# Argument Principle

The logarithmic derivative is defined as \begin{align*} { {\partial}^{\scriptsize \log} }f \coloneqq{f' \over f} .\end{align*}

It converts all poles and zeros of meromorphic $$f$$ into simple poles of $${ {\partial}^{\scriptsize \log} }f$$. If $$z_0$$ is a root of $$f$$ of multiplicity $$m$$, write $$f(z) = (z-z_0)^m g(z)$$ with $$g$$ holomorphic and nonzero near $$z_0$$. Then take log derivatives: \begin{align*} { {\partial}^{\scriptsize \log} }f(z) &= { {\partial}^{\scriptsize \log} }(z-z_0)^m g(z) \\ &= { {\partial}^{\scriptsize \log} }(z-z_0)^m + { {\partial}^{\scriptsize \log} }g(z) \\ &= {m\over (z-z_0)} + {g'(z) \over g(z)} .\end{align*}

Then if $$g$$ is holomorphic and nonzero away from $$z_0$$, so is $$g'/g$$. So the only contribution to $$\mathop{\mathrm{Res}}_{z=z_0} { {\partial}^{\scriptsize \log} }f$$ is $$m$$.

Note that the logarithmic derivative picks up the $$p{\hbox{-}}$$adic valuation for $$\left\langle{x-p}\right\rangle \in {\mathbf{C}}[x]$$ a point: \begin{align*} d \qty{ \log(f) } = {f'\over f}\,dz\implies \mathop{\mathrm{Res}}_{z=p}(d \log(f) ) = v_p(f) .\end{align*}

For $$\gamma \subseteq \Omega$$ a closed curve not passing through a point $$z_0$$, the winding number of $$\gamma$$ about $$z_0$$ (or the index) is defined as \begin{align*} \mathop{\mathrm{Index}}_{z=z_0}(\gamma) \coloneqq{1\over 2\pi i} \int_\gamma {1\over \xi -z_0}\,d\xi .\end{align*}

For $$f$$ meromorphic in $$\Omega$$ with multisets of zeros $$Z_f \coloneqq\left\{{ z_j }\right\}$$ and poles $$P_f\coloneqq\left\{{ p_k }\right\}$$ (so repeated with multiplicity) for $$\gamma \coloneqq{{\partial}}\Omega$$ not intersecting any of the zeros/poles,

\begin{align*} {1\over 2\pi i} \int_\gamma { {\partial}^{\scriptsize \log} }f(z) \,dz \coloneqq{1\over 2\pi i} \int_\gamma {f'(z) \over f(z)} \,dz= &= {\sharp}Z_f - {\sharp}P_f ,\end{align*} where $${\sharp}Z_f$$ and $${\sharp}P_f$$ are the number of zeros and poles respectively, counted with multiplicity. If $$f$$ is holomorphic, then \begin{align*} {1\over 2\pi i} \oint_{{{\partial}}\Omega} {f'(z) \over f(z)}\,dz &= \sum_{z_k\in f^{-1}(0) \cap\Omega} \mathrm{mult}(f, z_k) \\ {1\over 2\pi i} \oint_{{{\partial}}\Omega} {zf'(z) \over f(z)}\,dz &= \sum_{z_k\in f^{-1}(0) \cap\Omega} f(z_k) \mathrm{mult}(f, z_k) \\ .\end{align*}

• If $$z_0$$ is a zero of $$f$$ of order $$m$$, write $$f(z) = (z-z_0)^m g(z)$$ with $$g(z)$$ holomorphic and nonzero on some neighborhood of $$z_0$$.

• Compute \begin{align*} { {\partial}^{\scriptsize \log} }f(z) &= \frac{m\left(z-z_{0}\right)^{m-1} g(z)+\left(z-z_{0}\right)^{m} g^{\prime}(z)}{\left(z-z_{0}\right)^{m} g(z)} \\ &= {m \over z-z_0} + { {\partial}^{\scriptsize \log} }g(z) ,\end{align*} so $$z_0$$ is a simple pole of $${ {\partial}^{\scriptsize \log} }f$$ and $$\operatorname{res}_{z=z_0} { {\partial}^{\scriptsize \log} }f = m$$.

• If $$z_0$$ is a pole of $$f$$ of order $$m$$, write $$f(z) = (z-z_0)^{-m} g(z)$$, then \begin{align*} { {\partial}^{\scriptsize \log} }f = {-m \over z-z_0} + { {\partial}^{\scriptsize \log} }g ,\end{align*} so $$z_0$$ is a simple pole and $$\mathop{\mathrm{Res}}_{z=z_0} {\partial}_{\log f} = -m$$.

