# Argument Principle

## Spring 2020 HW 3.12, Tie’s Extra Questions Fall 2015 (Root counting with argument principle) #complex/exercise/completed

Prove that $$f(z) = z^4 + 2z^3 -2z + 10$$ has exactly one root in each open quadrant.

Take a large semicircle

• $$\gamma_1 = [0, R]$$

• $$\gamma_2 = \left\{{Re^{it} {~\mathrel{\Big\vert}~}t\in [0, \pi/2]}\right\}$$

• $$\gamma_3 = i[0, R]$$

• $$\Delta\operatorname{Arg}(f, \gamma_1) = 0$$: The only way $$f\circ \gamma_1$$ can change argument is by changing sign, since it’s real valued. Use that $$f(0) = 10, f(1) = 11$$ and $$f'(t) = 4t^3+6t-2 > 0$$ so $$f$$ is increasing on $$[1, \infty)$$. Then by Rouché, on $${\left\lvert {z} \right\rvert} = 1$$ we have $${\left\lvert {z^4+2z^3-2} \right\rvert}\leq 5 < 10 = {\left\lvert {10} \right\rvert}$$, so $$f$$ has no zeros on $${\left\lvert {z} \right\rvert} \leq 1$$.

• $$\Delta\operatorname{Arg}(f, \gamma_2) = 2\pi$$: parameterize $$\gamma_2(t) = Re^{it}$$, then $$f(\gamma(t)) \sim R^4e^{it}$$ for large $$R$$, which changes argument by $$2\pi$$ for $$t$$ in $$[0, \pi/2]$$.

• $$\Delta\operatorname{Arg}(f, \gamma_3) = 0$$: check $$f(it) = t^4 + 10 + i(-2t^3-2t)$$ and we let $$t$$ range through $$[0, R]$$. For $$t>0$$, the real part is strictly positive, so this can not wind about the origin.

By the argument principle, $${\sharp}Z_f = {1\over 2\pi} \Delta\operatorname{Arg}(f, \Gamma) = 1$$.

It suffices to show there’s only one root in the open quadrant $$Q_1$$, since they come in conjugate pairs. Assume that there are no roots on $${\mathbf{R}}$$ or $$i{\mathbf{R}}$$. Since polynomials are entire, the argument principle can be used to count zeros: \begin{align*} Z_f = {1\over 2\pi i}\int_\gamma { {\partial}^{\scriptsize \log} }f(z)\,dz= \Delta_\gamma \operatorname{Arg}(f) .\end{align*} To take the curve $$\gamma$$ comprised of

• $$\gamma_1 = [0, R]$$,
• $$\gamma_2 = Re^{it}$$ for $$t\in [0, \pi/2]$$
• $$\gamma_3 = i[0, R]$$.

Then

• $$\Delta_{\gamma_1}\operatorname{Arg}(f) = 0$$, since $$f({\mathbf{R}}_{\geq 0}) \subseteq {\mathbf{R}}_{\geq 0}$$.
• $$\Delta_{\gamma_2}\operatorname{Arg}(f) = 4\cdot {\pi\over 2} = 2\pi$$ since $$f\sim z^4$$ for large $$R$$.
• $$\Delta_{\gamma_3}\operatorname{Arg}(f)$$: consider \begin{align*} f(it) = t^4 - it^3 -2it + 10 = t^4\qty{1 - it^{-1}-2it^{-2} +10t^{-4}} \\ \implies \operatorname{Arg}(f(it)) \sim \operatorname{Arg}(t^4) =0 .\end{align*}

So $$\Delta_\gamma \operatorname{Arg}(f) = 1$$, meaning there is one zero enclosed by $$\gamma$$ for $$R$$ large enough. As $$R\to \infty$$, this covers $$Q_1$$.

### $$n$$-to-one functions #complex/exercise/completed

Let $$f$$ be analytic in a domain $$D$$ and fix $$z_0 \in D$$ with $$w_0 \coloneqq f(z_0)$$. Suppose $$z_0$$ is a zero of $$f(z) - w_0$$ with finite multiplicity $$m$$. Show that there exists $$\delta >0$$ and $${\varepsilon}> 0$$ such that for each $$w$$ such that $$0 < {\left\lvert {w-w_0} \right\rvert} < {\varepsilon}$$, the equation $$f(z) - w = 0$$ has exactly $$m$$ distinct solutions inside the disc $${\left\lvert {z-z_0} \right\rvert} < \delta$$.

