Consider a disk ${D(z_0,r) := \{ z: |z-z_0| < r \}}$ in the complex plane. If one applies an affine-linear map ${f(z) = az+b}$ to this disk, one obtains

$\displaystyle f(D(z_0,r)) = D(f(z_0), |f'(z_0)| r).$

For maps that are merely holomorphic instead of affine-linear, one has some variants of this assertion, which I am recording here mostly for my own reference:

Theorem 1 (Holomorphic images of disks) Let ${D(z_0,r)}$ be a disk in the complex plane, and ${f: D(z_0,r) \rightarrow {\bf C}}$ be a holomorphic function with ${f'(z_0) \neq 0}$.
• (i) (Open mapping theorem or inverse function theorem) ${f(D(z_0,r))}$ contains a disk ${D(f(z_0),\varepsilon)}$ for some ${\varepsilon>0}$. (In fact there is even a holomorphic right inverse of ${f}$ from ${D(f(z_0), \varepsilon)}$ to ${D(z_0,r)}$.)
• (ii) (Bloch theorem) ${f(D(z_0,r))}$ contains a disk ${D(w, c |f'(z_0)| r)}$ for some absolute constant ${c>0}$ and some ${w \in {\bf C}}$. (In fact there is even a holomorphic right inverse of ${f}$ from ${D(w, c |f'(z_0)| r)}$ to ${D(z_0,r)}$.)
• (iii) (Koebe quarter theorem) If ${f}$ is injective, then ${f(D(z_0,r))}$ contains the disk ${D(f(z_0), \frac{1}{4} |f'(z_0)| r)}$.
• (iv) If ${f}$ is a polynomial of degree ${n}$, then ${f(D(z_0,r))}$ contains the disk ${D(f(z_0), \frac{1}{n} |f'(z_0)| r)}$.
• (v) If one has a bound of the form ${|f'(z)| \leq A |f'(z_0)|}$ for all ${z \in D(z_0,r)}$ and some ${A>1}$, then ${f(D(z_0,r))}$ contains the disk ${D(f(z_0), \frac{c}{A} |f'(z_0)| r)}$ for some absolute constant ${c>0}$. (In fact there is holomorphic right inverse of ${f}$ from ${D(f(z_0), \frac{c}{A} |f'(z_0)| r)}$ to ${D(z_0,r)}$.)

Parts (i), (ii), (iii) of this theorem are standard, as indicated by the given links. I found part (iv) as (a consequence of) Theorem 2 of this paper of Degot, who remarks that it “seems not already known in spite of its simplicity”; an equivalent form of this result also appears in Lemma 4 of this paper of Miller. The proof is simple:

Proof: (Proof of (iv)) Let ${w \in D(f(z_0), \frac{1}{n} |f'(z_0)| r)}$, then we have a lower bound for the log-derivative of ${f(z)-w}$ at ${z_0}$:

$\displaystyle \frac{|f'(z_0)|}{|f(z_0)-w|} > \frac{n}{r}$

(with the convention that the left-hand side is infinite when ${f(z_0)=w}$). But by the fundamental theorem of algebra we have

$\displaystyle \frac{f'(z_0)}{f(z_0)-w} = \sum_{j=1}^n \frac{1}{z_0-\zeta_j}$

where ${\zeta_1,\dots,\zeta_n}$ are the roots of the polynomial ${f(z)-w}$ (counting multiplicity). By the pigeonhole principle, there must therefore exist a root ${\zeta_j}$ of ${f(z) - w}$ such that

$\displaystyle \frac{1}{|z_0-\zeta_j|} > \frac{1}{r}$

and hence ${\zeta_j \in D(z_0,r)}$. Thus ${f(D(z_0,r))}$ contains ${w}$, and the claim follows. $\Box$

The constant ${\frac{1}{n}}$ in (iv) is completely sharp: if ${f(z) = z^n}$ and ${z_0}$ is non-zero then ${f(D(z_0,|z_0|))}$ contains the disk

$\displaystyle D(f(z_0), \frac{1}{n} |f'(z_0)| r) = D( z_0^n, |z_0|^n)$

but avoids the origin, thus does not contain any disk of the form ${D( z_0^n, |z_0|^n+\varepsilon)}$. This example also shows that despite parts (ii), (iii) of the theorem, one cannot hope for a general inclusion of the form

$\displaystyle f(D(z_0,r)) \supset D(f(z_0), c |f'(z_0)| r )$

for an absolute constant ${c>0}$.

