# Rouché’s Theorem

## Standard Applications

### Tie’s Extra Questions: Fall 2009, Fall 2011, Spring 2014 (FTA) #complex/exercise/completed

Use Rouche’s theorem to prove the Fundamental Theorem of Algebra.

Write $$f(z) = \sum_{k\leq n} c_k z^k$$. Big: $$M(z) = c_nz^n$$. Small: $$m(z) = f(z) - M(z) = \sum_{k\leq n-1} c_k z^k$$.

Now use that \begin{align*} {\left\lvert {m(z) \over M(z)} \right\rvert} &\coloneqq{\left\lvert {c_n^{-1}\sum_{k\leq n-1} c_k z^{k-n}} \right\rvert} \\ &= {\left\lvert {c_n^{-1}\qty{ {c_1\over z^n} + {c_2\over z^{n-1} } + \cdots + {c_{n-1}\over z} }} \right\rvert} \\ &\overset{{\left\lvert {z} \right\rvert}\to\infty}\longrightarrow 0 ,\end{align*} so choose $$R$$ large enough such that for $${\left\lvert {z} \right\rvert} \geq R$$, $${\left\lvert {M(z)\over m(z)} \right\rvert} < 1$$. Then on $${\left\lvert {z} \right\rvert} = R$$, \begin{align*} {\left\lvert {m(z) \over M(z)} \right\rvert} < 1 \implies {\left\lvert {m(z)} \right\rvert} < {\left\lvert {M(z)} \right\rvert} \implies {\sharp}n = {\sharp}Z_{M} = {\sharp}Z_{M+m} = {\sharp}Z_{f} ,\end{align*} since $$c_n z^n$$ has $$z=0$$ as a root with multiplicity $$n$$.

An estimate: write $$f(z) = \sum_{k\leq n} c_k z^k$$ with $$c_n = 1$$, then for $$R> 1$$, on $${\left\lvert {z} \right\rvert} = R$$ we have \begin{align*} {\left\lvert {f(z) - z^n} \right\rvert} &\leq \sum_{k\leq n-1} {\left\lvert { c_k z^k} \right\rvert} \\ &\leq \sum_{k\leq n-1} {\left\lvert { c_k} \right\rvert} R^k \\ &\leq \sum_{k\leq n-1} {\left\lvert { c_k} \right\rvert} R^{n-1} \\ &= R^{n-1} \sum_{k\leq n-1} {\left\lvert { c_k} \right\rvert} \\ &\coloneqq R^{n-1} C \\ &\leq R^n \\ &= {\left\lvert {z^n} \right\rvert} ,\end{align*} provided we can choose $$C<R$$, but this is possible since $$\sum_{k\leq n-1}{\left\lvert {c_k} \right\rvert}$$ is a constant. So $$n = {\sharp}Z_{z^n} = {\sharp}Z_f$$.

### Tie’s Extra Questions: Fall 2015 (Standard polynomial) #complex/exercise/completed

Find the number of roots of $$z^4 - 6z + 3 =0$$ in $$|z|<1$$ and $$1 < |z| < 2$$ respectively.

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

• Big: $$M(z) = -6z$$
• Small: $$m(z) = z^4 + 3$$

Then on $${\left\lvert {z} \right\rvert} = 1$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^4 + 3 = 4 < 6 = {\left\lvert {-6z} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so $$1 = Z_M = Z_f$$ here.

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

• Big: $$M(z) = z^4$$
• Small: $$m(z) = -6z+3$$.

Then on $${\left\lvert {z} \right\rvert} = 2$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {-6z + 3} \right\rvert} \leq 6{\left\lvert {z} \right\rvert} + 3 = 15 < 16 = 2^4 = {\left\lvert {z} \right\rvert}^4 = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so $$4 = Z_M = Z_f$$ here.

Thus there are $$4-1 = 3$$ zeros in $$1 \leq {\left\lvert {z} \right\rvert} \leq 2$$.

### Tie’s Extra Questions: Fall 2016 (Standard polynomial) #complex/exercise/completed

Prove that all the roots of the complex polynomial \begin{align*}f(z) = z^7 - 5 z^3 +12 =0\end{align*} lie between the circles $$|z|=1$$ and $$|z|=2$$.

