Prove that: $f$ is not a constant function on the open unit disc $\Delta (0,1)$ I have come across the following questions from my text book! However, I'm not sure how to go about answering... Any help (or hint) would be greatly appreciated. Thanks.
Suppose $F=\mathbb{C}^2$ with norm $\sup$.
For $f: \mathbb{C} \to F$, the function $f$ be defined by $f(z)=(1,z)$, $\forall z \in \mathbb{C}$.
Prove that:
a. $f  \in \mathcal{P}(\mathbb{C},F)$ 
b. $\left \|f  \right \|$ is constant on the open unit disc $\Delta (0,1)$
c. $f$ is not a constant function on the open unit disc $\Delta (0,1)$
d. $\left \|f  \right \|$ is not a constant on $\mathbb{C}$
 A: Hint: You have $||f(z)||= \max (1,|z|)$. So if $z \in \Delta(0,1)$, $|z| <1$ hence $||f(z)||=1$; if $z \notin \Delta(0,1)$, $|z| \geq 1$ hence $||f(z)||=|z|$.
A: If you conpute the norm using elements of the open disc, it will always be 1, for the definition of the disc $B_0(1)$ (I use this notation) which contains every element whose norm is less of 1. Then use the definition in the "hint" and you will be ok. D) follows easily for the same reason.
the other points are immediate.
A: Thank you very much! Seirios and Ric Ped.
With Seirios's hints, I will rewrite as follows:
We consider: $f: \mathbb{C} \to \mathbb{C}^2$, $f$ is defined by $f(z)=(1,z)$.
a. $f \in \mathcal{P}(\mathbb{C},\mathbb{C}^2)$.
it is obvious. Because, $f(z)=(f_1,f_2)=(1,z)$ is a polynomial and continuous, since $f_1=1$ and $f_2=z$ are polynomials and continuous.
b. $∥f∥$ is constant on the open unit disc $\Delta(0,1)$.
We have: $\Delta(0,1)=\{\lambda \in \mathbb{C}:|\lambda|<1\}$.
$||f||=\sup_{z}{|f(z)|}=\sup_{z} \{1,|z|\}$.
Hence, if we consider $f$ on $\Delta(0,1)$ $\implies ||f||=1$.
Therefore, $||f||=1$ on $\Delta(0,1)$.
c. $f$ is not a constant function on the open unit disc $\Delta(0,1)$
If $f$ is a constant function on the open unit disc $\Delta(0,1)$ then $f(z_1)=f(z_2)$, $\forall z_1,z_2 \in \Delta(0,1)$. 
I show that there exists $z_1=\dfrac{1}{2} \ne z_2=\dfrac{1}{3} \in \Delta(0,1)$ so $f(z_1) \ne f(z_2)$.
d. $∥f∥$ is not a constant on $\mathbb{C}$.
Since a. $f$ on $\Delta(0,1)$ $\implies ||f||=1$ 
If $z \notin \Delta(0,1)$, $|z| \ge 1$, hence $||f(z)||=|z|$.
Therefore, $||f||$ is not a constant on $\mathbb{C}$. :p
