Singularities, Zeros, and Poles

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 $g\text{.}$

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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.

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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 $\alpha\text{.}$
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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 $\alpha\text{.}$
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iii.
If $c_n \ne 0$ for infinitely many negative integers $n\text{,}$ then $f$ has an essential singularity at $\alpha\text{.}$
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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 $z\text{.}$

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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 $0\text{.}$

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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.

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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*}

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A zero of order 1 is called a simple zero.

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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*}

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Proof.

The conclusion follows immediately from Definition 7.6, because we have $c_n=\frac{f^{(n)}(\alpha)}{n!}$ according to Taylor’s theorem.

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Example 7.4.4.

From Theorem 7.4.3 we see that the function

\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*}

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Then, $f(0) =f\,'(0) = f\,''(0) = 0\text{,}$ but $f\,'''(0) = 6 \ne 0\text{.}$

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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 $g(\alpha) \ne 0\text{.}$

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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 $g(\alpha)=c_k \ne 0\text{.}$

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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 $\alpha\text{.}$

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An immediate consequence of Theorem 7.4.5 is Corollary 7.4.6. The proof is left as an exercise.

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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 $z=\alpha\text{.}$

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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 $f(z)\text{.}$

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Theorem 7.4.8 gives a useful way to characterize a pole.

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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 $h(\alpha) \ne 0\text{.}$

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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 $h(\alpha) \ne 0\text{.}$

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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 $\alpha\text{.}$

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The following corollaries \big( Corollary 7.4.9– Corollary 7.4.11\big) are useful in determining the order of a zero or a pole. The proofs follow easily from Theorems 7.4.3 and Theorem 7.4.8, and are left as exercises.

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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 $\alpha\text{.}$

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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 $\alpha\text{.}$

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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 $\alpha\text{.}$

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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 $\alpha\text{.}$
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ii.
If $m\lt n\text{,}$ then $h$ has a pole of order $n-m$ at $\alpha\text{.}$
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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 $h(\alpha) = \lim\limits_{z \to \alpha}h(z)\text{.}$
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Example 7.4.13.

Locate the zeros and poles of $h(z) = \frac{\tan z}{z}$ and determine their order.

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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 $n=\pm1,\,\pm2,\ldots\text{;}$
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ii.
$h$ has simple poles at $(n+\frac{1}{2}) \pi\text{,}$ where $n$ is an integer; and
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iii.
$h$ is analytic at $0$ if we define $h(0) = \lim\limits_{z \to0}h(z) =1\text{.}$
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Example 7.4.14.

Locate the poles of $g(z) =\frac{1}{5z^4+26z^2+5}$ and specify their order.

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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 $\pm \frac{i}{\sqrt{5}}\text{.}$

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Example 7.4.15.

Locate the poles of $g(z) =\frac{\pi\cot(\pi z)}{z^2}$ and specify their order.

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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 $\pm1,\,\pm2,\ldots$ .

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Exercises Exercises

1.

Locate the zeros of the following functions and determine their order.

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(a)

$(1+z^2)^4\text{.}$

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Solution.

Zeros of order 4 at $\pm i\text{.}$

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(b)

$\sin ^2z\text{.}$

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(c)

$z^2+2z+2\text{.}$

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Solution.

Simple zeros at $-1\pm i\text{.}$

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(d)

$\sin z^2\text{.}$

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(e)

$z^4+10z^2+9\text{.}$

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Solution.

Simple zeros at $\pm i$ and $\pm 3i\text{.}$

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(f)

$1+\exp z\text{.}$

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(g)

$z^{6}+1\text{.}$

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Solution.

Simple zeros at $\frac{\sqrt{3}\pm i}{2}\text{,}$ $\frac{-\sqrt{3}\pm i}{2}\text{,}$ and $\pm i\text{.}$

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(h)

$z^3\exp (z-1)\text{.}$

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(i)

$z^6+2z^3+1\text{.}$

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Solution.

Zeros of order 2 at $\frac{1\pm i\sqrt{3}}{2}$ and $-1\text{.}$

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(j)

$z^3\cos ^2z\text{.}$

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(k)

$z^8+z^4\text{.}$

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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.

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(l)

$z^2\cosh z\text{.}$

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2.

Locate the poles of the following functions and determine their order.

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(a)

$(z^2+1)^{-3}(z-1)^{-4}\text{.}$

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Solution.

Poles of order 3 at $\pm i\text{,}$ and a pole of order 4 at 1.

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(b)

$z^{-1}(z^2-2z+2)^{-2}\text{.}$

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(c)

$(z^{6}+1)^{-1}\text{.}$

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Solution.

Simple poles at $\frac{\sqrt{3}\pm i}{2}\text{,}$ $\frac{-\sqrt{3}\pm i}{2}\text{,}$ and $\pm i\text{.}$

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(d)

$(z^4+z^3-2z^2)^{-1}\text{.}$

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(e)

$(3z^4+10z^2+3)^{-1}\text{.}$

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Solution.

Simple poles at $\pm \sqrt{3}i$ and $\frac{\pm i}{\sqrt{3}}\text{.}$

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(f)

$(i+\frac{2}{z})^{-1}(3+\frac{4}{z})^{-1}\text{.}$

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(g)

$z\cot z\text{.}$

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Solution.

Simple poles at $z=n\pi$ for $n=\pm 1, \, \pm 2,\ldots$ .

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(h)

$z^{-5}\sin z\text{.}$

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(i)

$(z^2\sin z)^{-1}\text{.}$

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Solution.

Simple poles at $z=n\pi$ for $n=\pm 1, \, \pm 2,\ldots\text{,}$ and a pole of order 3 at the origin.

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(j)

$z^{-1}\csc z\text{.}$

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(k)

$(1-\exp z)^{-1}\text{.}$

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Solution.

Simple poles at $z=2n\pi$ for $n=0,\,\pm 1, \, \pm 2,\ldots\text{.}$

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(l)

$z^{-5}\sinh z\text{.}$

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3.

Locate the singularities of the following functions and determine their type.

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(a)

$\frac{z^2}{z-\sin z}\text{.}$

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Solution.

Removable singularity at the origin.

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(b)

$\sin (\frac{1}{z})\text{.}$

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(c)

$z\exp (\frac{1}{z})\text{.}$

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Solution.

Essential singularity at the origin.

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(d)

$\tan z\text{.}$

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(e)

$(z^2+z)^{-1}\sin z\text{.}$

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Solution.

Removable singularity at the origin, and a simple pole at $-1\text{.}$

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(f)

$\frac{z}{\sin z}\text{.}$

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(g)

$\frac{(\exp z) -1}{z}\text{.}$

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Solution.

Removable singularity at the origin.

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(h)

$\frac{\cos z-\cos (2z)}{z^4}\text{.}$

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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 $z_0\text{.}$

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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 $z_0\text{.}$

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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.

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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 $z_0\text{?}$

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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 $z_0$

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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.

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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}

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