Given $f(z)=z^2+c$. Prove that $|z|>|c|+1$ implies $|f(z)|>|z|$

Question

Consider the quadratic function $f(z)=z^2+c$. If $|z|>|c|+1$, show that $|f(z)|>|z|$.

Edit: This is not a homework problem. I found this in my textbook.

Answer 1

Two preliminary inequalities:

• By the triangular inequality, $z^2=f(z)-c$ implies $|z|^2\leqslant|f(z)|+|c|$.
• If $|z|\gt|c|+1$, then $|z|^2-|z|=|z|(|z|-1)\gt(|c|+1)|c|=|c|^2+|c|$.

These two inequalities together yield $|f(z)|-|z|\geqslant|z|^2-|z|-|c|\gt|c|^2\geqslant0$ hence $|f(z)|\gt|z|$.

Answer 2

The problem is based on the algorithms used to make pictures of Julia sets. If the sequence $(x,f(x),f(f(x)), \dots )$ ever enters the region $|z| > R_f$ it will never return, and will be drawn toward the point at infinity ($|z|$ will get larger and larger, without bound), for a value of $R_f$ that depends on $f$ but not $z$. The colors of points in the thermal plots are assigned based on the number of iterations for a point to enter the region, or some subset $|z| > M_f > R_f$ for which $M_f$ is easier to compute explicitly in terms of $f$.

$M_{z^2+c} =|c|+1$ is enough but its only purpose is to have a clean bound that excludes $z$ from the Julia set. The minimum sufficient $R_f$ is the larger solution of $R^2 - |c| = R$, which is close to $\sqrt{|c|}$. The extreme case is when $|z|=R$ and the vector of the complex number $z^2$ points in the opposite direction to $c$.