# Rouché

Let $$M, m$$ be meromorphic on $$\Omega$$ and write $$Z_M, Z_m, P_M, P_m$$ for the numbers of zeros and poles of $$M$$ and $$m$$ respectively. Suppose $$\gamma \subseteq \Omega$$ is a toy contour winding about each zero and pole of $$f$$ and $$g$$ precisely once. Then \begin{align*} {\left\lvert {m} \right\rvert} \leq {\left\lvert {M} \right\rvert} \text{ on } \gamma \implies \mathop{\mathrm{Index}}_{z=0}(M\circ \gamma)(z) &= \mathop{\mathrm{Index}}_{z=0}((M+m)\circ \gamma)(z) \\ \implies Z_M - P_M &= Z_{M+m} - P_{M+m} .\end{align*} In particular, if $$M, m$$ are holomorphic on $$\Omega$$, then $$M$$ and $$M+m$$ have the same number of zeros in $$\Omega$$, i.e. $$Z_M = Z_{M+m}$$.

The number of zeros/poles in a region is determined by a dominating function on the boundary. You can add a small perturbation $$m$$ to $$M$$ and preserve the number of zeros, where “small” means $${\left\lvert {m} \right\rvert} < {\left\lvert {M} \right\rvert}$$ on the boundary.

On how to use Rouché, and some common tricks:

• Given $$f$$ and a region, find a big part $$M$$ and a small part $$m \coloneqq f-M$$. Then show $${\left\lvert {m} \right\rvert} < {\left\lvert {M} \right\rvert}$$ to get $${\sharp}Z_M = {\sharp}Z_f$$.
• It should also be clear how many zeros $$M$$ has in the region!
• Given $$f$$, just find a large part $$M$$, and show $${\left\lvert {f-M} \right\rvert} < M$$.
• Given $${\left\lvert {m} \right\rvert} < {\left\lvert {M} \right\rvert}$$ with no ambient $$f$$, you can freely choose $$f$$ to be any of $$\pm (M \pm m)$$ to obtain $$Z_M = Z_f$$
• Given $$f$$ and $$g$$, show $${\left\lvert {f-g} \right\rvert} < {\left\lvert {f} \right\rvert}$$ to get $$Z_f = Z_g$$.
• This can be improved to $${\left\lvert {f-g} \right\rvert} < {\left\lvert {f} \right\rvert} + {\left\lvert {g} \right\rvert}$$ using the symmetric/extended version of the theorem.
• A common trick: show $${\left\lvert {f-g} \right\rvert} < 1$$ and either $${\left\lvert {f} \right\rvert} > 1$$ or $${\left\lvert {g} \right\rvert} > 1$$.
• For power series $$f_n(z) \to f(z)$$: find a lower bound $$L$$ for $$f$$ and an upper bound for the tail $$f - f_n$$ to get $${\left\lvert {f_n - f} \right\rvert} < U < L < {\left\lvert {f} \right\rvert}$$ to get $$Z_f = Z_{f_n}$$.

Idea: use argument principle on $$(f+g)/f$$. Alternatively, use that $$N(f+tg, \Omega)$$ is a continuous $${\mathbf{Z}}{\hbox{-}}$$valued function for all $$t\in [0, 1]$$.

# Exercises

Find the number of zeros in $${\left\lvert {z} \right\rvert} < 1$$ of \begin{align*} p(z) \coloneqq z^6 + 9z^4 + z^3 + 2z + 4 .\end{align*}

Strategy: bound the difference. Find the big and small term:

• Big: $$F(z) = 9z^4$$, so $${\left\lvert {F(z)} \right\rvert} = 9$$ on the boundary
• Small: $$g(z) = p(z) - F(z) = z^6 + z^3 + 2z + 4$$, so $${\left\lvert {g(z)} \right\rvert}\leq 1+1+2+4=8$$ on the boundary.

So $${\left\lvert {p-F} \right\rvert} \leq {\left\lvert {F} \right\rvert}$$ on $${\left\lvert {z} \right\rvert} = 2$$, meaning $$Z_{p} = Z_F = 4$$.

Find the number of zeros in $${\left\lvert {z} \right\rvert} < 2$$ of \begin{align*} h(z) \coloneqq z^5 + 3z + 1 .\end{align*}

Strategy: bound the difference.

• Big: $$F(z) \coloneqq z^5$$ so $${\left\lvert {F(z)} \right\rvert} = 2^5 = 32$$ on $${\left\lvert {z} \right\rvert} = 2$$
• Small: $$g(z) \coloneqq p(z) - F(z) = 3z+1$$, so $${\left\lvert {g(z)} \right\rvert} \leq 3{\left\lvert {z} \right\rvert}+ 1 = 7$$ on $${\left\lvert {z} \right\rvert} = 2$$.