• Now apply the residue theorem, and group residues according to sign: \begin{align*} {1\over 2\pi i } \int_{\gamma} {\partial}_{\log }f(z) \,dz &= \sum_{z_i \in P_{{ {\partial}^{\scriptsize \log} }f}} \mathop{\mathrm{Res}}_{z=z_i} { {\partial}^{\scriptsize \log} }f(z)\\ &= \sum_{z_k \in Z_f} \mathop{\mathrm{Res}}_{z=z_k} f(z) - \sum_{z_j \in P_f} \mathop{\mathrm{Res}}_{z=z_j} f(z) .\end{align*}

With the same setup as above, \begin{align*} {1\over 2\pi i} \int_\gamma { {\partial}^{\scriptsize \log} }f(z) \,dz &= \mathop{\mathrm{Index}}_{w=0}(f\circ \gamma)(w) .\end{align*}

Make the change of variables $$w = f(z)$$, then $$z=\gamma(t) \mapsto w = (f\circ \gamma)(t)$$ and $$\,dw= f'(z) \,dz$$, so \begin{align*} {1\over 2\pi i }\int_{\gamma} { {\partial}^{\scriptsize \log} }f(z) \,dz = {1\over 2\pi i} \int_{f\circ \gamma} {1\over w} \,dw\coloneqq\mathop{\mathrm{Index}}_{w=0} (f\circ \gamma)(w) .\end{align*}

Let $$f(z) = z^2 + z = z(z+1)$$.

• $$\gamma_1 \coloneqq\left\{{{\left\lvert {z} \right\rvert} = 2}\right\}$$ contains 2 zeros and 0 poles, so $$f\circ \gamma$$ winds twice around zero counterclockwise.
• $$\gamma_2 \coloneqq\left\{{{\left\lvert {z} \right\rvert} = {1\over 2}}\right\}$$ contains 1 zero and 0 poles, so $$f\circ \gamma$$ winds once.

You can track the change in argument by just breaking a curve up into sub-curves and evaluating a branch of the $$\arg$$ function at the endpoints. For example, in this picture, the change in argument is $$\pi$$ no matter what the curve does in $${\mathbb{H}}$$:

The integral function \begin{align*} F(w) \coloneqq{1\over 2\pi i} \oint_{{{\partial}}\Omega} {f'(z) \over f(z) - w} \,dw \end{align*} counts the number of solutions to $$f(z) = w$$ in $$\Omega$$, since it’s of the form $$\int_\gamma { {\partial}^{\scriptsize \log} }g_w(z)\,dz$$ for $$g_w(z) \coloneqq f(z) - w$$. This is continuous provided $$f(z) \neq w$$ on $${{\partial}}\Omega$$ and is $${\mathbf{Z}}{\hbox{-}}$$valued, thus constant on connected components.

Also useful: zeros of $$f$$ with multiplicity $$m\geq 2$$ are zeros of $$f'$$. This also holds for $$f(z) -a$$.

## Exercises

Show that $${ {\partial}^{\scriptsize \log} }(fg) = { {\partial}^{\scriptsize \log} }f + { {\partial}^{\scriptsize \log} }g$$, i.e.  \begin{align*} { (fg)' \over fg} = {f'\over f} + {g' \over g} .\end{align*}

\begin{align*} {(fg)' \over fg} = { f'g + fg' \over fg} = {f'g \over fg} + {fg' \over fg} = {f'\over f} + {g' \over g} .\end{align*}

Show that if $$z_0$$ is a zero of $$f'$$ of order $$n-1$$, then $$f$$ is $$n$$-to-one in a neighborhood of $$z_0$$.

Wlog, assume $$z_0 = 0$$. We want to show that there exists discs $$U = {\mathbb{D}}_r(0)$$ and $$W = {\mathbb{D}}_R(0)$$ such that the fiber of $$f:U\to W$$ has exactly $$n$$ distinct points. Since $$0$$ is a zero of order $$n$$, expand $$f$$ as $$\sum_{k\geq n} c_k z^k = z^n\sum_{k\geq 0} c_{k+n}z^k$$. By dividing coefficients through, we may assume $$c_n = 1$$, so \begin{align*} f(z) = z^n + \qty{ c_{n+1} z^{n+1} + c_{n+2}z^{n+2} + \cdots} = z^n + z^{n+1} \sum_{k\geq 0} c_{k+n+1}z^k \coloneqq z^n + g(z) .\end{align*}

By Rouché, $$f(z)$$ and $$z^n$$ have the same number of zeros in a small disc $${\mathbb{D}}_\rho(0)$$.

