Maximum of $\vert f(z)\vert$ on $|z|\leqslant1$, where $f(z)= 1+z^2$ 
Let $f(z)=z^2+1$. Determine the maximum of $\vert f(z)\vert$ on $\overline{D_1(0)}$

First I want to use the maximum theorem for a complex holomorphic function. But I thought this is an easier way:
$$\vert f(z)\vert = \vert 1+z^2\vert \leq 1+ \vert z^2\vert = 1+ \vert z\vert^2$$
Due to the fact that this function is defined on $\overline{D_1(0)}$ one can say, that $\vert z\vert \leq 1 \forall z\in\mathbb{C}$ $$\Rightarrow \max\limits_{\overline{D_1(0)}} \vert f(z)\vert = 2$$
Is that correct? Thank you!

After you show that $$|f(z)| \leq 2$$
Show a point $z$ that makes $|f(z)|$ attain that value, in particular we can let $z=1$. 
Hence the maximum value is indeed $2$.
