Problem: Let $\langle \cdot, \cdot \rangle _{\mathbb{R}} $ denote the usual inner product in $\mathbb{R} ^2$. For any $\mathbf{x} = (x_1, x_2)$ and $\mathbf{y} = (y_1, y_2)$, we have
\[
\langle \mathbf{x} ,\mathbf{y} \rangle_{\mathbb{R}} = x_1 y_1 + x_2 y_2.
\]
Similarly, we define a Hermitian inner product $\langle \cdot, \cdot \rangle $ in $\mathbb{C} $ by
\[
\langle z, w \rangle_{\mathbb{C}} = z \bar{w} .
\]
Show that
\[
\langle z, w \rangle _{\mathbb{R} } = \frac{1}{2} \left[ \langle z,w \rangle _{\mathbb{C} } + \langle w,z \rangle _{\mathbb{C} } \right] = \text{Re}\left( \langle z,w \rangle _{\mathbb{C} } \right) ,
\]
where we use the usual identification $z = x + \iota y \in \mathbb{C} $ with $(x,y) \in \mathbb{R} ^2$.
Solution: Let us write $z = z_1 + \iota z_2$ and $w = w_1 + \iota w_2$. Now we consider
\begin{align*}
\langle z,w \rangle _{\mathbb{R} } & = z_1 w_1 + z_2 w_2.
\end{align*}
Now we consider the complex inner products.
\begin{align*}
\langle z,w \rangle_{\mathbb{C} } = z \bar{w} \text{ and } \langle w,z \rangle _{\mathbb{C} } = w \bar{z}.
\end{align*}
Thus,
\begin{align*}
\frac{1}{2} \left[ \langle z,w \rangle _{\mathbb{C} } + \langle w,z \rangle _{\mathbb{C} } \right] & = \frac{z \bar{w} + w \bar{z} }{2} \\
& = \frac{z \bar{w} + \overline{z \bar{w} } }{2} \\
& = \text{Re}(z \bar{w} ) \\
& = \text{Re} \left( \langle z, w \rangle_{\mathbb{C} } \right).
\end{align*}
We proved the last two inequalities.
Now we observe that
\begin{align*}
\text{Re}\left( z \bar{w} \right) & = \text{Re} \left( z_1 w_1 + z_2 w_2 + \iota \left( z_2 w_1 - z_1 w_2 \right) \right) \\
& = z_1 w_1 + z_2 w_2 = \langle z, w \rangle _{\mathbb{R} }.
\end{align*}
Therefore,
\[
\langle z, w \rangle _{\mathbb{R} } = \frac{1}{2} \left[ \langle z,w \rangle _{\mathbb{C} } + \langle w,z \rangle _{\mathbb{C} } \right] = \text{Re}\left( \langle z,w \rangle _{\mathbb{C} } \right) ,
\]