Proving $\sup\limits_{\theta,\psi}\{||e^{i\theta}x+e^{i\psi}y||^2:\theta,\psi\in \Bbb R\}=||x||^2+||y||^2+2 Re$ where $x,y \in \mathbb{C}^n$ Q. Let $x,y \in \mathbb{C}^n$. Consider $f(x,y)=\sup\limits_{\theta,\psi}\{\lVert e^{i\theta}x+e^{i\psi}y\rVert^2\colon \theta,\psi\in \Bbb R\}.$ Which of the following is/are correct?

*

*$f(x,y)\leq\lVert x\rVert^2+\lVert y\rVert^2+2|\langle x,y\rangle|$


*$f(x,y)= \lVert x\rVert^2+\lVert y\rVert^2+2 Re\langle x,y\rangle$


*$f(x,y)=\lVert x\rVert^2+\lVert y\rVert^2+2\lvert\langle x,y\rangle\vert$


*$f(x,y)>\lVert x\rVert^2+\lVert y\rVert^2+2\lvert\langle x, y\rangle\rvert$
I can easily reject the option $(4)$ by simply assuming $y=0 \in \Bbb C^n$, How can we deal others by connecting the quantity $\lVert x+y\rVert^2$ with $f(x,y)$?
How to continue with $\lVert e^{i\theta}x+e^{i\psi}y\rVert^2=\langle e^{i\theta}x+e^{i\psi}y, \overline{e^{i\theta}x+e^{i\psi}y}\rangle$ ?
 A: By expanding the inner product we get
$$\|e^{i\theta}x+e^{i\psi}y\|^2 = \|e^{i\theta}x\|^2 + \|e^{i\psi}y\|^2 + 2\operatorname{Re} \langle e^{i\theta}x,e^{i\psi}y\rangle = \|x\|^2+\|y\|^2 + 2 \operatorname{Re}e^{i(\theta-\psi)}\langle x,y\rangle.$$
Now this can be majorized by
$$\|x\|^2+\|y\|^2 + 2 |e^{i(\theta-\psi)}\langle x,y\rangle| \le \|x\|^2+\|y\|^2 + 2 |\langle x,y\rangle|.$$
Moreover, writing the complex number $\langle x,y\rangle$ in polar form, we have
$$\langle x,y \rangle = |\langle x,y\rangle| e^{i\eta}, \quad \text{ for some } \eta \in \Bbb{R}$$
so by setting $\theta = 0, \psi = \eta$, we have
\begin{align}
\|x\|^2+\|y\|^2 + 2 \operatorname{Re}e^{i(\theta-\psi)}\langle x,y\rangle &= \|x\|^2+\|y\|^2 + 2 \operatorname{Re}e^{-i\eta}\langle x,y\rangle \\
&= \|x\|^2+\|y\|^2 + 2 \operatorname{Re}|\langle x,y\rangle|\\
&= \|x\|^2+\|y\|^2 + 2|\langle x,y\rangle|.
\end{align}
We can conclude that $f(x,y) = \|x\|^2+\|y\|^2 + 2|\langle x,y\rangle|$ so $1$ and $3$ are correct.
To see that $2$ isn't correct in general, pick any two vectors $x,y \in \Bbb{C}^n$ such that $\operatorname{Re} \langle x,y\rangle < |\langle x,y\rangle|$.