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

• Big: $$M(z) = 12$$
• Small: $$m(z) = z^6 - 5z^3$$

For $${\left\lvert {z} \right\rvert} = 1$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} \coloneqq{\left\lvert {z^7-5z^3} \right\rvert}\leq {\left\lvert {z} \right\rvert}^7 + 5{\left\lvert {z} \right\rvert}^3 = 6 < 12 \coloneqq{\left\lvert {M(z)} \right\rvert} ,\end{align*} so $$0 = Z_M = Z_{f}$$.

On $${\left\lvert {z} \right\rvert} \leq 2$$,

• Big: $$M(z) = z^7$$
• Small: $$m(z) = -5z^3 + 12$$

On $${\left\lvert {z} \right\rvert} = 2$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} \coloneqq{\left\lvert {-5z^3 + 12} \right\rvert} \leq 5{\left\lvert {z} \right\rvert}^2 + 12 = 32 < 128 = 2^7 \coloneqq{\left\lvert {M(z)} \right\rvert} ,\end{align*} so $$7 = Z_M = Z_{f}$$.

So $$f$$ has 7 zeros in $$1 \leq {\left\lvert {z} \right\rvert} \leq 2$$.

### Spring 2020 HW 3.11 (Standard polynomial) #complex/exercise/completed

Find the number of roots of $$p(z) = z^4 - 6z + 3$$ in $${\left\lvert {z} \right\rvert} < 1$$ and $$1 < {\left\lvert {z} \right\rvert} < 2$$ respectively.

Note: the original problem used $$4z^4-6z+3$$, but I don’t think it’s possible to use Rouché on that at all!

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

• Small: $$m(z) = z^4+3$$
• Big: $$M(z) = -6z$$

On $${\left\lvert {z} \right\rvert} = 1$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {z^4+3} \right\rvert} \leq {\left\lvert {z} \right\rvert}^4 + 3 = 4 < 6 = {\left\lvert {-6z} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so $$Z_f = Z_M = 1$$.

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

• Small: $$-6z+3$$
• Big: $$z^4$$

On $${\left\lvert {z} \right\rvert} = 2$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} = \leq 6 + 3 = 9 < 2^4 = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so $$Z_f = Z_M = 4$$.

Thus there are $$4-1=3$$ zeros in $$1 \leq {\left\lvert {z} \right\rvert} \leq 2$$.

### Standard polynomial #complex/exercise/completed

How many roots does the following polynomial have in the open disc $${\left\lvert {z} \right\rvert} < 1$$? \begin{align*} f(z) = z^7 - 4z^3 - 1 .\end{align*}

Big: $$M(z) = -4z^3$$. Small: $$m(z) = z^7 - 1$$. Then on $${\left\lvert {z} \right\rvert} = 1$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {z^7-1} \right\rvert} \leq {\left\lvert {z} \right\rvert}^7 + 1 = 2 < 4 = {\left\lvert {-4z^4} \right\rvert} ,\end{align*} so $$f$$ and $$M$$ have the same number of zeros: three.

### Spring 2020 HW 1.3 (Standard polynomial) #complex/exercise/completed

Prove that the following polynomial has its roots outside of the unit circle: \begin{align*} p(z) = z^3 + 2z + 4 .\end{align*}

Hint: What is the maximum value of the modulus of the first two terms if $${\left\lvert {z} \right\rvert} \leq 1$$?

Big: $$M(z) = 4$$ Small: $$m(z) = z^3 + 2z$$. On $${\left\lvert {z} \right\rvert} = 1$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^3 + 2{\left\lvert {z} \right\rvert} = 1+2=3 < 4 = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so $$0 = Z_M = Z_{M+m} = Z_f$$ in $${\mathbb{D}}$$.

