$\textbf{Exercise 3.}$ Suppose $V$ is a complex vector space and $\varphi \in V^*$. Define $\sigma: V \to \mathbb{R}$ by $\sigma(v) = \operatorname{Re} \varphi(v)$ for each $v \in V$. Show that $\begin{equation*} \varphi(v) = \sigma(v) - i\sigma(iv) \end{equation*}$ for all $v \in V$. $\textbf{Solution 3.}$ If $v \in V$, then $\begin{align*} \operatorname{Im} \varphi(v) &= - \operatorname{Re}(i\varphi(v)) \\ &= - \operatorname{Re} \varphi(iv) \\ &= -\sigma(iv). \end{align*}$ Thus $\varphi(v) = \sigma(v) - i\sigma(iv)$. ---  --- Commentary: $\textbf{Exercise 3.}$ Suppose $V$ is a complex vector space and $\varphi \in V^*$. Define $\sigma: V \to \mathbb{R}$ by $\sigma(v) = \operatorname{Re} \varphi(v)$ for each $v \in V$. Show that $\begin{equation*} \varphi(v) = \sigma(v) - i\sigma(iv) \end{equation*}$ for all $v \in V$. $\textbf{Solution 3.}$ If $v \in V$, then $\begin{align*} \operatorname{Im} \varphi(v) &= - \operatorname{Re}(i\varphi(v)) \\ &= - \operatorname{Re} \varphi(iv) \\ &= -\sigma(iv). \end{align*}$ Thus $\varphi(v) = \sigma(v) - i\sigma(iv)$. $\textit{Commentary:}$ This exercise shows how to recover a complex-linear functional from its real part. Given a complex vector space $V$ and a complex-linear functional $\varphi$, we can define a real-linear functional $\sigma$ by taking the real part of $\varphi$. The exercise shows that $\varphi$ can be recovered from $\sigma$ by the formula $\varphi(v) = \sigma(v) - i\sigma(iv)$. The proof is a simple calculation using the linearity of $\varphi$ and the properties of complex numbers. This result is useful in the study of complex vector spaces and their duals, as it allows us to reduce some problems about complex-linear functionals to problems about real-linear functionals. $\textit{Example:}$ Let $V = \mathbb{C}^2$ and define $\varphi: \mathbb{C}^2 \to \mathbb{C}$ by $\varphi(z_1, z_2) = z_1 + iz_2$. The real part of $\varphi$ is the functional $\sigma: \mathbb{C}^2 \to \mathbb{R}$ given by $\begin{equation*} \sigma(z_1, z_2) = \operatorname{Re} \varphi(z_1, z_2) = \operatorname{Re}(z_1 + iz_2) = \operatorname{Re} z_1 - \operatorname{Im} z_2. \end{equation*}$ For any $(z_1, z_2) \in \mathbb{C}^2$, we have $\begin{align*} \sigma(z_1, z_2) - i\sigma(i(z_1, z_2)) &= \sigma(z_1, z_2) - i\sigma(iz_1, -z_2) \\ &= (\operatorname{Re} z_1 - \operatorname{Im} z_2) - i(\operatorname{Im} z_1 + \operatorname{Re} z_2) \\ &= \operatorname{Re} z_1 + i\operatorname{Im} z_1 + \operatorname{Re} z_2 + i\operatorname{Im} z_2 \\ &= z_1 + iz_2 \\ &= \varphi(z_1, z_2), \end{align*}$ as predicted by the theorem.