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Let $V$ be a finite-dimensional complex vector space. Then we can regard $V$ also as a real vector space; let $V_\Bbb R$ denote the real vector space. Suppose $\mu:V_\Bbb R\to \Bbb R$ is a positive definite ($\mu(v)>0$ for all nonzero $v\in V_\Bbb R$) quadratic function, and suppose $\mu(iv)=\mu(v)$ for all $v\in V$. I want to show that there is a unique complex inner product $V\times V\to \Bbb C$ satisfying $\langle v,v\rangle =\mu(v)$. I have shown existence: define $$ \langle v,w\rangle=\frac{1}{2}(\mu(v+w)-\mu(v)-\mu(w))+\frac{i}{2}(\mu(v+iw)-\mu(v)-\mu(iw)). $$ Then a straightforward (but little bit long) computation shows that this is indeed a complex inner product. But how can we show uniqueness?

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Assuming your formula is correct: suppose that $\langle \cdot,\cdot\rangle_1$ and $\langle \cdot, \cdot \rangle_2$ induce the same quadratic form $\mu$. It follows that for all $u,v$, we have $$ \langle v,w\rangle_1 = \langle v,w \rangle_2 = \frac{1}{2}(\mu(v+w)-\mu(v)-\mu(w))+\frac{i}{2}(\mu(v+iw)-\mu(v)-\mu(iw)). $$

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