Write $$m(z) = \sum_{k\geq 0}c_{k+n}z^k$$ and $$M(z) = z^n$$; then if $${\left\lvert {m(z)} \right\rvert} < {\left\lvert {M(z)} \right\rvert}$$ for any circle $${\left\lvert {z} \right\rvert} = \rho$$ with $$\rho< 1$$ then $$M$$ and $$m+M = f$$ will have the same number of zeros ($$n$$ with multiplicity).

Bounding $$m$$, the tail of the Laurent series of $$f$$: by Cauchy’s integral formula, on a disc of radius $$R$$, \begin{align*} c_k = {f^{(n)}(z_0) \over n!} = {1\over 2\pi i} \oint_{{\left\lvert {\xi} \right\rvert} = R} { f(\xi) \over (\xi - z_0)^{n+1} } \,d\xi ,\end{align*} so \begin{align*} {\left\lvert {c_k} \right\rvert} \leq \max_{{\left\lvert {\xi} \right\rvert} = R}{\left\lvert {f(\xi)} \right\rvert} R^{-k} \coloneqq{M_R \over R^{k}} .\end{align*}

We can now estimate $$g$$: \begin{align*} {\left\lvert {g(z)} \right\rvert} &= {\left\lvert {z^n \sum_{k\geq 0} c_{k+n+1} z^k} \right\rvert} \\ &\leq {\left\lvert {z} \right\rvert}^n \sum_{k\geq 0} {\left\lvert { c_{k+n+1}} \right\rvert} {\left\lvert {z} \right\rvert}^k \\ &\leq \sum_{k\geq 0} {M_R \over R^{k+n+1}} \rho^k \\ &= {\left\lvert {z} \right\rvert}^n {M_R \over R^{n+1}} \sum_{k\geq 0} \qty{\rho\over R}^k \\ &= {\left\lvert {z} \right\rvert}^n {M_R \over R^{n+1}} \qty{1\over 1- {\rho \over R}} \\ &= {\left\lvert {z} \right\rvert}^n {M_R \over R^{n+1}} {R\over R-\rho} \\ &= {\left\lvert {z} \right\rvert}^n \qty{ {M_R\over R^n( R-\rho)} } \\ &\coloneqq{\left\lvert {z} \right\rvert}^n C_{R, \rho} ,\end{align*} and $$R, \rho$$ can be chosen such that $$C_{R, \rho} < 1$$.

Thus on $${\left\lvert {z} \right\rvert} = \rho$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {g(z) } \right\rvert} \leq C_{R, \rho} {\left\lvert {z} \right\rvert} < {\left\lvert {z} \right\rvert} = {\left\lvert {M(z)} \right\rvert} .\end{align*}

So the fiber above $$z=0$$ is of size $$n$$, the claim is that this is also true in a neighborhood of zero. The above estimate also shows that for $$0 < {\left\lvert {z} \right\rvert}\leq \rho$$, $${\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {z^n} \right\rvert}$$, and so \begin{align*} {\left\lvert {g(z)} \right\rvert} = {\left\lvert {g(z) - z^n + z^n} \right\rvert} \geq {\left\lvert { {\left\lvert {g(z) - z^n} \right\rvert} - {\left\lvert {z^n} \right\rvert} } \right\rvert} > 0 ,\end{align*} so $$g$$ is nonzero on $${\mathbb{D}}_\rho(0)\setminus\left\{{0}\right\}$$. For the zero-counting function \begin{align*} F(w) \coloneqq{1\over 2\pi i} \oint_{{\left\lvert {\xi} \right\rvert} = \rho'} {f'(\xi) \over f(\xi) - w }\,d\xi .\end{align*} Taking $$\rho ' < \min_{{\left\lvert {\xi} \right\rvert} = \rho} {\left\lvert {f(z)} \right\rvert}$$ makes this a holomorphic function of $$w$$ on $${\mathbb{D}}_{\rho'}(0)$$, and as a continuous $${\mathbf{Z}}{\hbox{-}}$$valued function it is constant. Since $$F(0) = n$$, this forces $$F(w) = n$$ for all $${\left\lvert {w} \right\rvert} < \rho'$$, so there are $$n$$ solutions to $$f(z) = w$$ in these discs. After shrinking these discs if necessary, $$f'\neq 0$$ is nonvanishing on a punctured disc, so $$f$$ is injective there and these solutions are distinct.

#complex/exercise/completed