KaviRamaMurthy Sorry, I am not an expert but I mean that when we approach $z_0$ on $Re z$ line from positive or negative zero we get different signs for infinity after taking the limit.
    $endgroup$
    – Veetaha
    Dec 19 '18 at 23:40
















1












1








1


1



$begingroup$


As I have seen on the Internet each definition of a complex function pole says, that a pole is a point such that $$lim_{zto z_0} f(z) = infty$$
But what if this limit doesn't exist? What if
$$
begin{cases}
limlimits_{z to z_0 + 0} f(z) = infty \
limlimits_{z to z_0 - 0} f(z) = -infty end{cases}
$$

Does this mean, that $z_0$ is the pole of $f(z)$?
For example when I investigate
$$
f(z)=frac{1}{(z^2-4)(z^4-1)}
$$

All $limlimits_{zto z_k} f(z)$ where $z_k in { pm i, pm1, pm2}$ do not exist (as they give $pminfty$ depending on the way we approach those points), but doing a query on WolframAlpha says that all those points are function's poles.
Wolfram's output










share|cite|improve this question











$endgroup$




As I have seen on the Internet each definition of a complex function pole says, that a pole is a point such that $$lim_{zto z_0} f(z) = infty$$
But what if this limit doesn't exist? What if
$$
begin{cases}
limlimits_{z to z_0 + 0} f(z) = infty \
limlimits_{z to z_0 - 0} f(z) = -infty end{cases}
$$

Does this mean, that $z_0$ is the pole of $f(z)$?
For example when I investigate
$$
f(z)=frac{1}{(z^2-4)(z^4-1)}
$$

All $limlimits_{zto z_k} f(z)$ where $z_k in { pm i, pm1, pm2}$ do not exist (as they give $pminfty$ depending on the way we approach those points), but doing a query on WolframAlpha says that all those points are function's poles.
Wolfram's output







complex-analysis limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 20:17







Veetaha

















asked Dec 19 '18 at 20:10









VeetahaVeetaha

84




84












  • $begingroup$
    What do you meany by $ z to $z_0 -$ and $ z to z_0 +$?
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 23:36












  • $begingroup$
    @KaviRamaMurthy Sorry, I am not an expert but I mean that when we approach $z_0$ on $Re z$ line from positive or negative zero we get different signs for infinity after taking the limit.
    $endgroup$
    – Veetaha
    Dec 19 '18 at 23:40




















  • $begingroup$
    What do you meany by $ z to $z_0 -$ and $ z to z_0 +$?
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 23:36












  • $begingroup$
    @KaviRamaMurthy Sorry, I am not an expert but I mean that when we approach $z_0$ on $Re z$ line from positive or negative zero we get different signs for infinity after taking the limit.
    $endgroup$
    – Veetaha
    Dec 19 '18 at 23:40


















$begingroup$
What do you meany by $ z to $z_0 -$ and $ z to z_0 +$?
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 23:36






$begingroup$
What do you meany by $ z to $z_0 -$ and $ z to z_0 +$?
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 23:36














$begingroup$
@KaviRamaMurthy Sorry, I am not an expert but I mean that when we approach $z_0$ on $Re z$ line from positive or negative zero we get different signs for infinity after taking the limit.
$endgroup$
– Veetaha
Dec 19 '18 at 23:40






$begingroup$
@KaviRamaMurthy Sorry, I am not an expert but I mean that when we approach $z_0$ on $Re z$ line from positive or negative zero we get different signs for infinity after taking the limit.
$endgroup$
– Veetaha
Dec 19 '18 at 23:40












2 Answers
2






active

oldest

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$begingroup$

On the complex plane it does not make sense to talk about $-infty$ because there is no ordering on the complex numbers. Indeed, assume $i > 0$ then $i^2=-1>0$ and similarly for the other side. So in a sense $infty$ can be seen as what any unbounded complex functions converges to.



