Section 7.4 Singularities, Zeros, and Poles
Recall that the point \(\alpha\) is called a singular point, or singularity, of the complex function \(f\) if \(f\) is not analytic at the point \(\alpha\text{,}\) but every neighborhood \(D_{R}(\alpha)\) of \(\alpha\) contains at least one point at which \(f\) is analytic. For example, the function \(f(z) =\frac{1}{1-z}\) is not analytic at \(\alpha=1\) but is analytic for all other values of \(z\text{.}\) Thus the point \(\alpha=1\) is a singular point of \(f\text{.}\) As another example, consider the function \(g(z) =\) Log \(z\text{.}\) We showed in Section 5.2 that \(g\) is analytic for all \(z\) except at the origin and at the points on the negative real axis. Thus the origin and each point on the negative real axis are singularities of \(g\text{.}\)
The point \(\alpha\) is called an isolated singularity of a complex function \(f\) if \(f\) is not analytic at \(\alpha\) but there exists a real number \(R>0\) such that \(f\) is analytic everywhere in the punctured disk \(D_{R}^{\ast }(\alpha)\text{.}\) The function \(f(z) =\frac{1 }{1-z}\) has an isolated singularity at \(\alpha =1\text{.}\) The function \(g(z) =\) Log \(z\text{,}\) however, has a singularity at \(\alpha =0\) (or at any point of the negative real axis) that is not isolated, because any neighborhood of \(\alpha\) contains points on the negative real axis, and \(g\) is not analytic at those points. Functions with isolated singularities have a Laurent series because the punctured disk \(D_{R}^{\ast }(\alpha)\) is the same as the annulus \(A(\alpha ,0,R)\text{.}\) We now look at this special case of Laurent’s theorem in order to classify three types of isolated singularities.
Definition 7.4.1. Classification of singularities.
Let \(f\) have an isolated singularity at \(\alpha\) with Laurent series
\begin{equation*}
f(z) = \sum_{n=- \infty}^{\infty}c_n(z-\alpha)^n, \text{ valid for all } z \in A(\alpha,0,R)\text{.}
\end{equation*}
Then we distinguish the following types of singularities at \(\alpha\text{.}\)
-
i.
If \(c_n=0\text{,}\) for \(n=-1,\,-2,\,-3,\ldots\text{,}\) then \(f\) has a removable singularity at \(\alpha\text{.}\)
-
ii.
If \(k\) is a positive integer such that \(c_{-k} \ne 0\text{,}\) but \(c_n=0\) for \(n\lt -k\text{,}\) then \(f\) has a pole of order \(k\) at \(\alpha\text{.}\)
-
iii.
If \(c_n \ne 0\) for infinitely many negative integers \(n\text{,}\) then \(f\) has an essential singularity at \(\alpha\text{.}\)
Let’s investigate some examples of these three cases.
-
i.
If \(f\) has a removable singularity at \(\alpha\text{,}\) then it has a Laurent series
\begin{equation*}
f(z) =\sum\limits_{n=0}^{\infty}c_n(z-\alpha)^n, \text{ valid for all } z \in A(\alpha,0,R)\text{.}
\end{equation*}
Theorem 4.4.7 implies that the power series for
\(f\) defines an analytic function in the disk
\(D_{R}(\alpha)\text{.}\) If we use this series to define
\(f(\alpha) =c_0\text{,}\) then the function
\(f\) becomes analytic at
\(z=\alpha\text{,}\) removing the singularity. For example, consider the function
\(f(z) =\frac{\sin z}{z}\text{.}\) It is undefined at
\(z=0\) and has an isolated singularity at
\(z=0\text{,}\) as the Laurent series for
\(f\) is
\begin{align*}
f(z) \amp = \frac{\sin z}{z}=\frac{1}{z}\left(z-\frac{z^3}{3!} + \frac{z^{5}}{5!}-\frac{z^{7}}{7!}+\cdots\right)\\
\amp =1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^{6}}{7!}+\cdots, \text{ valid for } |z| > 0\text{.}
\end{align*}
We can remove this singularity if we define
\(f(0)=1\text{,}\) for then
\(f\) will be analytic at
\(0\) in accordance with
Theorem 4.4.7. Another example is
\(g(z)=\frac{\cos z-1}{z^2}\text{,}\) which has an isolated singularity at the point
\(0\text{,}\) as the Laurent series for
\(g\) is
\begin{align*}
g(z) \amp = \frac{1}{z^2}\left(-\frac{z^2}{2!}+\frac{z^4}{4!} - \frac{z^{6}}{6!}+\cdots\right)\\
\amp = -\frac{1}{2}+\frac{z^2}{4!}-\frac{z^4}{6!}+\cdots, \text{ valid for } |z| >0\text{.}
\end{align*}
If we define \(g(0) =-\frac{1}{2}\text{,}\) then \(g\) will be analytic for all \(z\text{.}\)
-
ii.
