Show that all the zeros of $(az+1)^m-b (z+1)^m$ lie in $|z|<1,$ where $a>1$ and $|b|\leq 1.$

Question

If $f(z)=(z+1)^m$ then $f(az)=(az+1)^m,$ where $m\in\mathbb{N}.$ I can prove that $|f(z)|<|f(az)|$ for $|z|=1$ and $a>1.$ Therefore, by Rouche's theorem, all the zeros of $(az+1)^m-b (z+1)^m$ lie in $|z|\leq (1/a)<1.$

But I don't want to use Rouche's theorem or any result from analysis to prove this result. I am looking for some elementary solution, like properties of Binomial theorem. Help appreciated in advance.

Answer 1

You can show that $|az + 1| > |z + 1|$ if $|z| > 1$. Indeed, $|az + 1|^2 = a^2|z|^2 + 2a\Re z + 1$ and $|z + 1|^2 = |z|^2 + 2\Re z + 1$. Hence $$ |az + 1| > |z + 1| \Longleftrightarrow |z|^2(a^2 - 1) + 2\Re z(a - 1) > 0 \Longleftrightarrow |z|^2(a + 1) > -2\Re z, $$ which holds if $|z| > 2/(a + 1)$. Hence, for $|z| \ge 1$, $|(az + 1)^m| > |(z + 1)^m| \geq |b(z + 1)^m|$ and $z$ cannot be a root.