Argument Principle

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 \({\mathbb{R}}\) or \(i{\mathbb{R}}\). 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 \(f({\mathbb{R}}_{\geq 0}) \subseteq {\mathbb{R}}_{\geq 0}\).
• \(\Delta_{\gamma_2}\operatorname{Arg}(f) = 4\cdot {\pi\over 2} = 2\pi\) 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 \(Q_1\).

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 \(m\).

Note that \(f\) is holomorphic on \({\mathbb{D}}\) and \(S^1\), since the poles are at \(1/\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5mu\) and if \({\left\lvert {a_l} \right\rvert} < 1\) then \({\left\lvert {\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5mu} \right\rvert} > 1\). 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-\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5mu z)^2} \qty{z-a_k \over 1-\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5mu 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-\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5muz) } \,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-\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5muz) } \,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 \(1/\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5mu\in {\mathbb{D}}^c\). 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.