If \(f\) has a pole of order \(k\) at \(\alpha\text{,}\) the Laurent series for \(f\) is
\begin{equation*}
f(z) =\sum\limits_{n=-k}^{\infty}c_n(z-\alpha)^n, \text{ valid for all } z \in A(\alpha,0,R)\text{,}
\end{equation*}
where \(c_{-k} \ne 0\text{.}\) For example,
\begin{equation*}
f(z) = \frac{\sin z}{z^3}=\frac{1}{z^2}-\frac{1}{3!}+\frac{z^2}{5!}-\frac{z^4}{7!}+\cdots
\end{equation*}
has a pole of order 2 at \(0\text{.}\) If \(f\) has a pole of order 1 at \(\alpha\text{,}\) we say that \(f\) has a simple pole at \(\alpha\text{.}\) For example,
\begin{equation*}
f(z) = \frac{1}{z}e^z=\frac{1}{z}+1+\frac{z}{2!}+\frac{z^2}{3!} + \cdots\text{,}
\end{equation*}
which has a simple pole at \(0\text{.}\)
-
iii.
If infinitely many negative powers of \((z-\alpha)\) occur in the Laurent series, then \(f\) has an essential singularity at \(\alpha\text{.}\) For example,
\begin{equation*}
f(z) = z^2\sin\frac{1}{z}=z-\frac{1}{3!}z^{-1}+\frac{1}{5!}z^{-3}-\frac{1}{7!}z^{-5}+\cdots
\end{equation*}
has an essential singularity at the origin.
Definition 7.4.2. Zero of order $k$.
A function \(f\) analytic in \(D_{R}(\alpha)\) has a zero of order \(k\) at the point \(\alpha\) iff
\begin{equation*}
f^{(n)}(\alpha)=0, \text{ for } n=0,\,1,\ldots , k-1, \text{ but } f^{(k)}(\alpha) \ne 0\text{.}
\end{equation*}
A zero of order 1 is called a simple zero.
Theorem 7.4.3.
A function \(f\) analytic in \(D_{R}(\alpha)\) has a zero of order \(k\) at the point \(\alpha\) iff its Taylor series given by \(f(z) = \sum\limits_{n=0}^{\infty}c_n(z-\alpha)^n\) has
\begin{equation*}
c_0 = c_1 = \cdots = c_{k-1} = 0, \text{ but } c_k \ne 0\text{.}
\end{equation*}
Proof.
The conclusion follows immediately from Definition 7.6, because we have \(c_n=\frac{f^{(n)}(\alpha)}{n!}\) according to Taylor’s theorem.
Example 7.4.4.
\begin{equation*}
f(z) = z\sin z^2 = z^3-\frac{z^{7}}{3!} + \frac{z^{11}}{5!} - \frac{z^{15}}{7!} + \cdots
\end{equation*}
has a zero of order 3 at \(z=0\text{.}\) Definition 7.6 confirms this fact because
\begin{align*}
f\,'(z) \amp = 2z^2\cos z^2+\sin z^2;\\
f\,''(z) \amp = 6z\cos z^2-4z^3\sin z^2\\
f\,'''(z) \amp = 6\cos z^2-8z^4\cos z^2-24z^2\sin z^2
\end{align*}
Then, \(f(0) =f\,'(0) = f\,''(0) = 0\text{,}\) but \(f\,'''(0) = 6 \ne 0\text{.}\)
Theorem 7.4.5.
Suppose that the function \(f\) is analytic in \(D_{R}(\alpha)\text{.}\) Then \(f\) has a zero of order \(k\)\ at the point \(\alpha\) iff \(f\) can be expressed in the form
\begin{equation}
f(z) = (z-\alpha)^kg(z)\text{,}\tag{7.4.1}
\end{equation}
where \(g\) is analytic at the point \(\alpha\) and \(g(\alpha) \ne 0\text{.}\)
Proof.
