theorem (Rouché's Theorem):

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. $Z_M = Z_{M+m}$ .

slogan:

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.

remark:

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 $Z_f = Z_{f_n}$ .

proof (of Rouché):

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

figures/2021-12-10_22-23-58.png

proof (of Rouché, alternative):

figures/2021-12-14_16-25-41.png

figures/2021-12-15_02-24-10.png

Exercises

exercise (Explicit Rouché,

$\mathbb{D}$ ):

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

#complex/exercise/completed

solution:

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 $Z_{p} = Z_F = 4$ .

exercise (Explicit Rouche,

$2\mathbb{D}$ ):

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

#complex/exercise/completed

solution:

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 ${\left\lvert {z} \right\rvert} = 2$ , $Z_{p} = Z_F = 5$ .

exercise (Arbitrary Rouché,

$R\mathbb{D}$ ):

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).

#complex/exercise/completed

solution:

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 $R{\mathbb{D}}$ .

exercise (Explicit Rouché, half-plane):

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 $\left\{{ \Re(z) \leq 0}\right\}$ .

#complex/exercise/completed

solution:

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.

solution (Alternative):

Use the following region:

figures/2021-07-29_20-39-31.png

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 $\Re(z) \leq 0$ .

exercise (?):

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

#complex/exercise/completed

solution:

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.

exercise (?):

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

#complex/exercise/completed

solution:

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 $f+g$ .

exercise (Unique fixed points):

Show that if $f$ is holomorphic on ${\mathbb{D}}$ and continuous on $\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu$ with $f(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu) \subseteq {\mathbb{D}}$ , then $f$ has a unique fixed point in ${\mathbb{D}}$ .

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

#complex/exercise/completed

solution:

Continuous images of compact sets are compact, so $f(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu)$ 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 ${\mathbb{D}}$ .

exercise (?):

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.

#complex/exercise/completed

solution:

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 $3+5 = 8$ .