Proving that linear isometries preserve angles without the polarization identity

Question

Let $V$ be an inner product space, and let $T : V \to V$ be a linear isometry, that is $$ \|Tv\| = \|v\| \quad \text{for all $v \in V$}. $$ I would like to prove that $T$ also preserves inner products, that is $$ \tag{1} \langle Tu, Tv \rangle = \langle u, v \rangle \quad \text{for all $u, v \in V$}. $$ The usual way to prove this is to use the polarization identity, which allows us to write an inner product $\langle u, v \rangle$ as a linear combination of vector norms. However, the polarization identity takes different forms depending on if $V$ is a real or complex inner product space, so, to me, this proof feels a bit lacking in generality. Is there a way to prove (1) without treating the real and complex case separately? If not, is there some suggestion that every proof must somewhere involve splitting up the real and complex case?