Suppose that
\(f\) has a zero of order
\(k\) at the point
\(\alpha\text{,}\) and that
\(f(z) = \sum\limits_{n=0}^{\infty}c_n(z-\alpha)^n\) for
\(z \in D_{R}(\alpha)\text{.}\) Theorem 7.4.3 assures us that
\(c_n=0\) for
\(0 \le n\le k-1\text{,}\) and that
\(c_k \ne 0\text{,}\) so that we can write
\(f\) as
\begin{align}
f(z) \amp = \sum_{n=k}^{\infty}c_n(z-\alpha)^n\notag\\
\amp = \sum_{n=0}^{\infty}c_{n+k}(z-\alpha)^{n+k}\notag\\
\amp = (z-\alpha)^k\sum\limits_{n=0}^{\infty}c_{n+k}(z-\alpha)^n\text{,}\tag{7.4.2}
\end{align}
where
\(c_k \ne 0\text{.}\) The series on the right side of Equation
(7.4.2) defines a function, which we denote by
\(g\text{.}\) That is,
\begin{equation*}
g(z) = \sum_{n=0}^{\infty}c_{n+k}(z-\alpha)^n = c_k+\sum\limits_{n=1}^{\infty}c_{n+k}(z-\alpha)^n, \text{ valid for all } z \in D_{R}(\alpha)\text{.}
\end{equation*}
By
Theorem 4.4.7,
\(g\) is analytic in
\(D_{R}(\alpha)\text{,}\) and
\(g(\alpha)=c_k \ne 0\text{.}\)
Conversely, suppose that
\(f\) has the form given by Equation
(7.4.1). Since
\(g\) is analytic at
\(\alpha\text{,}\) it has the power series representation
\(g(z)=\sum\limits_{n=0}^{\infty}b_n(z-\alpha)^n\text{,}\) where
\(g(\alpha)=b_0 \ne 0\) by assumption. If we multiply both sides of the expression defining
\(g(z)\) by
\((z-\alpha)^k\) we get
\begin{equation*}
f(z) = g(z)(z-\alpha)^k = \sum_{n=0}^{\infty}b_n(z-\alpha)^{n+k} = \sum\limits_{n=k}^{\infty}b_{n-k}(z-\alpha)^n\text{.}
\end{equation*}
By
Theorem 7.4.3,
\(f\) has a zero of order
\(k\) at the point
\(\alpha\text{.}\)
Corollary 7.4.6.
If \(f(z)\) and \(g(z)\) are analytic at \(z=\alpha\) and have zeros of orders \(m\) and \(n\text{,}\) respectively, at \(z=\alpha\text{,}\) then their product \(h(z) =f(z) g(z)\) has a zero of order \(m+n\) at \(z=\alpha\text{.}\)
Example 7.4.7.
Let \(f(z) =z^3\sin z\text{.}\) Then \(f(z)\) can be factored as the product of \(z^3\) and \(\sin z\text{,}\) which have zeros of orders \(m=3\) and \(n=1\text{,}\) respectively, at \(z=0\text{.}\) Hence \(z=0\) is a zero of order 4 of \(f(z)\text{.}\)
Theorem 7.4.8.
A function \(f\) analytic in the punctured disk \(D_{R}^*(\alpha)\) has a pole of order \(k\) at the point \(\alpha\) iff \(f\) can be expressed in the form
\begin{equation}
f(z) = \frac{h(z)}{(z-\alpha)^k}\text{,}\tag{7.4.3}
\end{equation}
where the function \(h\) is analytic at the point \(\alpha\text{,}\) and \(h(\alpha) \ne 0\text{.}\)
Proof.
