proof of $\langle z,w\rangle =\overline{\langle w,z\rangle}$ Here is an property of inner product space V: $\langle z,w\rangle =\overline{\langle w,z\rangle}$ (conjugate symmetry).
I need a reference on the proof of this property or general idea in proving this?
 A: As pointed out in the comments, the relation
$\langle z, w \rangle = \overline{\langle w, z \rangle } \tag 1$
is usually taken as an axiom; however, it is motivated by, and derives from, the corresponding property for the standard (hermitian) inner product on $\Bbb C^n$:
$\langle z, w \rangle = \displaystyle \sum_1^n \bar z_j w_j; \tag 2$
for this inner product we have
$\overline{\langle w, z \rangle} = \overline{ \displaystyle \sum_1^n \bar w_j z_j } = \sum_1^n  \bar{\bar w_j} \bar z_j = \sum_1^n w_j \bar z_j = \langle z, w \rangle. \tag 3$
The axiom (1) abstracts this to more general contexts in which $\Bbb C^n$ may not be directly available.
Also, see the comment by Arturo Magidin below.
A: Suppose $w, z \in \mathbb{C}^n$. Using the additive properties of complex conjugation,
$$
\langle z, w \rangle
 = \sum_{i = 1}^{n} z_i \overline{w}_i
 = \overline{\overline{\sum_{i = 1}^{n} z_i \overline{w}_i}}
 = \overline{\sum_{i = 1}^n \overline{z_i \overline{w}_i}}
 = \overline{\sum_{i = 1}^{n} \overline{z}_i w_i}
 = \overline{\sum_{i = 1}^{n} w_i \overline{z}_i}
 = \overline{\langle w, z \rangle}
$$
