Calculate $\sup\{|x|^2+3|y|^2;\;\;(x,y)\in \mathbb{C}^2,\;|x|^2+|y|^2=1\}$

Question

I want to calculate the following supremum $$c=\sup\{|x|^2+3|y|^2;\;\;(x,y)\in \mathbb{C}^2,\;|x|^2+|y|^2=1\}.$$

We have $$c=\sup\{1+2|y|^2;\;\;(x,y)\in \mathbb{C}^2,\;|x|^2+|y|^2=1\}.$$

Answer 1

Let $t,s \in \mathbb R$. If $t^2+s^2=1$, then $t^2+3s^2=t^2+3(1-t^2)=3-2t^2 \le 3$ and we have $3-2t^2=3 \iff t=0$.

Thus we have

$\sup\{t^2+3s^2;\;\;(t,s)\in \mathbb{R}^2,\;t^2+s^2=1\}=\max\{t^2+3s^2;\;\;(t,s)\in \mathbb{R}^2,\;t^2+s^2=1\}=3$.

Since $|x|^2 \in \mathbb R$ for $x \in \mathbb C$, we derive $c=3$.

Answer 2

You want to find $$c=\sup\{1+2|y|^2;\;\;(x,y)\in \mathbb{C}^2,\;|x|^2+|y|^2=1\}.$$

Since the maximum value for $|y|^2$ on the circle is $1$ we get $$c=\sup\{1+2|y|^2;\;\;(x,y)\in \mathbb{C}^2,\;|x|^2+|y|^2=1\}=1+2=3$$