A root of $z^n + f(z)$ within $B_1(0)$, $f$ entire

Question

Suppose we're given an entire function $f$, with the property that $|z| < 1 \implies |f(z)| < 1$.

We're asked to show that for any $n \geq 1$, $f(z) + z^n$ has a root within $B_1(0)$.

Further I cannot use Rouche's theorem, or the argument principle or the residue theorem :).

We can use Cauchy's theorems,

or the dog walking theorem:

If $\gamma_1, \gamma_2$ are closed curves s.t $|\gamma_1(t) - \gamma_2(t)| < |\gamma_1(t)|$ for any $t \in [a,b]$ then $n(\gamma_1,0) = n(\gamma_2,0)$.

I've spent some times with this: tried by contradiction (it is hinted to try something along these lines) which yields that for any $r \in [0,1)$, $\gamma_1(t) := (re^{\pi it})^n + f(re^{\pi it})$, $0 \notin Image(\gamma_1)$. This gives, through a homotopy to the constant function $f(0)$, that $n(\gamma_1, 0) = 0$.

Then from here I've been trying to define a meaningful $\gamma_2$ either by simply $\gamma_2(t) = (re^{\pi it})^n$, or by a using the entirety of $f$ to define it as a finite combination of parts of its power series.

None of these yielded successful estimates. I'd like a hint please, first.

Edit 1:

By the above discussion if we can find $r \in [0,1)$ s.t $|f(re^{\pi it})| < r^n$ we would be done. Tried this by contradiction, which yields a sequence within the ball whose values under $f$ converge to a point in $S^1$. From here I don't see much more to do. Btw the above condition would be required to use Rouche's theorem as well, no?

Edit 2

Ok so we've been given a correction; we have the above question with the following now: $|z| \leq 1 \implies |f(z)| < 1$. I will post my suggested answer as well, alongside Daniel's great answer.

Answer 1

If $B_1(0)$ denotes the open unit disk the assertion is wrong. It's not hard to construct examples where all zeros of $f(z) + z^n$ lie on the unit circle. Let $0 < \varepsilon < 1$, and $f(z) = \varepsilon(1 - z^n) - 1$. Then $\lvert f(z)\rvert < 1$ for $\lvert z\rvert < 1$ and $f(z) + z^n = (1-\varepsilon)(z^n-1)$, so the zeros are just the $n^{\text{th}}$ roots of unity.

If $B_1(0)$ denotes the closed unit disk, use what you know to deduce that for $0 < \varepsilon < 1$ the function $g_{\varepsilon}(z) = (1-\varepsilon)f(z) + z^n$ has a zero in the open unit disk, and from that conclude that $f(z) + z^n$ has a zero in the closed unit disk.

If $h \colon \overline{\mathbb{D}} \to \mathbb{C}$ is continuous and $\lvert h(z)\rvert < 1$ for $\lvert z\rvert = 1$, applying the "dog walking theorem" to $\gamma_1(t) = e^{int}$ and $\gamma_2(t) = h(e^{it}) + e^{int}$ yields $n(\gamma_2,0) = n(\gamma_1,0) = n > 0$, so $h(z) + z^n$ must have a zero in the open unit disk (for otherwise $\gamma_2$ would be nullhomotopic in $\mathbb{C}\setminus \{0\}$). Now if $f(z) + z^n$ had no zero in the closed unit disk $\overline{\mathbb{D}}$, there'd be a $\delta \in (0,1]$ with $\lvert f(z) + z^n\rvert \geqslant \delta$ for all $z \in \overline{\mathbb{D}}$. Then, since $\bigl\lvert \bigl(f(z) + z^n\bigr) - g_{\varepsilon}(z)\bigr\rvert = \varepsilon \lvert f(z)\rvert \leqslant \varepsilon$ for $z \in \overline{\mathbb{D}}$, it follows that $g_{\varepsilon}$ has no zeros in the open unit disk for $0 < \varepsilon < \delta$, contradicting the previous result (with $h(z) = (1-\varepsilon)f(z)$).

Answer 2

In light of a correction, as noted under edit 2, I'd like to post my attempt:

Suppose on the contrary that $g(z) = f(z) + z^n$ has no roots in the open unit ball.

Then $\forall r \in [0,1)$ we have that $ 0 \notin Range(\gamma_{1,r}(t))$ for $\gamma_{1,r}(t) := g(re^{2\pi i t})$.

It follows that through the homotopy $H:[0,1] \times [0,1] \to \mathbb{C} -\{0\}$, for any $r_1, r_2 \in [0,1)$

$H(s,t) = g([(1-s)r_1 + sr_2]e^{2\pi i t})$, $\gamma_{1,r}$ is null-homotopic, and then $n(\gamma_{1,r}, 0) = 0$.

$*[(1-s)r_1 + sr_2] \in [0,1)*$.

Defining $\gamma_{2,r}(t) = r^ne^{2n\pi i t}$ we have that $|\gamma_{1,r}(t) - \gamma_{2,r}(t)| = |f(re^{2\pi i t})|$. By the 'dog walking theorem' it suffices to show that there is some $r \in [0,1)$ with $|f(re^{2\pi i t})| < |\gamma_{2,r}(t)| = r^n$ $\forall t \in [0,1]$, to get a contradiction, as the index of the $\gamma_{2,r}$ is $n$.

Again assume that no such $r$ exists. Take $r_m = \frac{m-1}{m}$, so that $r_m^n \underset{m \to \infty}{\to} 1$.

We have $|f(r_m e^{2\pi i t_m})| \geq r_m^n$, for some $t_m \in [0,1]$. Take some convergent subsequence $r_{m,k} e^{2\pi i t_{m,k}} \underset{{m,k} \to \infty}{\to} r_{\infty} \in S^1$, to get that $|f(r_\infty)| = 1$, a contradiction.

There's no need to assume $f$ is entire, if this is correct.