Norm of linear operator

Question

Given two real numbers $\alpha$ and $\beta$, consider the linear operator $T:\mathbb{C}\rightarrow \mathbb{R}$ defined by $T(x+iy)=\alpha x +\beta y$.

I am trying to figure out the norm of this operator:

$$\| T\|=\sup\limits_{z \in \mathbb{C}\neq 0}\left\{\frac{\|Tz \|}{\| z\|} \right\}=\sup\limits_{(x,y)\neq (0,0)}\left\{ \left| \frac{\alpha x +\beta y }{ x+iy}\right|\right\}=\sup\limits_{(x,y)\neq (0,0)}\left\{ \frac{\left|\alpha x +\beta y\right| }{ \sqrt{x^2+y^2}}\right\} $$

I thought maybe using polar coordinates to simplify the ratio as follows:

$$ \frac{\left|\alpha x +\beta y\right| }{ \sqrt{x^2+y^2}}=\frac{\left|\alpha r \cos(\theta) +\beta r \sin(\theta)\right| }{ r}= \left|\alpha \cos(\theta) +\beta \sin(\theta)\right| $$

By using derivatives, this function reaches its maximum when $\tan{\theta}=\frac{\beta}{\alpha}$, i.e. : $$ \| T\| = \left|\alpha \cos(\arctan{\frac{\beta}{\alpha}}) +\beta \sin(\arctan{\frac{\beta}{\alpha}})\right|=\left| \alpha\frac{1}{\sqrt{1+\left(\frac{\beta}{\alpha}\right)^2}}+\beta\frac{ \frac{\beta}{\alpha}}{\sqrt{1+\left(\frac{\beta}{\alpha}\right)^2}} \right|=\sqrt{\alpha^2+\beta^2} $$

Any other approach is welcome.

Answer 1

If $x^2+y^2=1$, then by Cauchy Schwarz, $|\alpha x + \beta y| \leq (\alpha^2+\beta^2)^{1/2}$.

On the other hand, $T\big(\frac{\alpha}{(\alpha^2+\beta^2)^{1/2}}+i\frac{\beta}{(\alpha^2+\beta^2)^{1/2}}\big)= (\alpha^2+\beta^2)^{1/2}$.

So, $\|T\| = (\alpha^2+\beta^2)^{1/2}$.