### Polynomials with parameters #complex/exercise/completed

Assume that $${\left\lvert {b} \right\rvert} < 1$$ and show that the following polynomial has exactly two roots (counting multiplicity) in $${\left\lvert {z} \right\rvert} < 1$$: \begin{align*} f(z) \coloneqq z^3 + 3z^2 + bz + b^2 .\end{align*}

Big: $$M(z) = 3z^2$$. Small: $$m(z) = z^3+bz + b^2$$. Then on $${\left\lvert {z} \right\rvert} = 1$$: \begin{align*} {\left\lvert {m(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^3 + b{\left\lvert {z} \right\rvert} + b^2 = 1 + b + b^2 < 3 = {\left\lvert {M(z)} \right\rvert} ,\end{align*} and $$M(z)$$ has exact two roots in $${\mathbb{D}}$$.

### Tie’s Extra Questions: Spring 2015 (Power series) #complex/exercise/completed

Let $$0<r<1$$. Show that polynomials $$P_n(z) = 1 + 2z + 3 z^2 + \cdots + n z^{n-1}$$ have no zeros in $$|z|<r$$ for all sufficiently large $$n$$’s.

Key observation: \begin{align*} P_n(z) = \sum_{1\leq k\leq n-1} kz^{k-1} = {\frac{\partial }{\partial z}\,}Q_n(z) \qquad Q(z) \coloneqq\sum_{0\leq k \leq n} z^k .\end{align*} Note that $$Q(z) \to \sum_{k\geq 0} z^k = {1\over 1-z}$$ uniformly on $${\left\lvert {z} \right\rvert} \leq R < 1$$ since this power series has radius of convergence 1. Similarly $$P_n(z)$$ converges uniformly to $${\frac{\partial }{\partial z}\,}{1\over 1-z} = {1\over (1-z)^2}$$, so let $$P(z) \coloneqq{1\over (1-z)^2}$$. Note that $$P$$ is nonvanishing in $${\mathbb{D}}$$.

Strategy:

• Small: $$m(z) = P(z) - P_n(z)$$, look for an upper bound $${\left\lvert {m(z)} \right\rvert} < U$$
• Big: $$P(z)$$, look for a lower bound $$L$$ with $${\left\lvert {P(z)} \right\rvert} > L > U$$.

Just by considering the geometry of circles of radius $$R < 1$$ and $$1$$ and measuring distances to the point $$1$$, we can estimate \begin{align*} 0 < 1-R \leq {\left\lvert {1-z} \right\rvert} < 1+R < 2 \implies {\left\lvert {P(z)} \right\rvert} = {1\over {\left\lvert {1-z} \right\rvert}^2} \geq {1\over 2^2} = {1\over 4} .\end{align*}

Now fix $${\varepsilon}< {1\over 4}$$ and use uniform convergence of $$P_n\to P$$ to produce an $$N$$ such that $$n\geq N$$ implies $${\left\lVert {P-P_n} \right\rVert}_\infty < {\varepsilon}$$ in $${\left\lvert {z} \right\rvert} \leq R$$. Then on $${\left\lvert {z} \right\rvert} = R$$, for $$n\geq N$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} \coloneqq{\left\lvert {P(z) - P_n(z)} \right\rvert} \leq {\left\lVert {P - P_n} \right\rVert}_\infty < {\varepsilon}< {1\over 4} \leq {\left\lvert {P(z)} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so $$0 = Z_P = Z_{P_n}$$ by Rouché.

## Exponentials

### UMN Fall 2009 (Solutions as zeros) #complex/exercise/completed

Find the number of solutions to the following equation on $${\left\lvert {z} \right\rvert} < 1$$: \begin{align*} 6z^3 + 1 = -e^z .\end{align*}

Write $$f(z) \coloneqq 6z^3 + 1 + e^z$$.

• Small: $$m(z) = e^z + 1$$
• Large: $$M(z) = 6z^3$$
• The estimate: \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {e^z + 1} \right\rvert} \leq e^{\Re(z)} + 1 \leq e^{{\left\lvert {z} \right\rvert}} + 1 = e+1 < 6 = {\left\lvert {6z^3} \right\rvert} ,\end{align*} so $$3 = Z_M = Z_f$$.