Part (v) is implicit in the standard proof of Bloch’s theorem (part (ii)), and is easy to establish:

Proof: (Proof of (v)) From the Cauchy inequalities one has ${f''(z) = O(\frac{A}{r} |f'(z_0)|)}$ for ${z \in D(z_0,r/2)}$, hence by Taylor’s theorem with remainder ${f(z) = f(z_0) + f'(z_0) (z-z_0) (1 + O( A \frac{|z-z_0|}{r} ) )}$ for ${z \in D(z_0, r/2)}$. By Rouche’s theorem, this implies that the function ${f(z)-w}$ has a unique zero in ${D(z_0, 2cr/A)}$ for any ${w \in D(f(z_0), cr|f'(z_0)|/A)}$, if ${c>0}$ is a sufficiently small absolute constant. The claim follows. $\Box$

Note that part (v) implies part (i). A standard point picking argument also lets one deduce part (ii) from part (v):

Proof: (Proof of (ii)) By shrinking ${r}$ slightly if necessary we may assume that ${f}$ extends analytically to the closure of the disk ${D(z_0,r)}$. Let ${c}$ be the constant in (v) with ${A=2}$; we will prove (iii) with ${c}$ replaced by ${c/2}$. If we have ${|f'(z)| \leq 2 |f'(z_0)|}$ for all ${z \in D(z_0,r/2)}$ then we are done by (v), so we may assume without loss of generality that there is ${z_1 \in D(z_0,r/2)}$ such that ${|f'(z_1)| > 2 |f'(z_0)|}$. If ${|f'(z)| \leq 2 |f'(z_1)|}$ for all ${z \in D(z_1,r/4)}$ then by (v) we have

$\displaystyle f( D(z_0, r) ) \supset f( D(z_1,r/2) ) \supset D( f(z_1), \frac{c}{2} |f'(z_1)| \frac{r}{2} )$

$\displaystyle \supset D( f(z_1), \frac{c}{2} |f'(z_0)| r )$

and we are again done. Hence we may assume without loss of generality that there is ${z_2 \in D(z_1,r/4)}$ such that ${|f'(z_2)| > 2 |f'(z_1)|}$. Iterating this procedure in the obvious fashion we either are done, or obtain a Cauchy sequence ${z_0, z_1, \dots}$ in ${D(z_0,r)}$ such that ${f'(z_j)}$ goes to infinity as ${j \rightarrow \infty}$, which contradicts the analytic nature of ${f}$ (and hence continuous nature of ${f'}$) on the closure of ${D(z_0,r)}$. This gives the claim. $\Box$

Here is another classical result stated by Alexander (and then proven by Kakeya and by Szego, but also implied to a classical theorem of Grace and Heawood) that is broadly compatible with parts (iii), (iv) of the above theorem:

Proposition 2 Let ${D(z_0,r)}$ be a disk in the complex plane, and ${f: D(z_0,r) \rightarrow {\bf C}}$ be a polynomial of degree ${n \geq 1}$ with ${f'(z) \neq 0}$ for all ${z \in D(z_0,r)}$. Then ${f}$ is injective on ${D(z_0, \sin\frac{\pi}{n})}$.

The radius ${\sin \frac{\pi}{n}}$ is best possible, for the polynomial ${f(z) = z^n}$ has ${f'}$ non-vanishing on ${D(1,1)}$, but one has ${f(\cos(\pi/n) e^{i \pi/n}) = f(\cos(\pi/n) e^{-i\pi/n})}$, and ${\cos(\pi/n) e^{i \pi/n}, \cos(\pi/n) e^{-i\pi/n}}$ lie on the boundary of ${D(1,\sin \frac{\pi}{n})}$.

If one narrows ${\sin \frac{\pi}{n}}$ slightly to ${\sin \frac{\pi}{2n}}$ then one can quickly prove this proposition as follows. Suppose for contradiction that there exist distinct ${z_1, z_2 \in D(z_0, \sin\frac{\pi}{n})}$ with ${f(z_1)=f(z_2)}$, thus if we let ${\gamma}$ be the line segment contour from ${z_1}$ to ${z_2}$ then ${\int_\gamma f'(z)\ dz}$. However, by assumption we may factor ${f'(z) = c (z-\zeta_1) \dots (z-\zeta_{n-1})}$ where all the ${\zeta_j}$ lie outside of ${D(z_0,r)}$. Elementary trigonometry then tells us that the argument of ${z-\zeta_j}$ only varies by less than ${\frac{\pi}{n}}$ as ${z}$ traverses ${\gamma}$, hence the argument of ${f'(z)}$ only varies by less than ${\pi}$. Thus ${f'(z)}$ takes values in an open half-plane avoiding the origin and so it is not possible for ${\int_\gamma f'(z)\ dz}$ to vanish.

To recover the best constant of ${\sin \frac{\pi}{n}}$ requires some effort. By taking contrapositives and applying an affine rescaling and some trigonometry, the proposition can be deduced from the following result, known variously as the Grace-Heawood theorem or the complex Rolle theorem.

Proposition 3 (Grace-Heawood theorem) Let ${f: {\bf C} \rightarrow {\bf C}}$ be a polynomial of degree ${n \geq 1}$ such that ${f(1)=f(-1)}$. Then ${f'}$ contains a zero in the closure of ${D( 0, \cot \frac{\pi}{n} )}$.

