Zeros and Poles

Let $f$ be analytic in $D_R(\alpha)$ and $k$ be a positive integer. $f$ has a zero of order $k$ at $z = \alpha$ iff

A zero of order 1 is called a simple zero.

Zeros

Let the Taylor series of $f$ centered at $\alpha$ be $\sum_{n=0}^{\infty} c_n (z-\alpha)^n$. $f$ has a zero of order $k$ at $z = \alpha$ iff

Equivalently, $f$ has a zero of order $k$ at $z = \alpha$ iff

where $g$ is analytic at $\alpha$ and $g(\alpha) \neq 0$.

If $f$ and $g$ have zeros of orders $m$ and $n$ at $z = \alpha$, then $fg$ has a zero of order $m + n$ at $z = \alpha$.

Examples:

• $z \sin(z^2)$ has a zero of order 3 at $z = 0$

Poles

Let $f$ be analytic in the punctured disk $D_R^*(\alpha)$. $f$ has a pole of order $k$ at $z = \alpha$ iff

where $h$ is analytic at $\alpha$ and $h(\alpha) \neq 0$.

If $f$ and $g$ have poles of orders $m$ and $n$ at $z = \alpha$, then $fg$ has a pole of order $m + n$ at $z = \alpha$.

Reciprocals

• If $f$ is analytic at $\alpha$ with a zero of order $k$, then $\frac{1}{f}$ has a pole of order $k$ at $\alpha$.
• If $f$ has a pole of order $k$ at $\alpha$, then $\frac{1}{f}$ has a removable singularity at $\alpha$. After redefining the value at $\alpha$, the resulting analytic function has a zero of order $k$ at $\alpha$.

Quotients

Let $f$ and $g$ be analytic at $\alpha$ with zeros of orders $m$ and $n$ respectively. Let $h = \frac{f}{g}$.

• If $m > n$
  $h$ has a removable singularity at $\alpha$. After redefining $h(\alpha) = 0$, $h$ has a zero of order $m - n$.
• If $m < n$
  $h$ has a pole of order $n - m$ at $\alpha$.
• If $m = n$
  $h$ has a removable singularity at $\alpha$. Redefining $h(\alpha) = \lim_{z \to \alpha} h(z)$ makes $h$ analytic at $\alpha$.

Examples:

• $\frac{\tan z}{z}$ has a removable singularity at $z = 0$, simple zeros at $z = k\pi$ for $k \neq 0$, and simple poles at $z = \frac{\pi}{2} + k\pi$
• $\frac{1}{5z^4 + 26z^2 + 5}$ has simple poles at $z = \pm \frac{i}{\sqrt{5}}$ and $z = \pm i\sqrt{5}$

Worked Examples

Zero by Derivatives

Find the order of the zero of $f(z) = 1 - \cos z$ at $z = 0$.

$f$ has a zero of order 2.

Zero by Taylor Series

Show that $z \sin(z^2)$ has a zero of order 3 at $z = 0$.

$c_0 = c_1 = c_2 = 0$ and $c_3 = 1 \neq 0$, so the zero has order 3.

In factored form, $z \sin(z^2) = z^3 g(z)$ where $g(z) = \frac{\sin(z^2)}{z^2}$ is analytic at $0$ with $g(0) = 1$.

Product of Zeros

Find the order of the zero of $(1 - \cos z)\sin z$ at $z = 0$.

$1 - \cos z$ has a zero of order 2 and $\sin z$ has a zero of order 1. The product has a zero of order $2 + 1 = 3$.

Pole by Factored Form

Show that $\frac{\sin z}{z^3}$ has a pole of order 2 at $z = 0$.

$h$ is analytic at $0$ with $h(0) = 1 \neq 0$, so the pole has order 2.

Product of Poles

Find the order of the pole of $\frac{e^z}{z} \cdot \frac{1}{z^2}$ at $z = 0$.

$\frac{e^z}{z}$ has a pole of order 1 and $\frac{1}{z^2}$ has a pole of order 2. The product $\frac{e^z}{z^3}$ has a pole of order $1 + 2 = 3$.

Reciprocal

Find the singularity of $\frac{1}{1 - \cos z}$ at $z = 0$.

$1 - \cos z$ has a zero of order 2 at $0$, so $\frac{1}{1 - \cos z}$ has a pole of order 2.

Conversely, $\frac{1}{z^2}$ has a pole of order 2, so its reciprocal $z^2$ has a zero of order 2.

Quotient of Zeros

Let $f$ have a zero of order $m$ and $g$ a zero of order $n$ at $0$.

• For $f = z^3$ and $g = \sin z$, $m = 3 > n = 1$. $\frac{z^3}{\sin z}$ has a removable singularity. After setting the value to $0$, it has a zero of order $2$.
• For $f = \sin z$ and $g = z^3$, $m = 1 < n = 3$. $\frac{\sin z}{z^3}$ has a pole of order $2$.
• For $f = \sin z$ and $g = z$, $m = n = 1$. $\frac{\sin z}{z}$ has a removable singularity. Setting the value to $\lim_{z \to 0} \frac{\sin z}{z} = 1$ makes it analytic.

Locating Zeros and Poles

Locate the zeros and poles of $h(z) = \frac{\tan z}{z} = \frac{\sin z}{z \cos z}$.

• At $z = 0$, the numerator and denominator both have simple zeros. $h$ has a removable singularity with $h(0) = 1$.
• At $z = k\pi$ with $k \neq 0$, the numerator has a simple zero and the denominator is non-zero. $h$ has a simple zero.
• At $z = \frac{\pi}{2} + k\pi$, the numerator equals $\pm 1$ and $\cos z$ has a simple zero. $h$ has a simple pole.

Locate the zeros and poles of $f(z) = \frac{z^2 + 1}{z^2 (z-1)^3}$.

$z^2 + 1 = (z-i)(z+i)$, so $f$ has simple zeros at $z = \pm i$. The numerator is non-zero at $0$ and $1$, so $f$ has a pole of order 2 at $z = 0$ and a pole of order 3 at $z = 1$.

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

The 4 roots $z = \pm \frac{i}{\sqrt{5}}$ and $z = \pm i\sqrt{5}$ are distinct, so each is a simple zero of the denominator. $g$ has simple poles at all 4 points.