We could use linearity in the first argument, homogeneity in first argument, and conjugate symmetry properties of the dot product. So this was my attempt at proving this:

We know that $\langle v+v',w \rangle = \langle v, w \rangle + \langle v',w \rangle$ and we also know that $\langle v,w \rangle = \overline{\langle w,v \rangle}$ (conjugate). Therefore

$\langle v+v',w \rangle = \overline{\langle w, v+v' \rangle}$

$\langle v+w \rangle + \langle v'+w \rangle = \overline{\langle w,v \rangle}+ \overline{\langle w+v' \rangle}$

Then $\overline{\langle w, v+v' \rangle}= \overline{\langle w,v \rangle}+ \overline{\langle w+v' \rangle}$

and taking the conjugate ${\langle w, v+v' \rangle}={\langle w,v \rangle}+ {\langle w+v' \rangle}$

Not sure if my argument makes sense (I'm not very familiar with conjugates)


$$\begin{align*} \langle w,v+v' \rangle &= \overline{\langle v+v',w \rangle}\\ &= \overline{\langle v,w \rangle + \langle v',w \rangle}\\ &= \overline{\langle v,w \rangle} + \overline{\langle v',w \rangle}\\ &= \langle w,v \rangle + \langle w,v' \rangle \end{align*}$$


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