If 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.

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.

If 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 $|z|<1,\psi\in\mathbb{R}.$

I know at least two solution of this problem one by using Luguerre's theorem concerning the polar derivative of a polynomial and another by using a known result. But I want to find its direct solution with using known results.

If 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.

If 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 $|z|<1,\psi\in\mathbb{R}.$

I know at least two solution of this problem one by using Luguerre's theorem concerning the polar derivative of a polynomial and another by using a known result. But I want to find its direct solution with using known results.

If 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 $|z|<1,\psi\in\mathbb{R}.$

I know at least two solution of this problem one by using Luguerre's theorem concerning the polar derivative of a polynomial and another by using a known result. But I want to find its direct solution with using known results.