Let us denote the maximum norm on $\mathbb{C}^n$ as $\lVert \cdot \rVert_{\infty}$ and the normal Euclidean norm on $\mathbb{C}^n$ (arising from the inner product) as $\lVert \cdot \rVert_2$. It is a well-known result in several complex variables that the unit ball $B = \{ z \in \mathbb{C}^n \ | \ \lVert z \rVert_2 < 1 \}$ and the unit polydisc $D = \{ z \in \mathbb{C}^n \ | \ \lVert z \rVert_{\infty} < 1\}$ are not biholomorphically equivalent. However, they are homeomorphic to one another which is shown via the homeomorphism $f : B \to D$ given by $$ f(z) = \begin{cases} \frac{\lVert z\rVert_2}{\lVert z \rVert_{\infty}}z & \text{ if } z \neq 0 \\ 0 & \text{ if } z = 0 \\ \end{cases} $$
Despite the famed result, I'm having trouble understanding why the homeomorphism $f$ isn't a (bi)holomorphic map. My explicit calculations don't seem to be working out how I expect they should (i.e., I expect to get a limit I can't calculate) and every time I think I've shown $f$ (or $f^{-1}$) is not holomorphic I question my working or find an error. I'm probably way overthinking it as it's getting late. Help would be greatly appreciated.
Let's take two variables for simplicity and consider the map from $\Omega=|z| < \frac{1}{2}$ to $\mathbb{C}$ given as $g=p \circ f \circ i$, where $i:\Omega \to B, i(z)=(z,\frac{1}{2}), p:D \to \mathbb{C}, p(z,w)=z$. Clearly $i,p$ are holomorphic, so if $f$ were holomorphic, $g$ would be too.
A simple computation shows that $g(z)=2z\sqrt{|z|^2+\frac{1}{4}}$ which is highly non-holomorphic on $\Omega$