### UMN Spring 2009 (Checking the equality case) #complex/exercise/completed

Find the number of roots on $${\left\lvert {z} \right\rvert} \leq 1$$ of \begin{align*} f(z)=z^{6}+4 z^{2} e^{z+1}-3 .\end{align*}

• Small: $$m(z) = z^6-3$$
• Big: $$M(z) = 4z^2 e^{z+1}$$, which has two such zeros

Now estimate $$m$$ from above: \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {z^6 - 3} \right\rvert} \leq {\left\lvert {z} \right\rvert}^6 + 3 = 4 .\end{align*} and $$M$$ from below: \begin{align*} {\left\lvert {M(z)} \right\rvert} = {\left\lvert {4z^2 e^{z+1}} \right\rvert} = 4e{\left\lvert {z} \right\rvert}^4 e^{\Re(z)} = 4e e^{\Re(z)} \geq 4e e^{-1} = 4 ,\end{align*} which unfortunately isn’t quite enough. But equality occurs iff $$\Re(z) = -1$$ on $$S^1$$, so $$z=-1$$, in which case $${\left\lvert {m(-1)} \right\rvert} = {\left\lvert {1-3} \right\rvert} = 2$$, so the inequality is in fact strict. So $$2 = Z_M = Z_f$$.

### Right half-plane estimate #complex/exercise/completed

Find the number of zeros $$z$$ with $$\Re(z) > 0$$ for the following function: \begin{align*} f(z) \coloneqq z^3-z+1 .\end{align*}

Take a contour $$\gamma_1 \coloneqq\left\{{it {~\mathrel{\Big\vert}~}t\in {\mathbf{R}}}\right\}$$ and $$\gamma_2\coloneqq\left\{{Re^{it} {~\mathrel{\Big\vert}~}t\in [-\pi, \pi]}\right\}$$.

• Big: $$M(z) = z^3 + 1$$
• Small: $$m(z) = -z$$

On $$\gamma_2$$, we have $${\left\lvert {z} \right\rvert} = R$$, so take $$R$$ large enough that the following estimate holds: \begin{align*} {\left\lvert {M(z)} \right\rvert} = {\left\lvert {z^3 + 1} \right\rvert} \geq {\left\lvert { {\left\lvert {z} \right\rvert}^3 - 1} \right\rvert} = R^3 - 1 > R = {\left\lvert {m(z)} \right\rvert} = R .\end{align*} In particular, this works for $$R> 1$$.

On $$\gamma_1$$, note

• $${\left\lvert {M(z)} \right\rvert} = {\left\lvert { (it)^3 + 1 } \right\rvert} = {\left\lvert {1-it^3} \right\rvert}$$
• $${\left\lvert {m(z)} \right\rvert} = {\left\lvert {it} \right\rvert}$$

These can be interpreted geometrically: the former is the hypotenuse of a triangle and the latter is a leg, so $${\left\lvert {M(z)} \right\rvert} \geq {\left\lvert {m(z)} \right\rvert}$$ will hold:

Now note that $$z^3 + 1$$ has roots $$\omega_3, \omega_3^2, \omega_3^3=-1$$ for $$\omega_k \coloneqq e^{i\pi\over k}$$, and the first two are in the right half-plane. So $$2 = {\sharp}Z_M = {\sharp}Z_f$$ by Rouché.

### Zeros of $$e^z$$#complex/exercise/completed

Prove that for every $$n\in {\mathbf{Z}}^{\geq 0}$$ the following polynomial has no roots in the open unit disc: \begin{align*} f_n(z) \coloneqq\sum_{k=0}^n {z^k \over k!} .\end{align*}

Hint: check $$n=1,2$$ directly.

For the $$n=1$$ case, $$f_1(z) = 0 \iff 1+z = 0 \iff z=-1$$, so this has no roots in $${\mathbb{D}}$$. For $$n=2$$, factor \begin{align*} f_2(z) = 1 + z + z^2 = (z-\zeta_3^2)(z-\zeta_3^{-2}) ,\end{align*} using that \begin{align*} \zeta_3^2\cdot \zeta_3^{-2} = 1,\qquad -(\zeta_3^2 + \zeta_3^{-2}) = -2\Re(\zeta_3^2) = -2\cos\qty{2\pi \over 3} = 1 .\end{align*} Now use that $${\left\lvert {\zeta_3^{k}} \right\rvert} = 1$$, which is not in $${\mathbb{D}}$$.