Yes, your $f(z)$ does have a pole at your described points. In a sense having a pole of order $N$ at $z_0$ can be interpolated that the functions behaves like begin{equation} frac{1}{(z-z_0)^N} end{equation} near $z_0$ which can be made precise by saying begin{equation} f(z) = frac{1}{(z-z_0)^N} g(z) end{equation} where $g(z)$ is holomorphic on a small neighborhood of $z_0$. This can be done to your $f(z)$ by polynomial division.



But it seems like you're confused about isolated singularities. So I'll give a brief overview of the three types:



A removable singularity at $z_0$ is one where $lim_{z rightarrow z_0} (z-z_0)f(z) = 0$ In particular, this is called removable because one can redefine a $f(z)$ to be holomorphic at $z_0$ using Cauchy's Integral Formula representation. So this is the most mild type of singularity because essentially it's not one. For example $f(z) = z$ on $B(varepsilon,0) - 0$ But defining $f(0)=0$, we see $f$ extends to be holomorphic on the entire ball.



A pole is what I described above, which can be characterized by $lim_{z rightarrow z_0} f(z) = infty$ (recall $-infty$ does not make sense on $mathbb{C}$) An example is any rational function with a zero on the denominator that is not canceled out by the numerator's zero.



The worst type of singularity is an essential singularity which can be seen as an isolated singularity that is neither a pole or removable, which is equivalent to begin{equation} limsup_{z rightarrow z_0} f(z) = infty end{equation} begin{equation} liminf_{z rightarrow z_0} |f(z)| < infty end{equation}
Some powerful theorems are that if $f$ is holomorphic on the punctured disk of $z_0$ then it's image on any $delta$ neighborhood is dense in $mathbb{C}$. A stronger refinement of this theorem is Picard's Big Theorem which states if $f$ is holomorphic on the punctured disk with an essential singularity at $z_0$ then it must touch every point on $mathbb{C}$ except at most one infinitely often. An example is $f(z)=e^{1/z}$. In fact every holomorphic function that is entire and non-polynomial has an essential singularity at $infty$.



Another way to look at them is their Laurent Series. A removable singularity has no $a_{-n}$ terms in its Laurent series, a pole has finitely many $a_{-n}$ terms, while an essential singularity has infinitely many negative $a_{-n}$ terms.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Do not trust random people from the internet, trust me instead!



    A singularity is a point where a function is not defined. In complex analysis, we can say a lot more about singularities though.



    A holomorphic function is a complex differentiable function $f:Utomathbb{C}$ for some open set
    $Usubsetmathbb{C}$, meaning it is complex differentiable at every point in $U$. For example $frac1z$ is not holomorphic on $mathbb{C}$, but it is on $mathbb{C}setminus{0}$. If a function is holomorphic on a set of the form $Usetminus S$ for some open set $Usubsetmathbb{C}$ and some discrete set $Ssubsetmathbb{C}$, then we say the function is meromorphic on $U$. The set $S$ is then commonly refered to as the set of singularities of the function.



    A singularity of a meromorphic function always falls into exactly one of the following three categories. A removable singularity, a pole, or an essential singularity.



    A removable singularity means that, even though we did not define our function at that point, we can extend the definition of our function to include this point in its domain such that it is still meromorphic.



    A singularity $z_0$ of the meromorphic function $f$ is called a pole if it is not removable, but there exists $n$ such that $z_0$ is removable for $(z-z_0)^nf(z)$. In particular, if $n=1$ suffices, we call this a simple pole, which has some nice properties.



    Finally, any other singularity is called an essential singularity. I think these are the coolest, even though they are generally the hardest to work with. An easy example is $f(z)=e^{1/z}$. Just look at the Laurent series at $z=0$ to find out this is an essential singularity.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      $begingroup$

      On the complex plane it does not make sense to talk about $-infty$ because there is no ordering on the complex numbers. Indeed, assume $i > 0$ then $i^2=-1>0$ and similarly for the other side. So in a sense $infty$ can be seen as what any unbounded complex functions converges to.



