Suppose that a polynomial $p(z)$ of degree $n$ does not assume the value $w$ for $|z|<1$, that is $p(z)\neq w$ for $|z|<1.$ Show that $p(z)-\dfrac{(1-e^{i\psi})}{n}zp^{\prime}(z)\neq w$ for $|z|<1,\psi\in\mathbb{R}.$

The polar derivative of a polynomial $p(z)$ is defined as $$ D_\alpha p(z):=np(z)+(\alpha -z)p^{\prime}(z) \qquad \alpha\in\mathbb{C}. $$ If all the zeros of $p(z)$ lie inside a circular region $\mathcal{C},$ then by Laguerre's Separation theorem]1 the zero $w$ of $D_\alpha p(z)$ and the point $\alpha$ cannot lie both outside $\mathcal{C}.$ Above result follows by applying Laguerre's theorem to $p(z)-w.$

I am looking for a proof which is independent of Laguerre's theorem.

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    $\begingroup$ This is not true: take p(z)=z$, $\psi=\pi$, $w=1$. $\endgroup$ – Alexandre Eremenko Mar 23 '15 at 3:07
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    $\begingroup$ <strike>Huh? What is the value omited by $p(z)=z$ ?</strike> Sorry, I had only read the title $\endgroup$ – Feldmann Denis Mar 23 '15 at 5:41
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    $\begingroup$ @ Alexandre: There was a typo error in expression $p(z)-\dfrac{(1-e^{i\psi})}{n}zp^{\prime}(z).$ $\endgroup$ – Suhail Mar 23 '15 at 13:23

We can assume that $w=0$ (replace $p$ by $q=p-w$). Then $p=c\prod (z-a_j)$ with $|a_j|\ge 1$. Since $$ \frac{p'}{p} =\sum_{j=1}^n \frac{1}{z-a_j} , $$ we want to show that $$ 1 - \frac{1-e^{i\psi}}{n} \sum_{j=1}^n \frac{z}{z-a_j} $$ does not take the value $0$ on $|z|<1$, or, equivalently, the equation $$ \frac{1}{1-e^{i\psi}} = \frac{1}{n} \sum_{j=1}^n \frac{1}{1-a_j/z} \quad\quad\quad (1) $$ can not be solved with a $|z|<1$.

The map $z\mapsto 1/z$ maps the circle of radius $1$ with center $z_0=1$ to the vertical line $L$ given by $\textrm{Re}\, z=1/2$. Now $1-e^{i\psi}$ is on this circle while the $1-a_j/z$ are outside (recall that $|a_j|\ge 1$, $|z|<1$). Thus what we are trying to do in (1) is to satisfy $$ u = \frac{1}{n} \sum_{j=1}^n v_j $$ with $u$ on $L$ and all $v_j$'s on one side of $L$. Clearly this is impossible.

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