For $$n\geq 3$$: toward applying Rouche’s theorem, let $$M(z) = 1 + z$$ and $$m(z) = {1\over 2}z^2 + \cdots + {1\over n!}z^n$$. Note that on $${\left\lvert {z} \right\rvert} = 1$$, $${\left\lvert {m(z)} \right\rvert} = 2$$, and \begin{align*} {\left\lvert {m(z)} \right\rvert} &= {\left\lvert {\sum_{k\geq n+1} {z^k\over k!} } \right\rvert} \\ &\leq \sum_{k\geq n+1} { {\left\lvert {z} \right\rvert}^k \over k!} \\ &\leq \sum_{k\geq n+1} { 1 \over k!} \\ &= e^1 - \sum_{k\leq n} {1\over k!} .\end{align*} Suppose $$n\geq 3$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} < e - (1 + 1 + \cdots) \approx 0.718 < 2 ,\end{align*} then Rouché applies directly and \begin{align*} 0 = {\sharp}Z_M({\mathbb{D}}) ={\sharp}Z_{M+m}({\mathbb{D}}) \coloneqq{\sharp}Z_f({\mathbb{D}}) ,\end{align*} noting that $$M(z) = 0 \iff z= -1$$, which isn’t contained in the open disc $${\mathbb{D}}$$.

### More $$e^z$$#complex/exercise/completed

Let $$n\in {\mathbf{Z}}^{\geq 0}$$ and show that the equation \begin{align*} e^z = az^n \end{align*} has $$n$$ solutions in the open unit disc if $${\left\lvert {a} \right\rvert} > e$$, and no solutions if $${\left\lvert {a} \right\rvert} < {1\over e}$$.

Note that $${\left\lvert {e^z} \right\rvert} = e^{\Re(z)}$$, which is maximizes on $$S^1$$ at $$z=1 \in {\mathbf{R}}$$ and minimized at $$z=-1$$. Write $$f(z) = e^z-az^n$$, so solution correspond to zeros of $$f$$.

Case 1: suppose $${\left\lvert {a} \right\rvert} > e$$. Big: $$M(z) = az^n$$. Small: $$m(z) = e^z$$. On $${\left\lvert {z} \right\rvert} = 1$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {e^z} \right\rvert} = e^\Re(z) \leq e^1 < {\left\lvert {a} \right\rvert} = {\left\lvert {az^n} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so $$f$$ has $${\sharp}Z_M = n$$ zeros.

Case 2: suppose $${\left\lvert {a} \right\rvert} < 1/e$$. Big: $$M(z) = e^z$$. Small: $$m(z) = az^n$$. On $${\left\lvert {z} \right\rvert} = 1$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {az^n} \right\rvert} = {\left\lvert {a} \right\rvert} < e^{-1}\leq e^{\Re(z)} = {\left\lvert {e^z} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*} and $$M$$ has no zeros in $${\mathbb{D}}$$ (and in fact none in $${\mathbf{C}}$$), so neither does $$f$$.

### Zeros of partial sums of exponential #complex/exercise/completed

For each $$n\in {\mathbf{Z}}^{\geq 1}$$, let \begin{align*} P_n(z) = 1 + z + {1\over 2!} z^2 + \cdots + {1\over n!}z^n .\end{align*} Show that for sufficiently large $$n$$, the polynomial $$P_n$$ has no zeros in $${\left\lvert {z} \right\rvert} < 10$$, while the polynomial $$P_n(z) - 1$$ has precisely 3 zeros there.

More is true: this will hold for any disc of arbitrary radius $$R$$, with $$n$$ depending on $$R$$. Fix $$R$$, then use that $$P_n(z) \overset{n\to\infty}\longrightarrow e^z$$ uniformly on the compact disc $${\left\lvert {z} \right\rvert} \leq R$$. Consequently, setting $$g_n(z) \coloneqq{P_n(z)\over e^z}$$, we have $$g_n(z) \to 1$$ uniformly on this disc, for any $${\varepsilon}> 0$$ this can be used to produce an $$n\gg 1$$ such that $${\left\lvert { g_n(z) - 1 } \right\rvert} < {\varepsilon}$$ for all $${\left\lvert {z} \right\rvert} \leq R$$.