      Yes, your $f(z)$ does have a pole at your described points. In a sense having a pole of order $N$ at $z_0$ can be interpolated that the functions behaves like begin{equation} frac{1}{(z-z_0)^N} end{equation} near $z_0$ which can be made precise by saying begin{equation} f(z) = frac{1}{(z-z_0)^N} g(z) end{equation} where $g(z)$ is holomorphic on a small neighborhood of $z_0$. This can be done to your $f(z)$ by polynomial division.



      But it seems like you're confused about isolated singularities. So I'll give a brief overview of the three types:



      A removable singularity at $z_0$ is one where $lim_{z rightarrow z_0} (z-z_0)f(z) = 0$ In particular, this is called removable because one can redefine a $f(z)$ to be holomorphic at $z_0$ using Cauchy's Integral Formula representation. So this is the most mild type of singularity because essentially it's not one. For example $f(z) = z$ on $B(varepsilon,0) - 0$ But defining $f(0)=0$, we see $f$ extends to be holomorphic on the entire ball.



      A pole is what I described above, which can be characterized by $lim_{z rightarrow z_0} f(z) = infty$ (recall $-infty$ does not make sense on $mathbb{C}$) An example is any rational function with a zero on the denominator that is not canceled out by the numerator's zero.



      The worst type of singularity is an essential singularity which can be seen as an isolated singularity that is neither a pole or removable, which is equivalent to begin{equation} limsup_{z rightarrow z_0} f(z) = infty end{equation} begin{equation} liminf_{z rightarrow z_0} |f(z)| < infty end{equation}
      Some powerful theorems are that if $f$ is holomorphic on the punctured disk of $z_0$ then it's image on any $delta$ neighborhood is dense in $mathbb{C}$. A stronger refinement of this theorem is Picard's Big Theorem which states if $f$ is holomorphic on the punctured disk with an essential singularity at $z_0$ then it must touch every point on $mathbb{C}$ except at most one infinitely often. An example is $f(z)=e^{1/z}$. In fact every holomorphic function that is entire and non-polynomial has an essential singularity at $infty$.



      Another way to look at them is their Laurent Series. A removable singularity has no $a_{-n}$ terms in its Laurent series, a pole has finitely many $a_{-n}$ terms, while an essential singularity has infinitely many negative $a_{-n}$ terms.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        On the complex plane it does not make sense to talk about $-infty$ because there is no ordering on the complex numbers. Indeed, assume $i > 0$ then $i^2=-1>0$ and similarly for the other side. So in a sense $infty$ can be seen as what any unbounded complex functions converges to.



        Yes, your $f(z)$ does have a pole at your described points. In a sense having a pole of order $N$ at $z_0$ can be interpolated that the functions behaves like begin{equation} frac{1}{(z-z_0)^N} end{equation} near $z_0$ which can be made precise by saying begin{equation} f(z) = frac{1}{(z-z_0)^N} g(z) end{equation} where $g(z)$ is holomorphic on a small neighborhood of $z_0$. This can be done to your $f(z)$ by polynomial division.



        But it seems like you're confused about isolated singularities. So I'll give a brief overview of the three types:



        A removable singularity at $z_0$ is one where $lim_{z rightarrow z_0} (z-z_0)f(z) = 0$ In particular, this is called removable because one can redefine a $f(z)$ to be holomorphic at $z_0$ using Cauchy's Integral Formula representation. So this is the most mild type of singularity because essentially it's not one. For example $f(z) = z$ on $B(varepsilon,0) - 0$ But defining $f(0)=0$, we see $f$ extends to be holomorphic on the entire ball.



        A pole is what I described above, which can be characterized by $lim_{z rightarrow z_0} f(z) = infty$ (recall $-infty$ does not make sense on $mathbb{C}$) An example is any rational function with a zero on the denominator that is not canceled out by the numerator's zero.