Suppose that \(f\) has a pole of order \(k\) at the point \(\alpha\text{.}\) We can then write the Laurent series for \(f\) as
\begin{equation*}
f(z) = \frac{1}{(z-\alpha)^k}\sum\limits_{n=0}^{\infty}c_{n-k}(z-\alpha)^n\text{,}
\end{equation*}
where \(c_{-k} \ne 0\text{.}\) The series on the right side of this equation defines a function, which we denote by \(h(z)\text{.}\) That is,
\begin{equation*}
h(z) = \sum_{n=0}^{\infty}c_{n-k}(z-\alpha)^n, \text{ for all } z \in D_{R}^{\ast }(\alpha) = \{z:0\lt |z-\alpha|\lt R\}\text{.}
\end{equation*}
If we specify that \(h(\alpha)=c_{-k}\text{,}\) then \(h\) is analytic in all of \(D_{R}(\alpha)\text{,}\) with \(h(\alpha) \ne 0\text{.}\)
Conversely, suppose that Equation
(7.4.3) is satisfied. Because
\(h\) is analytic at the point
\(\alpha\) with
\(h(\alpha) \ne 0\text{,}\) it has a power series representation
\begin{equation*}
h(z) = \sum_{n=0}^{\infty}b_n(z-\alpha)^n\text{,}
\end{equation*}
where \(b_0 \ne 0\text{.}\) If we divide both sides of this equation by \((z-\alpha)^k\text{,}\) we obtain the following Laurent series representation for \(f\text{:}\)
\begin{align*}
f(z) \amp = \sum_{n=0}^{\infty}b_n(z-\alpha)^{n-k}\\
\amp = \sum_{n=-k}^{\infty}b_{n+k}(z-\alpha)^n\\
\amp = \sum_{n=-k}^{\infty}c_n(z-\alpha)^n\text{,}
\end{align*}
where \(c_n=b_{n+k}\text{.}\) Since \(c_{-k}=b_0 \ne 0\text{,}\) \(f\) has a pole of order \(k\) at \(\alpha\text{.}\)
Corollary 7.4.9.
If \(f\) is analytic and has a zero of order \(k\) at the point \(\alpha\text{,}\) then \(g(z) =\frac{1}{f(z)}\) has a pole of order \(k\) at \(\alpha\text{.}\)
Corollary 7.4.10.
If \(f\) has a pole of order \(k\) at the point \(\alpha\text{,}\) then \(g(z)=\frac{1}{f(z)}\) has a removable singularity at \(\alpha\text{.}\) If we define \(g(\alpha)=0\text{,}\) then \(g(z)\) has a zero of order \(k\) at \(\alpha\text{.}\)
Corollary 7.4.11.
If \(f\) and \(g\) have poles of orders \(m\) and \(n\text{,}\) respectively, at the point \(\alpha\text{,}\) then their product \(h(z)=f(z)g(z)\) has a pole of order \(m+n\) at \(\alpha\text{.}\)
Corollary 7.4.12.
Let \(f\) and \(g\) be analytic with zeros of orders \(m\) and \(n\text{,}\) respectively, at \(\alpha\text{.}\) Then their quotient \(h(z) = \frac{f(z)}{g(z)}\) has the following behavior:
-
i.
If \(m>n\text{,}\) then \(h\) has a removable singularity at \(\alpha\text{.}\) If we define \(h(\alpha)=0\text{,}\) then \(h\) has a zero of order \(m-n\) at \(\alpha\text{.}\)
-
ii.
If \(m\lt n\text{,}\) then \(h\) has a pole of order \(n-m\) at \(\alpha\text{.}\)
-
iii.
If \(m=n\text{,}\) then \(h\) has a removable singularity at \(\alpha\) and can be defined so that \(h\) is analytic at \(\alpha\)\ by \(h(\alpha) = \lim\limits_{z \to \alpha}h(z)\text{.}\)
Example 7.4.13.
Locate the zeros and poles of \(h(z) = \frac{\tan z}{z}\) and determine their order.
Solution.
In
Section 5.4 we saw that the zeros of
\(f(z) =\sin z\) occur at the points
\(n\pi\text{,}\) where
\(n\) is an integer. Because
\(f\,'(n\pi) = \cos n\pi \ne 0\text{,}\) the zeros of
\(f\) are simple. Similarly, the function
\(g(z) = z\cos z\) has simple zeros at the points
\(0\) and
\((n+\frac{1}{2}) \pi\text{,}\) where
\(n\) is an integer. From the information given, we find that
\(h(z) = \frac{f(z)}{g(z)}\) behaves as follows:
-
i.
\(h\) has simple zeros at \(n\pi\text{,}\) where \(n=\pm1,\,\pm2,\ldots\text{;}\)
-
ii.
\(h\) has simple poles at \((n+\frac{1}{2}) \pi\text{,}\) where \(n\) is an integer; and
-
iii.
