I want to calculate
$$K=\sup\left\{\left|y\overline{x}+|y|^2\right|^2+1;\;\;(x,y)\in \mathbb{C}^2,\;|x|^2+|y|^2=1\right\}.$$
I try to solve the problem as follows:
let $x=r_1e^{i\theta_1}$ and $y=r_2e^{i\theta_2}$ then
$$ |x|^2+ |y|^2=1 \quad \Rightarrow \quad r^2_1+r_2^2 =1.$$
and
$$y\overline{x}+|y|^2= r_2^2+r_1r_2e^{i(\theta_2-\theta_1)}.$$
so
$$|r_2^2+r_1r_2e^{i(\theta_2-\theta_1)}|^2\leq r_2^4+r_1^2r_2^2.$$
Hence
$$K\leq\sup\{r_2^4+r_1^2r_2^2;\;\;r_1,r_2\geq0,\;r^2_1+r_2^2 =1\}.$$
Best Answer
You can simplify the problem a little first: Note that $|y \bar{x} +|y|^2| = |y (\bar{x}+\bar{y})| = |y| |x+y|$. We can choose the phase of $x,y$ arbitrarily, and $|x+y| \le |x|+|y|$ with equality when they are colinear. so we can choose $x,y$ to be real.
In particular, the problem reduces to $\max \{ y (x+y) | x,y \in \mathbb{R}, x^2+y^2 = 1 \}$.
You can solve the equivalent problem $ \max_t \sin t (\cos t + \sin t) = \max_t {1 \over 2} (1+\sin (2t) - \cos (2t)) = \max_t {1 \over 2}(1+\sqrt{2}\sin (2x - {\pi \over 4}))$.