        The worst type of singularity is an essential singularity which can be seen as an isolated singularity that is neither a pole or removable, which is equivalent to begin{equation} limsup_{z rightarrow z_0} f(z) = infty end{equation} begin{equation} liminf_{z rightarrow z_0} |f(z)| < infty end{equation}
        Some powerful theorems are that if $f$ is holomorphic on the punctured disk of $z_0$ then it's image on any $delta$ neighborhood is dense in $mathbb{C}$. A stronger refinement of this theorem is Picard's Big Theorem which states if $f$ is holomorphic on the punctured disk with an essential singularity at $z_0$ then it must touch every point on $mathbb{C}$ except at most one infinitely often. An example is $f(z)=e^{1/z}$. In fact every holomorphic function that is entire and non-polynomial has an essential singularity at $infty$.



        Another way to look at them is their Laurent Series. A removable singularity has no $a_{-n}$ terms in its Laurent series, a pole has finitely many $a_{-n}$ terms, while an essential singularity has infinitely many negative $a_{-n}$ terms.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          On the complex plane it does not make sense to talk about $-infty$ because there is no ordering on the complex numbers. Indeed, assume $i > 0$ then $i^2=-1>0$ and similarly for the other side. So in a sense $infty$ can be seen as what any unbounded complex functions converges to.



          Yes, your $f(z)$ does have a pole at your described points. In a sense having a pole of order $N$ at $z_0$ can be interpolated that the functions behaves like begin{equation} frac{1}{(z-z_0)^N} end{equation} near $z_0$ which can be made precise by saying begin{equation} f(z) = frac{1}{(z-z_0)^N} g(z) end{equation} where $g(z)$ is holomorphic on a small neighborhood of $z_0$. This can be done to your $f(z)$ by polynomial division.



          But it seems like you're confused about isolated singularities. So I'll give a brief overview of the three types:



          A removable singularity at $z_0$ is one where $lim_{z rightarrow z_0} (z-z_0)f(z) = 0$ In particular, this is called removable because one can redefine a $f(z)$ to be holomorphic at $z_0$ using Cauchy's Integral Formula representation. So this is the most mild type of singularity because essentially it's not one. For example $f(z) = z$ on $B(varepsilon,0) - 0$ But defining $f(0)=0$, we see $f$ extends to be holomorphic on the entire ball.



          A pole is what I described above, which can be characterized by $lim_{z rightarrow z_0} f(z) = infty$ (recall $-infty$ does not make sense on $mathbb{C}$) An example is any rational function with a zero on the denominator that is not canceled out by the numerator's zero.



          The worst type of singularity is an essential singularity which can be seen as an isolated singularity that is neither a pole or removable, which is equivalent to begin{equation} limsup_{z rightarrow z_0} f(z) = infty end{equation} begin{equation} liminf_{z rightarrow z_0} |f(z)| < infty end{equation}
          Some powerful theorems are that if $f$ is holomorphic on the punctured disk of $z_0$ then it's image on any $delta$ neighborhood is dense in $mathbb{C}$. A stronger refinement of this theorem is Picard's Big Theorem which states if $f$ is holomorphic on the punctured disk with an essential singularity at $z_0$ then it must touch every point on $mathbb{C}$ except at most one infinitely often. An example is $f(z)=e^{1/z}$. In fact every holomorphic function that is entire and non-polynomial has an essential singularity at $infty$.



          Another way to look at them is their Laurent Series. A removable singularity has no $a_{-n}$ terms in its Laurent series, a pole has finitely many $a_{-n}$ terms, while an essential singularity has infinitely many negative $a_{-n}$ terms.






          share|cite|improve this answer









          $endgroup$



          On the complex plane it does not make sense to talk about $-infty$ because there is no ordering on the complex numbers. Indeed, assume $i > 0$ then $i^2=-1>0$ and similarly for the other side. So in a sense $infty$ can be seen as what any unbounded complex functions converges to.



          Yes, your $f(z)$ does have a pole at your described points. In a sense having a pole of order $N$ at $z_0$ can be interpolated that the functions behaves like begin{equation} frac{1}{(z-z_0)^N} end{equation} near $z_0$ which can be made precise by saying begin{equation} f(z) = frac{1}{(z-z_0)^N} g(z) end{equation} where $g(z)$ is holomorphic on a small neighborhood of $z_0$. This can be done to your $f(z)$ by polynomial division.