\(h\) is analytic at \(0\) if we define \(h(0) = \lim\limits_{z \to0}h(z) =1\text{.}\)
Example 7.4.14.
Locate the poles of \(g(z) =\frac{1}{5z^4+26z^2+5}\) and specify their order.
Solution.
The roots of the quadratic equation
\(5z^2+26z+5=0\) occur at the points
\(-5\) and
\(-\frac{1}{5}\text{.}\) If we replace
\(z\) with
\(z^2\) in this equation, the function
\(f(z) =5z^4+26z^2+5\) has simple zeros at the points
\(\pm i\sqrt{5}\) and
\(\pm\frac{i}{\sqrt{5}}\text{.}\) Corollary 7.4.9 implies that
\(g\) has simple poles at
\(\pm i\sqrt{5}\) and
\(\pm \frac{i}{\sqrt{5}}\text{.}\)
Example 7.4.15.
Locate the poles of \(g(z) =\frac{\pi\cot(\pi z)}{z^2}\) and specify their order.
Solution.
The function
\(f(z) = z^2\sin\pi z\) has a zero of order
\(3\) at
\(z=0\) and simple zeros at the points
\(z=\pm1,\,\pm2,\ldots\) .
Corollary 7.4.9 implies that
\(g\) has a pole of order
\(3\) at the point
\(0\) and simple poles at the points
\(\pm1,\,\pm2,\ldots\) .
Exercises Exercises
1.
Locate the zeros of the following functions and determine their order.
(a)
\((1+z^2)^4\text{.}\)
Solution.
Zeros of order 4 at \(\pm i\text{.}\)
(b)
\(\sin ^2z\text{.}\)
(c)
\(z^2+2z+2\text{.}\)
Solution.
Simple zeros at \(-1\pm i\text{.}\)
(d)
\(\sin z^2\text{.}\)
(e)
\(z^4+10z^2+9\text{.}\)
Solution.
Simple zeros at \(\pm i\) and \(\pm 3i\text{.}\)
(f)
\(1+\exp z\text{.}\)
(g)
\(z^{6}+1\text{.}\)
Solution.
Simple zeros at \(\frac{\sqrt{3}\pm i}{2}\text{,}\) \(\frac{-\sqrt{3}\pm i}{2}\text{,}\) and \(\pm i\text{.}\)
(h)
\(z^3\exp (z-1)\text{.}\)
(i)
\(z^6+2z^3+1\text{.}\)
Solution.
Zeros of order 2 at \(\frac{1\pm i\sqrt{3}}{2}\) and \(-1\text{.}\)
(j)
\(z^3\cos ^2z\text{.}\)
(k)
\(z^8+z^4\text{.}\)
Solution.
Simple zeros at \(\frac{1\pm i}{\sqrt{2}}\) and \(\frac{-1\pm i}{\sqrt{2}}\text{,}\) and a zero of order 4 at the origin.
(l)
\(z^2\cosh z\text{.}\)
2.
Locate the poles of the following functions and determine their order.
(a)
\((z^2+1)^{-3}(z-1)^{-4}\text{.}\)
Solution.
Poles of order 3 at \(\pm i\text{,}\) and a pole of order 4 at 1.
(b)
\(z^{-1}(z^2-2z+2)^{-2}\text{.}\)
(c)
\((z^{6}+1)^{-1}\text{.}\)
Solution.
Simple poles at \(\frac{\sqrt{3}\pm i}{2}\text{,}\) \(\frac{-\sqrt{3}\pm i}{2}\text{,}\) and \(\pm i\text{.}\)
(d)
\((z^4+z^3-2z^2)^{-1}\text{.}\)
(e)
\((3z^4+10z^2+3)^{-1}\text{.}\)
Solution.
Simple poles at \(\pm \sqrt{3}i\) and \(\frac{\pm i}{\sqrt{3}}\text{.}\)
(f)
\((i+\frac{2}{z})^{-1}(3+\frac{4}{z})^{-1}\text{.}\)
(g)
\(z\cot z\text{.}\)
Solution.
Simple poles at \(z=n\pi\) for \(n=\pm 1, \, \pm 2,\ldots\) .
(h)
\(z^{-5}\sin z\text{.}\)
(i)
\((z^2\sin z)^{-1}\text{.}\)
Solution.
Simple poles at \(z=n\pi\) for \(n=\pm 1, \, \pm 2,\ldots\text{,}\) and a pole of order 3 at the origin.