So take $${\varepsilon}\coloneqq 1$$ and define $$h(z) \coloneqq 1$$, then for $${\left\lvert {z} \right\rvert} = R$$ \begin{align*} {\left\lvert {g_n(z) - 1} \right\rvert} < 1 = {\left\lvert {h(z)} \right\rvert} ,\end{align*} so by Rouché, \begin{align*} 0 = {\sharp}Z_{h} = {\sharp}Z_{h + (g_n - 1)} = {\sharp}Z_{g_n} ,\end{align*} since $$h$$ has no zeros at all. Take $$R=10$$ to get the stated result.

For $$P_n(z) - 1$$, note that $$e^z-1=0$$ has three solutions in $${\left\lvert {z} \right\rvert} < 10$$, namely $$z=0, \pm 2\pi i$$. We similarly have $$P_n(z)-1\to e^z-1$$ uniformly, so on a disc of radius $$R$$ choose $$n$$ large enough so that \begin{align*} {\left\lvert {{P_n(z) -1 \over e^z - 1} - 1} \right\rvert} &< 1 \\ \implies {\left\lvert { (P_n(z) - 1) - (e^z-1) \over e^z-1} \right\rvert} &< 1 \\ \implies {\left\lvert { (P_n(z) - 1) - (e^z-1)} \right\rvert} &< {\left\lvert {e^z-1} \right\rvert} \\ \coloneqq{\left\lvert {m(z)} \right\rvert} &< {\left\lvert {M(z)} \right\rvert} ,\end{align*} so \begin{align*} 3 = {\sharp}Z_M = {\sharp}Z_{M+m} = {\sharp}Z_{P_n - 1} .\end{align*}

## Working for the estimate

### Max of a polynomial on $$S^1$$#complex/exercise/completed

Prove that \begin{align*} \max_{{\left\lvert {z} \right\rvert} = 1} {\left\lvert {a_0 + a_1 z + \cdots + a_{n-1}z^{n-1} + z^n} \right\rvert} \geq 1 .\end{align*}

Hint: the first part of the problem asks for a statement of Rouche’s theorem.

Write $$p(z) \coloneqq a_0 + \cdots + z^n$$. Toward a contradiction, suppose not so that $${\left\lvert {p(z)} \right\rvert} < 1$$ on $${\left\lvert {z} \right\rvert} = 1$$. Then \begin{align*} {\left\lvert {f(z)} \right\rvert} < 1 = {\left\lvert {z} \right\rvert}^n \qquad \text{ on } {\left\lvert {z} \right\rvert} = 1 .\end{align*} Taking $$m(z) \coloneqq f(z)$$ and $$M(z) \coloneqq-z^n$$, we have \begin{align*} n = {\sharp}Z_M = {\sharp}Z_{M+m} = {\sharp}Z_{f(z) - z^n} \leq n-1 ,\end{align*} since $$f(z) - z^n$$ is degree at most $$n-1$$, a contradiction.

### Fixed points #complex/exercise/completed

Let $$c\in {\mathbf{C}}$$ with $${\left\lvert {c} \right\rvert} < {1\over 3}$$. Show that on the open set $$\left\{{z\in {\mathbf{C}}{~\mathrel{\Big\vert}~}\Re(z) < 1}\right\}$$, the function $$f(z) \coloneqq ce^z$$ has exactly one fixed point.

The boundary region is $$\left\{{1+it{~\mathrel{\Big\vert}~}t\in {\mathbf{R}}}\right\}$$, write $$g(z) = ce^z-z$$ so that fixed points of $$f$$ are zeros of $$g$$.

Big: $$M(z) = z$$. Small: $$m(z) = ce^z$$. Then for $$z=1+it$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {c} \right\rvert}e^{\Re(z)} < ce < 1 \leq \sqrt{1^2+t^2} = {\left\lvert {1+it} \right\rvert} = {\left\lvert {z} \right\rvert} ,\end{align*} so $$M$$ and $$g$$ have the same number of zeros, and $$M$$ has a unique zero.