Then $${\left\lvert {g} \right\rvert}\leq {\left\lvert {F} \right\rvert}$$ on $${\left\lvert {z} \right\rvert} = 2$$, $$Z_{p} = Z_F = 5$$.

Find the number of zeros in $${\left\lvert {z} \right\rvert} < R$$ of \begin{align*} p(z) \coloneqq z^d + a_1z^{d-1} + \cdots + a_d ,\end{align*} supposing that $${\left\lvert {a_k} \right\rvert}< {R^k \over d}$$ for every $$k$$ (noting the strict inequality).

Strategy: bound the difference. Find the big and small term:

• Big: $$F(z) = z^d$$, so $${\left\lvert {F(z)} \right\rvert} = R^d$$ on $${\left\lvert {z} \right\rvert} = R$$.
• Small: $$g(z) = p(z) - F(z) = a_1 z^{d-1} + \cdots + a_d$$, so \begin{align*} {\left\lvert {g(z)} \right\rvert} &\leq {\left\lvert {a_1} \right\rvert} R^{d-1} + {\left\lvert {a_2} \right\rvert} R^{d-2} + \cdots + {\left\lvert {a_d} \right\rvert} \\ &< {R\over d} \cdot R^{d-1} + {R^2 \over d} \cdot R^{d-2} + \cdots + {R^{d-1} \over d} \cdot R + {R^{d} \over d} \\ &= d {R^d\over d} = R^d ,\end{align*} so $${\left\lvert {g} \right\rvert} < R^d = {\left\lvert {F} \right\rvert}$$, meaning $$Z_{p-F} = Z_F = d$$ in $$R{\mathbb{D}}$$.

Find the number of solutions in $$\left\{{\Re(z) \leq 0}\right\}$$ of \begin{align*} -2e^z = z+3 .\end{align*}

Hint: show $$h(z) = z + 3 + 2e^z$$ has one root in $$\left\{{ \Re(z) \leq 0}\right\}$$.

Note that $${\left\lvert {e^z} \right\rvert} = e^{\Re(z)} \leq e^{0} = 1$$ since $$\Re(z) \leq 0$$, so if the equality holds then \begin{align*} {\left\lvert {2e^z} \right\rvert} = {\left\lvert {z+3} \right\rvert} \implies {\left\lvert {z+3} \right\rvert}\leq 2 .\end{align*} So apply Rouché to $$\Omega$$ the circle of radius 2 centered at $$z=-3$$. Write $$p(z) \coloneqq z+3 + 2e^z$$, then

• Big: $$F(z) = z+3$$, so $${\left\lvert {F(z)} \right\rvert} = 2$$ on $${{\partial}}\Omega$$.
• Small: $$g(z) = 2e^z$$, so $${\left\lvert {g(z)} \right\rvert} = 2e^{\Re(z)} < 2$$ in $$\Omega$$.

Then $$Z_p = Z_F = 1$$, and any such zero is a solution to the original equation.

Use the following region:

Consider $$p(z) \coloneqq z+3+2e^z$$, take $$F(z) \coloneqq z+3$$ and $$h(z) \coloneqq 2e^z$$ for the perturbation. On $$C_1, z=it$$ for $$t\in [-R, R]$$, so \begin{align*} {\left\lvert {F(z)} \right\rvert} &= {\left\lvert {3+it} \right\rvert} \geq 3 \\ {\left\lvert {h(z)} \right\rvert} &= 2e^{\Re(iy)}=2 ,\end{align*} so $${\left\lvert {h} \right\rvert} < {\left\lvert {F} \right\rvert}$$ here. On $${\left\lvert {z} \right\rvert} = R$$, $${\left\lvert {h(z)} \right\rvert} < 2e^{\Re(z)} < 2$$ since $$\Re(z) < 0$$, and $${\left\lvert {F(z)} \right\rvert} = { \mathsf{O}}(R)$$, so for $$R\gg 1$$ we have $${\left\lvert {F} \right\rvert} > {\left\lvert {h} \right\rvert}$$ here too.

Thus $$Z_{h+F} = Z_f = 1$$ in this region, and taking $$R\to\infty$$ covers all of $$\Re(z) \leq 0$$.

Show that $$P(z) \coloneqq z^4 + 6z + 3$$ has 3 zeros in $$\left\{{1\leq {\left\lvert {z} \right\rvert} \leq 2}\right\}$$.

