Let $E$ be a $\complex$-[[Vector Space|vector space]] and $\phi \in E^*$ be a [[Linear Function|linear functional]], then $ \angles{x, \phi} = \re{\angles{x, {\phi}}} - i\angles{ix, \re{\phi}} $ The conjugate-linear functional $ \anglesn{x, \ol \phi} = \angles{x, \re{\phi}} + i\angles{ix, \re{\phi}} $ is an **antilinear functional**, and $\overline{E}^*$ is the **antilinear dual** of $E$. The map $\phi \mapsto \overline{\phi}$ is an isomorphism. If $E$ is normed, then $\phi \mapsto \overline{\phi}$ is an isometry. *Proof*. $ \begin{align*} \angles{x, \phi} &= \angles{x, \re{\phi}} + i\angles{x, \im{\phi}} \\ &= \angles{x, \re{\phi}} - i\angles{ix, \re{\phi}} \end{align*} $ For any $\lambda = a + bi \in \complex$, $ \begin{align*} \anglesn{\lambda x, \overline \phi} &= \anglesn{ax + bix, \overline \phi} \\ &= \anglesn{ax, \overline{\phi}} + \angles{bix, \re{\phi}} + i\angles{bix, \re{\phi}} \\ &= \anglesn{ax, \overline{\phi}} -\angles{bx, \im{\phi}} - i\angles{bx, \im{\phi}} \\ &= \end{align*} $