Antisymmetric rank one operator on a Hilbert space Let $(H, \langle \cdot, \cdot \rangle)$ be a complex Hilbert space. Suppose $\langle \cdot, \cdot \rangle$ is antilinear in the first argument and linear in the second. We know that every bounded rank one linear operator $T : H \to H$ has the form $T(u) = \langle \varphi, u \rangle y$ for some $\varphi$ and $y \in H$.

If we also know that $T$ is antisymmetric, i.e., $\langle v, Tu \rangle = -\langle Tv, u \rangle$, then may we conclude that there exists $f \in H$ so that $T(u) = i \langle f, u \rangle f$?

Note that any map of the proposed form is antisymmetric:
$$ \langle v, i\langle f, u \rangle f \rangle = i \langle v, f \rangle \langle f, u \rangle = -\overline{i \langle f,v \rangle} \langle f, u \rangle = -\langle i\langle f,v \rangle f, u \rangle.$$
For general antisymmetric rank one $T$, I computed, for all $u \in H$,
$$\overline{ \langle \varphi, u \rangle \langle u, y \rangle} = -  \langle \varphi, u \rangle \langle u, y \rangle. $$
Setting $u = y$ and dividing by $\|y\|^2$ gives that the real part of $\langle \varphi, y\rangle$ is zero. I'm not sure where to go from here. Hints or solutions are greatly appreciated.

Let $Tu=\langle \varphi, u\rangle y.$ If $T$ is antisymmetric we get
$$\|\varphi\|^2\langle y,u\rangle =\langle T\varphi,u\rangle= -\langle \varphi, Tu\rangle =-\langle \varphi,u\rangle \langle \varphi,y\rangle$$ for every $u.$ For $u=y$ we obtain $$\|\varphi\|^2\|y\|^2=-\langle \varphi,y\rangle^2\qquad (*)$$ Thus we have equality in the Cauchy-Schwarz inequality. Hence $\varphi =\lambda y.$ Identity $(*)$ implies that $|\lambda|^2=-\bar{\lambda}^2.$  Therefore $\lambda = -it$ for a real number $t.$ Summarizing $$Tu=it\langle y, u\rangle y\qquad t\in \mathbb{R}\setminus \{0\}$$
Remark We cannot expect that $$Tu=i\langle y, u\rangle y$$ as such operator has additional property $$i\langle Tu,u\rangle =|\langle y,u\rangle |^2\ge 0$$