          But it seems like you're confused about isolated singularities. So I'll give a brief overview of the three types:



          A removable singularity at $z_0$ is one where $lim_{z rightarrow z_0} (z-z_0)f(z) = 0$ In particular, this is called removable because one can redefine a $f(z)$ to be holomorphic at $z_0$ using Cauchy's Integral Formula representation. So this is the most mild type of singularity because essentially it's not one. For example $f(z) = z$ on $B(varepsilon,0) - 0$ But defining $f(0)=0$, we see $f$ extends to be holomorphic on the entire ball.



          A pole is what I described above, which can be characterized by $lim_{z rightarrow z_0} f(z) = infty$ (recall $-infty$ does not make sense on $mathbb{C}$) An example is any rational function with a zero on the denominator that is not canceled out by the numerator's zero.



          The worst type of singularity is an essential singularity which can be seen as an isolated singularity that is neither a pole or removable, which is equivalent to begin{equation} limsup_{z rightarrow z_0} f(z) = infty end{equation} begin{equation} liminf_{z rightarrow z_0} |f(z)| < infty end{equation}
          Some powerful theorems are that if $f$ is holomorphic on the punctured disk of $z_0$ then it's image on any $delta$ neighborhood is dense in $mathbb{C}$. A stronger refinement of this theorem is Picard's Big Theorem which states if $f$ is holomorphic on the punctured disk with an essential singularity at $z_0$ then it must touch every point on $mathbb{C}$ except at most one infinitely often. An example is $f(z)=e^{1/z}$. In fact every holomorphic function that is entire and non-polynomial has an essential singularity at $infty$.



          Another way to look at them is their Laurent Series. A removable singularity has no $a_{-n}$ terms in its Laurent series, a pole has finitely many $a_{-n}$ terms, while an essential singularity has infinitely many negative $a_{-n}$ terms.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 20:41









          Story123Story123

          18519




          18519























              0












              $begingroup$

              Do not trust random people from the internet, trust me instead!



              A singularity is a point where a function is not defined. In complex analysis, we can say a lot more about singularities though.



              A holomorphic function is a complex differentiable function $f:Utomathbb{C}$ for some open set
              $Usubsetmathbb{C}$, meaning it is complex differentiable at every point in $U$. For example $frac1z$ is not holomorphic on $mathbb{C}$, but it is on $mathbb{C}setminus{0}$. If a function is holomorphic on a set of the form $Usetminus S$ for some open set $Usubsetmathbb{C}$ and some discrete set $Ssubsetmathbb{C}$, then we say the function is meromorphic on $U$. The set $S$ is then commonly refered to as the set of singularities of the function.



              A singularity of a meromorphic function always falls into exactly one of the following three categories. A removable singularity, a pole, or an essential singularity.



              A removable singularity means that, even though we did not define our function at that point, we can extend the definition of our function to include this point in its domain such that it is still meromorphic.



              A singularity $z_0$ of the meromorphic function $f$ is called a pole if it is not removable, but there exists $n$ such that $z_0$ is removable for $(z-z_0)^nf(z)$. In particular, if $n=1$ suffices, we call this a simple pole, which has some nice properties.



              Finally, any other singularity is called an essential singularity. I think these are the coolest, even though they are generally the hardest to work with. An easy example is $f(z)=e^{1/z}$. Just look at the Laurent series at $z=0$ to find out this is an essential singularity.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Do not trust random people from the internet, trust me instead!



                A singularity is a point where a function is not defined. In complex analysis, we can say a lot more about singularities though.



                A holomorphic function is a complex differentiable function $f:Utomathbb{C}$ for some open set
                $Usubsetmathbb{C}$, meaning it is complex differentiable at every point in $U$. For example $frac1z$ is not holomorphic on $mathbb{C}$, but it is on $mathbb{C}setminus{0}$. If a function is holomorphic on a set of the form $Usetminus S$ for some open set $Usubsetmathbb{C}$ and some discrete set $Ssubsetmathbb{C}$, then we say the function is meromorphic on $U$. The set $S$ is then commonly refered to as the set of singularities of the function.