This is in turn implied by a remarkable and powerful theorem of Grace (which we shall prove shortly). Given two polynomials ${f,g}$ of degree at most ${n}$, define the apolar form ${(f,g)_n}$ by

$\displaystyle (f,g)_n := \sum_{k=0}^n (-1)^k f^{(k)}(0) g^{(n-k)}(0). \ \ \ \ \ (1)$

Theorem 4 (Grace’s theorem) Let ${C}$ be a circle or line in ${{\bf C}}$, dividing ${{\bf C} \backslash C}$ into two open connected regions ${\Omega_1, \Omega_2}$. Let ${f,g}$ be two polynomials of degree at most ${n \geq 1}$, with all the zeroes of ${f}$ lying in ${\Omega_1}$ and all the zeroes of ${g}$ lying in ${\Omega_2}$. Then ${(f,g)_n \neq 0}$.

(Contrapositively: if ${(f,g)_n=0}$, then the zeroes of ${f}$ cannot be separated from the zeroes of ${g}$ by a circle or line.)

Indeed, a brief calculation reveals the identity

$\displaystyle f(1) - f(-1) = (f', g)_{n-1}$

where ${g}$ is the degree ${n-1}$ polynomial

$\displaystyle g(z) := \frac{1}{n!} ((z+1)^n - (z-1)^n).$

The zeroes of ${g}$ are ${i \cot \frac{\pi j}{n}}$ for ${j=1,\dots,n-1}$, so the Grace-Heawood theorem follows by applying Grace’s theorem with ${C}$ equal to the boundary of ${D(0, \cot \frac{\pi}{n})}$.

The same method of proof gives the following nice consequence:

Theorem 5 (Perpendicular bisector theorem) Let ${f: {\bf C} \rightarrow C}$ be a polynomial such that ${f(z_1)=f(z_2)}$ for some distinct ${z_1,z_2}$. Then the zeroes of ${f'}$ cannot all lie on one side of the perpendicular bisector of ${z_1,z_2}$. For instance, if ${f(1)=f(-1)}$, then the zeroes of ${f'}$ cannot all lie in the halfplane ${\{ z: \mathrm{Re} z > 0 \}}$ or the halfplane ${\{ z: \mathrm{Re} z < 0 \}}$.

I’d be interested in seeing a proof of this latter theorem that did not proceed via Grace’s theorem.

Now we give a proof of Grace’s theorem. The case ${n=1}$ can be established by direct computation, so suppose inductively that ${n>1}$ and that the claim has already been established for ${n-1}$. Given the involvement of circles and lines it is natural to suspect that a Möbius transformation symmetry is involved. This is indeed the case and can be made precise as follows. Let ${V_n}$ denote the vector space of polynomials ${f}$ of degree at most ${n}$, then the apolar form is a bilinear form ${(,)_n: V_n \times V_n \rightarrow {\bf C}}$. Each translation ${z \mapsto z+a}$ on the complex plane induces a corresponding map on ${V_n}$, mapping each polynomial ${f}$ to its shift ${\tau_a f(z) := f(z-a)}$. We claim that the apolar form is invariant with respect to these translations:

$\displaystyle ( \tau_a f, \tau_a g )_n = (f,g)_n.$

Taking derivatives in ${a}$, it suffices to establish the skew-adjointness relation

$\displaystyle (f', g)_n + (f,g')_n = 0$

but this is clear from the alternating form of (1).

Next, we see that the inversion map ${z \mapsto 1/z}$ also induces a corresponding map on ${V_n}$, mapping each polynomial ${f \in V_n}$ to its inversion ${\iota f(z) := z^n f(1/z)}$. From (1) we see that this map also (projectively) preserves the apolar form:

$\displaystyle (\iota f, \iota g)_n = (-1)^n (f,g)_n.$

More generally, the group of Möbius transformations on the Riemann sphere acts projectively on ${V_n}$, with each Möbius transformation ${T: {\bf C} \rightarrow {\bf C}}$ mapping each ${f \in V_n}$ to ${Tf(z) := g_T(z) f(T^{-1} z)}$, where ${g_T}$ is the unique (up to constants) rational function that maps this a map from ${V_n}$ to ${V_n}$ (its divisor is ${n(T \infty) - n(\infty)}$). Since the Möbius transformations are generated by translations and inversion, we see that the action of Möbius transformations projectively preserves the apolar form; also, we see this action of ${T}$ on ${V_n}$ also moves the zeroes of each ${f \in V_n}$ by ${T}$ (viewing polynomials of degree less than ${n}$ in ${V_n}$ as having zeroes at infinity). In particular, the hypotheses and conclusions of Grace’s theorem are preserved by this Möbius action. We can then apply such a transformation to move one of the zeroes of ${f}$ to infinity (thus making ${f}$ a polynomial of degree ${n-1}$), so that ${C}$ must now be a circle, with the zeroes of ${g}$ inside the circle and the remaining zeroes of ${f}$ outside the circle. But then

$\displaystyle (f,g)_n = (f, g')_{n-1}.$

By the Gauss-Lucas theorem, the zeroes of ${g'}$ are also inside ${C}$. The claim now follows from the induction hypothesis.