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Summary of the discussion: Using the triangle inequality, one sees that that $|f(z)|< f(|z|)$, and so the root of largest absolute value is the positive real root $z_k$. Differentiating $f(z)(z-1)$, one gets a bound:

$$z_k < \frac{1 + k}{1 + n^{-1}}.$$

When $k \rightarrow 1/n$, the largest real root approaches $1$ (by continuity, since $f(1) = 1 - nk$). Thus any bound must involve $n$.

The OP complains that he wants something better. It is pointed out that as $k \rightarrow 1/n$, the quantity $1 - z_k$ is asymptotic to

$$\frac{2(1 - kn)}{(1 + n)}.$$

The OP then complains that he wants a bound in $n$ and $k$ (which was already given). The OP askes whether the asymptotic above was found in the following way: "Are you simply using the fact that the root would occur roughly twice as far as the turning point?" No --- mathematics was used at this point.

The OP says that he simply wants an upper bound on the real part of each root. Since the real part of the real root $z_k$ is itself, this question has already been answered. The asymptotic result shows it is impossible to impove this bound significantly.

It's hard to tell if the problem with the OP's repeated questions involve English, Mathematics, or both. In either case, this has already wasted 15 minutes of my time. To paraphrase Zagier, that's the equivalent of 15 days of the OP's time. Feel free to edit this post to make it more "civil".

Summary of the discussion: Using the triangle inequality, one sees that that $|f(z)|\ge f(|z|)$, and so the root of largest absolute value is the positive real root $z_k$. Differentiating $f(z)(z-1)$, one gets a bound:

$$z_k < \frac{1 + k}{1 + n^{-1}}.$$

When $k \rightarrow 1/n$, the largest real root approaches $1$ (by continuity, since $f(1) = 1 - nk$). Thus any bound must involve $n$.

The OP complains that he wants something better. It is pointed out that as $k \rightarrow 1/n$, the quantity $1 - z_k$ is asymptotic to

$$\frac{2(1 - kn)}{(1 + n)}.$$

The OP then complains that he wants a bound in $n$ and $k$ (which was already given). The OP askes whether the asymptotic above was found in the following way: "Are you simply using the fact that the root would occur roughly twice as far as the turning point?" No --- mathematics was used at this point.

The OP says that he simply wants an upper bound on the real part of each root. Since the real part of the real root $z_k$ is itself, this question has already been answered. The asymptotic result shows it is impossible to impove this bound significantly.

It's hard to tell if the problem with the OP's repeated questions involve English, Mathematics, or both. In either case, this has already wasted 15 minutes of my time. To paraphrase Zagier, that's the equivalent of 15 days of the OP's time. Feel free to edit this post to make it more "civil".

If $k=1/n$ then there is a root at $z=1$. Thus, when $k\to 1/n$ from below, the continuity of the roots as function of the coefficients implies that some root converges to 1. Thus if there is an estimate $|z|<\rho(k)$ on the roots, then $\limsup_{k\to 1/n}\rho(k)\ge1$ for all $n$, which probably defeats the goal of the questioner.

However, the polynomial $g(z)=(z-1)f(z)=z^{n+1}-(1+k)z^n+k$ satisfies $g(1)=0$ and $g'(0)=1-nk>0$. Note that if $z$ is a root then $k=|z^{n+1}-(1+k)z^n|\ge(1+k)|z|^n-|z|^{n+1}$ which implies $g(|z|)=|z|^{n+1}-(1+k)|z|^n+k\ge0$, so there is a root between $|z|$ and 1. In other words, the root with the biggest absolute value is in the interval $(0,1)$. We also note that $g'(z)=0$ precisely for $z=(1+k)/(1+n^{-1})$, so there is no root between this point and 1. Hence $$|z|<\frac{1+k}{1+n^{-1}}$$ for any root. Perhaps not good enough for the asker's purposes.

Summary of the discussion: Using the triangle inequality, one sees that that $|f(z)|< f(|z|)$, and so the root of largest absolute value is the positive real root $z_k$. Differentiating $f(z)(z-1)$, one gets a bound:

$$z_k < \frac{1 + k}{1 + n^{-1}}.$$

When $k \rightarrow 1/n$, the largest real root approaches $1$ (by continuity, since $f(1) = 1 - nk$). Thus any bound must involve $n$.

The OP complains that he wants something better. It is pointed out that as $k \rightarrow 1/n$, the quantity $1 - z_k$ is asymptotic to

$$\frac{2(1 - kn)}{(1 + n)}.$$

The OP then complains that he wants a bound in $n$ and $k$ (which was already given). The OP askes whether the asymptotic above was found in the following way: "Are you simply using the fact that the root would occur roughly twice as far as the turning point?" No --- mathematics was used at this point.

The OP says that he simply wants an upper bound on the real part of each root. Since the real part of the real root $z_k$ is itself, this question has already been answered. The asymptotic result shows it is impossible to impove this bound significantly.

It's hard to tell if the problem with the OP's repeated questions involve English, Mathematics, or both. In either case, this has already wasted 15 minutes of my time. To paraphrase Zagier, that's the equivalent of 15 days of the OP's time. Feel free to edit this post to make it more "civil".

If $k=1/n$ then there is a root at $z=1$. Thus, when $k\to 1/n$ from below, the continuity of the roots as function of the coefficients implies that some root converges to 1. Thus if there is an estimate $|z|<\rho(k)$ on the roots, then $\limsup_{k\to 1/n}\rho(k)\ge1$ for all $n$, which probably defeats the goal of the questioner.

However, the polynomial $g(z)=(z-1)f(z)=z^{n+1}-(1+k)z^n+k$ satisfies $g(1)=0$ and $g'(0)=1-nk>0$. Note that if $z$ is a root then $k=|z^{n+1}-(1+k)z^n|\ge(1+k)|z|^n-|z|^{n+1}$ which implies $g(|z|)=|z|^{n+1}-(1+k)|z|^n+k\ge0$, so there is a root between $|z|$ and 1. In other words, the root with the biggest absolute value is in the interval $(0,1)$. We also note that $g'(z)=0$ precisely for $z=(1+k)/(1+n^{-1})$, so there is no root between this point and 1. Hence $$|z|<\frac{1+k}{1+n^{-1}}$$ for any root. Perhaps not good enough for the asker's purposes.

Lots of credit should go to FC for this answer (see the comments to the question). On second thought, so much of the credit belongs to FC I am making this answer community wiki so I am not stealing reputation that should be his.

If $k=1/n$ then there is a root at $z=1$. Thus, when $k\to 1/n$ from below, the continuity of the roots as function of the coefficients implies that some root converges to 1. Thus if there is an estimate $|z|<\rho(k)$ on the roots, then $\limsup_{k\to 1/n}\rho(k)\ge1$ for all $n$, which probably defeats the goal of the questioner.

However, the polynomial $g(z)=(z-1)f(z)=z^{n+1}-(1+k)z^n+k$ satisfies $g(1)=0$ and $g'(0)=1-nk>0$. Note that if $z$ is a root then $k=|z^{n+1}-(1+k)z^n|\ge(1+k)|z|^n-|z|^{n+1}$ which implies $g(|z|)=|z|^{n+1}-(1+k)|z|^n+k\ge0$, so there is a root between $|z|$ and 1. In other words, the root with the biggest absolute value is in the interval $(0,1)$. We also note that $g'(z)=0$ precisely for $z=(1+k)/(1+n^{-1})$, so there is no root between this point and 1. Hence $$|z|<\frac{1+k}{1+n^{-1}}$$ for any root. Perhaps not good enough for the asker's purposes.