Order of zero (complex analysis)

[Incand]

I'm confused about how this related to the roots of an equation. Here's the definition:

We say that $f$ has a zero of order $m$ at $z_0$ if
$f^{(k)}(z_0)=0$ for $k=0,\dots , m-1$, but $f^{(m)}(z_0)\ne 0$
or equivalently that
$f(z) = \sum_{k=m}^\infty a_k(z-z_0)^k$, $a_m \ne 0$.

For example the function $(z^2+z-2)^3$ has zeros of order $3$ for $z=2$ and $z=-1$ which is the multiple of the roots so the easiest way to determine the order seems to just note this instead of differentiating $4$ times.

However for a function like $z^2(1-\cos z)$ this doesn't work anymore. If you differentiate this you actually find a zero of order $4$ at $z=0$ and zeros of order $2$ at $z = 2\pi n, \; = \pm 1, \pm 2, \dots$ which isn't what I would expect just looking for roots to the equation.

So when determining the order of a zero is the approach always to differentiate? Since for simple functions like polynomials they seem to be the same as the roots.

 

[RUber]

I take it you mean for your example:
$f(z) = (z^2 + z -2) ^ 3$ has zeros at $z=-2, 1$.
For polynomials, you are correct that the multiplicity of the repeated roots is the same as the order of the zero at those roots.
Example:
$p(z)=(z-z_0)^m (z-z_1)^n$ has a root of multiplicity m at $z=z_0$ and a root of multiplicity n at $z = z_1$.

If you look at the derivatives up to the (m-1)th, you will notice that all the terms will include at least one power of $(z-z_0)$.
However, the mth derivative will have one term where the $(z-z_0)$ has been reduced to a constant times $(z-z_1)^n$.
Assuming that $z_0 \neq z_1$, this means that the order of the zero at $z=z_0$ is m.

Therefore, you can show, in general, for all polynomials that the multiplicity of a root is the same as the order of the zero at that root.If you differentiate your example, you get:
$f^1(z) = 3(z^2 + z -2)^2 (2z + 1)$ which as zeros at $z=-2, 1$ and also $z=-1/2$.
And again,
$f^2(z) = 6(z^2 + z -2) (2z + 1)^2 + 6(z^2 + z -2)^2$ which as zeros at $z=-2, 1$ and not $z=-1/2$.
Once more:
$f^3(z) = 6 (2z + 1)^3+24(z^2 + z -2) (2z + 1) + 12(z^2 + z -2)(2z+1)$ has a zero at $z=-1/2$ and not $z=-2, 1$
So, using the derivative test, you confirm that -2 and 1 are zeros of order 3.

 

[Incand]

Thanks! Writing it that way make it pretty obvious for polynomials.

From your example (and my second one) it also seems we have a product rule, that is if $f(z) = g(z)h(z)$ and $g(z)$ has a zero at $a_0$ of order $n$ and $h(z)$ have a zero of order $m$ at $a_0$ then $f(z)$ have a zero of order $n+m$ at $a_0$.

This seem to follow from Leibniz rule , $f^{(n+m-1)}=(gh)^{(n+m-1)} = \sum_{k=0}^{n+m-1} \binom{n+m-1}{k} g^{(k)}h^{(n+m-1-k)}$ , since $g^{(k)}= 0$ for all $k < n$ while $h^{(n+m-k)}=0$ for $n\le k \le (n+m-1)$ hence every term is zero. Similar applies for every higher derivative in between $0$ and $(n+m-1)$.
While for the next order $(n+m)$ we have the only non-zero term $f^{(n)}g^{(m)}$ left.

 

[Incand]

Never mind second half of my post, apparently there's a really obvious way to see this directly.
From the theorem stating that:
If $f$ has a zero of order $m$ at $z_0$, then $f(z) = (z-z_0)^mg(z)$, where $g$ is analytic in $D$ and $g(z_0) \ne 0$. This makes it really obvious when multiplying two functions, and also the polynomial part.

Anyway thanks again for the explanation, I believe I understand the order of zeros now!