• Take $$P(z) = z^4 + 6z + 3$$.
• On $${\left\lvert {z} \right\rvert} < 2$$:
• Set $$f(z) = z^4$$ and $$g(z) = 6z + 3$$, then $${\left\lvert {g(z)} \right\rvert} \leq 6{\left\lvert {z} \right\rvert} + 3 = 15 < 16= {\left\lvert {f(z)} \right\rvert}$$.
• So $$P$$ has 4 zeros here.
• On $${\left\lvert {z} \right\rvert} < 1$$:
• Set $$f(z) = 6z$$ and $$g(z) = z^4 + 3$$.
• Check $${\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^4 + 3 = 4 < 6 = {\left\lvert {f(z)} \right\rvert}$$.
• So $$P$$ has 1 zero here.

Show that $$\alpha z e^z = 1$$ where $${\left\lvert {\alpha} \right\rvert} > e$$ has exactly one solution in $${\mathbb{D}}$$.

• Set $$f(z) = \alpha z$$ and $$g(z) = e^{-z}$$.
• Estimate at $${\left\lvert {z} \right\rvert} =1$$ we have $${\left\lvert {g} \right\rvert} ={\left\lvert {e^{-z}} \right\rvert} = e^{-\Re(z)} \leq e^1 < {\left\lvert {\alpha} \right\rvert} = {\left\lvert {f(z)} \right\rvert}$$
• $$f$$ has one zero at $$z_0 = 0$$, thus so does $$f+g$$.

Show that if $$f$$ is holomorphic on $${\mathbb{D}}$$ and continuous on $$\overline{{\mathbb{D}}}$$ with $$f(\overline{{\mathbb{D}}}) \subseteq {\mathbb{D}}$$, then $$f$$ has a unique fixed point in $${\mathbb{D}}$$.

Note: this is subtle because $${\mathbb{D}}$$ is not compact!

Continuous images of compact sets are compact, so $$f(\overline{{\mathbb{D}}})$$ is a compact subset of $${\mathbb{D}}$$ and thus contained in some $${\mathbb{D}}_r(0)$$ with $$0<r<1$$. On this disc, \begin{align*} {\left\lvert {f(z)} \right\rvert} = {\left\lvert {f(z) - z + z} \right\rvert} < {\left\lvert {z} \right\rvert} .\end{align*} By Rouché, $$f(z)-z$$ and $$z$$ have the same number of zeros, which is one. This holds for any $$r'$$ with $$r<r'<1$$, and thus holds on $${\mathbb{D}}$$.

Find the number of zeros in $${\left\lvert {z} \right\rvert} \in (1, 2)$$ of \begin{align*} f(z) \coloneqq z^4 + 5z + 3 .\end{align*} Note the strict inequality.

On $${\left\lvert {z} \right\rvert} = 1$$:

• Big: $$M(z) = 5z$$
• Small: $$m(z) = z^4 + 3$$
• The estimate: \begin{align*} {\left\lvert {m(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^4 + 3 = 4 < 5 = {\left\lvert {M(z)} \right\rvert} \implies 1 = {\sharp}Z_M = {\sharp}Z_f .\end{align*}

On $${\left\lvert {z} \right\rvert} = 2$$:

• Big: $$M(z) = z^4$$
• Small: $$m(z) = 5z + 3$$
• The estimate: \begin{align*} {\left\lvert {m(z)} \right\rvert} \leq 5{\left\lvert {z} \right\rvert} + 3 = 13 < 16 = 2^4 = {\left\lvert {M(z)} \right\rvert} \implies 4 = {\sharp}Z_M = {\sharp}Z_f .\end{align*}

So $$f$$ has $$4-1=3$$ zeros on $${\left\lvert {z} \right\rvert} \in (0, 2)\setminus(0, 1) = [1, 2)$$. To get the strict inequality, it suffices to show $$f$$ has no zeros on $${\left\lvert {z} \right\rvert} = 1$$. Estimate a lower bound: \begin{align*} {\left\lvert {z^4 + 5z + 3} \right\rvert} \geq {\left\lvert {{\left\lvert {5z+3} \right\rvert} - {\left\lvert {z} \right\rvert}^4 } \right\rvert} = {\left\lvert {{\left\lvert {5z+3} \right\rvert} - 1 } \right\rvert} \geq {\left\lvert { 2-1} \right\rvert} > 0 .\end{align*} The nontrivial bound at the end comes from the fact that $$5z+3 = 3 + 5e^{it}$$ whose modulus ranges between $$3-5 = -2$$ and $$3+5 = 8$$.

#complex/exercise/completed