(j)
\(z^{-1}\csc z\text{.}\)
(k)
\((1-\exp z)^{-1}\text{.}\)
Solution.
Simple poles at \(z=2n\pi\) for \(n=0,\,\pm 1, \, \pm 2,\ldots\text{.}\)
(l)
\(z^{-5}\sinh z\text{.}\)
3.
Locate the singularities of the following functions and determine their type.
(a)
\(\frac{z^2}{z-\sin z}\text{.}\)
Solution.
Removable singularity at the origin.
(b)
\(\sin (\frac{1}{z})\text{.}\)
(c)
\(z\exp (\frac{1}{z})\text{.}\)
Solution.
Essential singularity at the origin.
(d)
\(\tan z\text{.}\)
(e)
\((z^2+z)^{-1}\sin z\text{.}\)
Solution.
Removable singularity at the origin, and a simple pole at \(-1\text{.}\)
(f)
\(\frac{z}{\sin z}\text{.}\)
(g)
\(\frac{(\exp z) -1}{z}\text{.}\)
Solution.
Removable singularity at the origin.
(h)
\(\frac{\cos z-\cos (2z)}{z^4}\text{.}\)
4.
Suppose that \(f\) has a removable singularity at \(z_0\text{.}\) Show that the function \(\frac{1}{f}\) has either a removable singularity or a pole at \(z_0\text{.}\)
5.
Let \(f\) be analytic and have a zero of order \(k\) at \(z_0\text{.}\) Show that \(f\,'\) has a zero of order \(k-1\) at \(z_0\text{.}\)
Solution.
By
Theorem 7.4.5,
\(f( z) =(z-z_0)^kh(z)\text{,}\) where
\(h\) is analytic at
\(z_0\) and
\(h(z_0) \ne 0\text{.}\) We compute
\begin{align*}
f\,'(z) \amp = k(z-z_0)^{k-1}h(z)+(z-z_0)^kh\,'(z)\\
\amp = (z-z_0)^{k-1}[kh(z)+(z-z_0) h\,'(z)]\\
\amp = (z-z_0)^{k-1}g(z)\text{,}
\end{align*}
where
\(g(z) = kh(z) + (z-z_0) h\,'(z)\text{.}\) Explain why
\(g(z_0) \ne 0\text{,}\) why
\(g\) is analytic at
\(z_0\text{,}\) and why
Theorem 7.4.5 now gives the conclusion.
6.
Let \(f\) and \(g\) be analytic at \(z_0\) and have zeros of order \(m\) and \(n\text{,}\) respectively, at \(z_0\text{.}\) What can you say about the zero of \(f+g\) at \(z_0\text{?}\)
7.
Let \(f\) and \(g\) have poles of order \(m\) and \(n\text{,}\) respectively, at \(z_0\text{.}\) Show that \(f+g\) has either a pole or a removable singularity at \(z_0\)
Solution.
If it so happens that \(m=n\text{,}\) and the coefficients in the Laurent expansions for \(f\) and \(g\) about \(z_0\) are negatives of each other, then \(f+g\) will have a Taylor series representation at \(z_0\text{,}\) making \(z_0\) a removable singularity (show the details for this). If \(m \ne n\text{,}\) then it is easy to show that \(f+g\) still has a pole. State why, and what the order of the pole is.
8.
Let \(f\) be analytic and have a zero of order \(k\) at \(z_0\text{.}\) Show that the function \(\frac{f\,'}{f}\) has a simple pole at \(z_0\text{.}\) \label {7.4.8}
9.
Let \(f\) have a pole of order \(k\) at \(z_0\text{.}\) Show that \(f\,'\) has a pole of order \(k+1\) at \(z_0\text{.}\)
10.
Prove the following corollaries.
(a)
(b)
(c)
(d)
(e)
11.
Find the singularities of the following functions.
(a)
\(\frac{1}{\sin (\frac{1}{z})}\text{.}\)
Solution.
Simple poles at \(z=\frac{1}{n\pi}\) for \(n=\pm 1, \, \pm 2,\ldots\text{,}\) and a nonisolated singularity at the origin.
(b)
\(\mathrm{Log}(z^2)\text{.}\)
(c)
\(\cot z - \frac{1}{z}\text{.}\)
12.
How are the definitions of singularity in complex analysis and asymptote in calculus different? How are they similar?