### $$z\sin(z)=1$$#complex/exercise/completed

Show that $$z\sin(z) = a$$ has only real solutions.

Consider $$f(z) \coloneqq z\sin(z) - a$$.

Big: $$M(z) \coloneqq z\sin(z)$$. Small: $$m(z) \coloneqq-a$$.

Use the following estimate:

\begin{align*} {\left\lvert {z\sin(z)} \right\rvert}^2 &= {\left\lvert {z\over 2} \right\rvert}^2 {\left\lvert {e^{iz} - e^{-iz}} \right\rvert}^2 \\ &\geq {\left\lvert {z\over 2} \right\rvert}^2 {\left\lvert { {\left\lvert {e} \right\rvert}^{iz} } \right\rvert} - {\left\lvert { e^{-iz} } \right\rvert} ^2 \\ &= {\left\lvert {z\over 2} \right\rvert}^2 {\left\lvert {e^{-\Im(z)} - e^{\Im(z)} } \right\rvert} \\ &\overset{\Im(z)\to\infty}\longrightarrow \infty ,\end{align*} and so in particular a radius $$R$$ can be chosen large enough so that $${\left\lvert {z\sin(z)} \right\rvert} > a$$ for any $$a$$. Thus for $${\left\lvert {z} \right\rvert} = R$$, \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {a} \right\rvert} \leq {\left\lvert {z\sin(z)} \right\rvert} < {\left\lvert {M(z)} \right\rvert} \implies {\sharp}Z_{M} = {\sharp}Z_{M+m} = {\sharp}Z_f .\end{align*} To count the number of zeros of $$z\sin(z)$$, note that this equals zero at $$z=0$$ with multiplicity two and $$z= k\pi$$ for $$k\in {\mathbf{Z}}$$. Choosing $$R = {\pi \over 2} + n\pi$$ for $$n$$ large enough, there are exactly $$2n+2$$ such zeros (with multiplicity) to $$z\sin(z)$$, and thus $$2n+2$$ zeros to $$z\sin(z) - a$$. Now using that $$z\sin(z) - a$$ has exactly $$2n+2$$ real roots (??), this must be all of them.

Unsure how to find any roots of this thing, real or not!

### Spring 2020 HW 3.13 #complex/exercise/work#stuck

Prove that for $$a> 0$$, $$z\tan z - a$$ has only real roots.

### UMN Spring 2011 (Constant coefficient trick) #complex/exercise/completed

Let $$a\in {\mathbf{C}}$$ and $$n\geq 2$$. Show that the following polynomial has one root in $${\left\lvert {z} \right\rvert} \leq 2$$: \begin{align*} f(z) = az^n + z + 1 .\end{align*}

The key step: getting the following inequality to work \begin{align*} {\left\lvert {az^n} \right\rvert} = {\left\lvert {a} \right\rvert}{\left\lvert {z} \right\rvert}^n < c \leq 1 = {\left\lvert { {\left\lvert {z} \right\rvert} - {\left\lvert {1} \right\rvert} } \right\rvert} \leq {\left\lvert {z+1} \right\rvert} .\end{align*} If this is true, then $$1 = {\sharp}Z_{z+1} = {\sharp}Z_f$$. If $${\left\lvert {a} \right\rvert} < 2^n$$, this holds because $${\left\lvert {a} \right\rvert}{\left\lvert {z} \right\rvert}^n < {1\over 2^n} 2^n = 1$$, so taking $$c\coloneqq 1$$ works.

Otherwise, suppose $${\left\lvert {a} \right\rvert} \geq 2^n$$. Letting $$z_k$$ be the roots of $$f$$ and considering the constant coefficient, we have \begin{align*} a\prod_{k\leq n} z_k = 1 \implies {\left\lvert { \prod_{k\leq n} z_k } \right\rvert} = {\left\lvert {1\over a} \right\rvert} \leq 2^n ,\end{align*} so not every $$z_k$$ can satisfy $${\left\lvert {z_k} \right\rvert} > 2$$ and at least one is in $${\left\lvert {z} \right\rvert} \leq 2$$.