My textbook says, if $\left\lvert c \right\rvert >2$ then $K_c$$\subset$ $D(0,\left\lvert c \right\rvert)$. With iterating $f(z)=z^2+c$.
So what i understand what this says is that, if c is bigger then 2 then the juliaset $K_c$ lies whitin the circle with radius $\left\lvert c \right\rvert$ with origin $0$. I have to proof that this is true, but don't really how to approach it.
If $|c| > 2$ and $|z| > |c|$ then $$ |f(z)| \ge |z|^2 - |c| > |c|^2 - |c| > |c| $$ so that $f$ maps $A = \{z: |z| > |c| \}$ into itself.
It follows that the family $(f_n)$ of iterates is normal in $A$ and therefore $A$ is a subset of the Fatou set of $f$. Consequently, the Julia set of $f$ is contained in $A^C = \{z: |z| \le |c| \}$.
Only little more effort is needed to show that the Julia set is strictly contained in the open ball of radius $|c|$: Define $r = (2 + |c|)/2$. Then $2 < r < |c|$ and $$ r^2 + r = r(r+1) = \frac 14 (2 + |c|)|c| > |c| \quad . $$ Then for $|z| > r$ $$ |f(z)| \ge r^2 - |c| > r $$ so that $f$ maps $A = \{z: |z| > r \}$ into itself.
As above it follows that the Julia set is contained in $\{z: |z| \le r \} \subset \{z: |z| < |c| \} $.