Complex inner product linearity

Question

Let $V$ be an inner product space over $\mathbb{C}$. Is the expression

$$ \newcommand{\<}{\langle} \newcommand{\>}{\rangle} \<v,\lambda u\> = \bar{\lambda}\<v,u\> = \bar{\lambda}\overline{\<u,v\>} = \overline{\<\bar{\lambda}u,v\>} = \<v,\bar{\lambda}u\> $$ true? I mean is every move "legal"?

Answer 1

No, the third equality does not hold. Note that

$$\overline{\lambda}\,\overline{\langle u, v\rangle} = \overline{\lambda\langle u, v\rangle} = \overline{\langle\lambda u, v\rangle}.$$

Answer 2

No, because this isn't justified:

$$\bar{\lambda}\overline{\<u,v\>} = \overline{\<\bar{\lambda}u,v\>} $$

It should be

$$\bar{\lambda}\overline{\<u,v\>} =\overline{\lambda \<u,v\>} = \overline{\<\lambda u,v\>} $$