Write $$g(z) \coloneqq f(z) - w_0$$, then $$g$$ is holomorphic on $$D$$ and thus $$w_0$$ is an isolated zero. Choose $$\delta$$ small enough so that $$g$$ is nonvanishing on $${\mathbb{D}}_\delta(z_0)\setminus\left\{{ z_0 }\right\}$$. Let \begin{align*} \gamma \coloneqq\left\{{{\left\lvert {\xi - z_0} \right\rvert} = \delta }\right\}= {{\partial}}{\mathbb{D}}_{\delta}(z_0) .\end{align*} Choose $${\varepsilon}< \inf\left\{{w\in f(\delta)}\right\}$$ so that $${\left\lvert {f(z) - w_0} \right\rvert} > {\varepsilon}$$ in $${\mathbb{D}}_{\varepsilon}(w_0)\setminus\left\{{ w_0 }\right\}$$ for every $$z\in \gamma$$. Let \begin{align*} \gamma' \coloneqq{{\partial}}{\mathbb{D}}_{{\varepsilon}}(w_0) = \left\{{{\left\lvert {z-w_0} \right\rvert} = {\varepsilon}}\right\} ,\end{align*} and define the solution counting function: \begin{align*} F(w) \coloneqq{1\over 2\pi i} \oint_{\gamma'} { {\partial}^{\scriptsize \log} }(g(z)) \,dz = {1\over 2\pi i } \oint_{\gamma'} {g'(z)\over g(z) }\,dz = {1\over 2\pi i} \oint_{\gamma'} {f'(z)\over f(z) - w} \,dz ,\end{align*} which counts the zeros of $$g$$ (since it has no poles) and consequently the number of solutions to $$f(z) = w$$ in $${\mathbb{D}}_{\varepsilon}(w_0)$$. This is now a continuous integer valued function on $${\mathbb{D}}_{\varepsilon}(w_0)$$, and is thus constant. Since $$f(z_0) = w_0$$ with $$z_0$$ enclosed by $$\gamma$$ and $$w_0$$ enclosed by $$\gamma'$$, the constant is exactly the multiplicity of the zero of $$f(z) - w_0$$ at $$z_0$$, which is $$m$$.

### Blaschke products are $$n$$ to one #complex/exercise/completed

For $$k=1,2,\cdots, n$$, suppose $${\left\lvert {a_k} \right\rvert} < 1$$ and \begin{align*} f(z) \coloneqq\qty{z - a_1 \over 1 - \overline{a}_1 z} \qty{z-a_2 \over 1 - \overline{a}_2 z} \cdots \qty{z - a_n \over 1 - \overline{a}_n z} .\end{align*} Show that $$f(z) = b$$ has $$n$$ solutions in $${\left\lvert {z} \right\rvert} < 1$$.

Note that $$f$$ is holomorphic on $${\mathbb{D}}$$ and $$S^1$$, since the poles are at $$1/\overline{a_k}$$ and if $${\left\lvert {a_l} \right\rvert} < 1$$ then $${\left\lvert {\overline{a_k}} \right\rvert} > 1$$. Fix $$b$$, then define $$g_w(z) \coloneqq f(z) - w$$ and form the solution counting function \begin{align*} F(w) \coloneqq{1\over 2\pi i}\oint_{S^1} { {\partial}^{\scriptsize \log} }g_w(z) \,dz = {1\over 2\pi i} \oint_{S^1} {f'(z) \over f(z)-w}\,dz .\end{align*} Start by computing $$F(0)$$. \begin{align*} F(0) &= {1\over 2\pi i }\oint_{S^1} { {\partial}^{\scriptsize \log} }\prod_{1\leq k\leq n} \psi_{a_k}(z) \,dz\\ &= {1\over 2\pi i }\oint_{S^1} \sum_{1\leq k\leq n} { {\partial}^{\scriptsize \log} }\psi_{a_k}(z) \,dz\\ &= {1\over 2\pi i }\oint_{S^1} \sum_{1\leq k\leq n} \qty{1-{\left\lvert {a_k} \right\rvert}^2 \over (1-\overline{a_k} z)^2} \qty{z-a_k \over 1-\overline{a_k} z}^{-1}\,dz\\ &= {1\over 2\pi i }\oint_{S^1} \sum_{1\leq k\leq n} {1-{\left\lvert {a_k} \right\rvert}^2 \over (z-a_k)( 1-\overline{a_k}z) } \,dz\\ &= {1\over 2\pi i } \sum_{1\leq k\leq n} \oint_{S^1} {1-{\left\lvert {a_k} \right\rvert}^2 \over (z-a_k)( 1-\overline{a_k}z) } \,dz\\ &= {1\over 2\pi i } \sum_{1\leq k\leq n} 2\pi i \\ &= n ,\end{align*} where we’ve used that the integrand has a simple pole at $$a_k$$ since $$1/\overline{a_k}\in {\mathbb{D}}^c$$. So the equation $$f(z) = 0$$ has $$n$$ solutions. Now use that $$F$$ is a continuous function of $$w$$ on $${\mathbb{D}}$$ and integer valued, thus constant. So $$F(w) = n$$ for any $$w$$, meaning $$f(z) = w$$ has $$n$$ solutions in $${\mathbb{D}}$$ for every $$w$$.

Alternative: $$F$$ continuously depends on the $$a_k$$, so send them all to zero to get $$f(z) = z^n$$ which trivially has $$n$$ zeros.

#complex/exercise/completed