                A singularity of a meromorphic function always falls into exactly one of the following three categories. A removable singularity, a pole, or an essential singularity.



                A removable singularity means that, even though we did not define our function at that point, we can extend the definition of our function to include this point in its domain such that it is still meromorphic.



                A singularity $z_0$ of the meromorphic function $f$ is called a pole if it is not removable, but there exists $n$ such that $z_0$ is removable for $(z-z_0)^nf(z)$. In particular, if $n=1$ suffices, we call this a simple pole, which has some nice properties.



                Finally, any other singularity is called an essential singularity. I think these are the coolest, even though they are generally the hardest to work with. An easy example is $f(z)=e^{1/z}$. Just look at the Laurent series at $z=0$ to find out this is an essential singularity.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Do not trust random people from the internet, trust me instead!



                  A singularity is a point where a function is not defined. In complex analysis, we can say a lot more about singularities though.



                  A holomorphic function is a complex differentiable function $f:Utomathbb{C}$ for some open set
                  $Usubsetmathbb{C}$, meaning it is complex differentiable at every point in $U$. For example $frac1z$ is not holomorphic on $mathbb{C}$, but it is on $mathbb{C}setminus{0}$. If a function is holomorphic on a set of the form $Usetminus S$ for some open set $Usubsetmathbb{C}$ and some discrete set $Ssubsetmathbb{C}$, then we say the function is meromorphic on $U$. The set $S$ is then commonly refered to as the set of singularities of the function.



                  A singularity of a meromorphic function always falls into exactly one of the following three categories. A removable singularity, a pole, or an essential singularity.



                  A removable singularity means that, even though we did not define our function at that point, we can extend the definition of our function to include this point in its domain such that it is still meromorphic.



                  A singularity $z_0$ of the meromorphic function $f$ is called a pole if it is not removable, but there exists $n$ such that $z_0$ is removable for $(z-z_0)^nf(z)$. In particular, if $n=1$ suffices, we call this a simple pole, which has some nice properties.



                  Finally, any other singularity is called an essential singularity. I think these are the coolest, even though they are generally the hardest to work with. An easy example is $f(z)=e^{1/z}$. Just look at the Laurent series at $z=0$ to find out this is an essential singularity.






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                  $endgroup$



                  Do not trust random people from the internet, trust me instead!



                  A singularity is a point where a function is not defined. In complex analysis, we can say a lot more about singularities though.



                  A holomorphic function is a complex differentiable function $f:Utomathbb{C}$ for some open set
                  $Usubsetmathbb{C}$, meaning it is complex differentiable at every point in $U$. For example $frac1z$ is not holomorphic on $mathbb{C}$, but it is on $mathbb{C}setminus{0}$. If a function is holomorphic on a set of the form $Usetminus S$ for some open set $Usubsetmathbb{C}$ and some discrete set $Ssubsetmathbb{C}$, then we say the function is meromorphic on $U$. The set $S$ is then commonly refered to as the set of singularities of the function.



                  A singularity of a meromorphic function always falls into exactly one of the following three categories. A removable singularity, a pole, or an essential singularity.



                  A removable singularity means that, even though we did not define our function at that point, we can extend the definition of our function to include this point in its domain such that it is still meromorphic.



                  A singularity $z_0$ of the meromorphic function $f$ is called a pole if it is not removable, but there exists $n$ such that $z_0$ is removable for $(z-z_0)^nf(z)$. In particular, if $n=1$ suffices, we call this a simple pole, which has some nice properties.



                  Finally, any other singularity is called an essential singularity. I think these are the coolest, even though they are generally the hardest to work with. An easy example is $f(z)=e^{1/z}$. Just look at the Laurent series at $z=0$ to find out this is an essential singularity.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 19 '18 at 20:40









                  SmileyCraftSmileyCraft

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