Is $ \sup\{|a|^4+|a\overline{b}|^2;\;(a,b) \in \mathbb{C}^2\;\;\hbox{and}\;|a|^2+|b|^2=1\}<1$?

Question

I want to compute the following supremum of at least prove that $$ \sup\left\{|a|^4+|a\overline{b}|^2;\;(a,b) \in \mathbb{C}^2\;\;\hbox{and}\;|a|^2+|b|^2=1\right\}<1.$$

I used the following inequalities $$|a\overline{b}|\leq \frac{1}{2}(|a|^2+|b|^2)\text{ and } |a|^2\leq |a|^2+|b|^2.$$ I get $$|a|^4+|a\overline{b}|^2\leq \frac{5}{4},$$ for every $(a,b) \in \mathbb{C}^2$ such that $|a|^2+|b|^2=1$. So $$ \sup\left\{|a|^4+|a\overline{b}|^2;\;(a,b) \in \mathbb{C}^2\;\;\hbox{and}\;|a|^2+|b|^2=1\right\}\leq \frac{5}{4}.$$ But this not solve my problem.

Answer 1

You have

$$|a|^4+|a\overline{b}|^2= \vert a \vert^4 + \vert a \vert^2 \vert b\vert^2= \vert a \vert^4 + \vert a \vert^2( 1- \vert a \vert^2)=\vert a \vert^2 \le 1$$

for all $a,b$ such that $\vert a \vert^2+ \vert b\vert^2=1$.