<s>[INST] How do I show that $\frac {|s|} {\sigma} \leq \sqrt {1 + \tan \left (\frac {\pi} {2} - \delta \right )}\ $? Show that in the region $\left \{s\ \big |\ \left |\arg \left (s \right ) \right | \leq \frac {\pi} {2} - \delta \right \}$ $$\frac {|s|} {\sigma} \leq \sqrt {1 + \tan \left (\frac {\pi} {2} - \delta \right )}$$ where $\mathfrak {R} (s) = \sigma$ and $0 \lt \delta \lt \frac {\pi} {2}.$
What I did is the following $:$
Let $\mathfrak {I} (s) = \tau.$ Then $$\begin {align*} \frac {|s|} {\sigma} & = \frac {\sqrt {\sigma^2 + \tau^2}} {\sigma} \\ & = \sqrt {1 + \left (\frac {\tau} {\sigma} \right )^2} \\ & \leq \sqrt {1 + \tan^2 \left (\frac {\pi} {2} - \delta \right )} \end{align*}$$ The last equality follows from the fact that $-\frac {\pi} {2} \lt - \frac {\pi} {2} + \delta \leq \arg \left (s \right ) \leq \frac {\pi} {2} - \delta \lt \frac {\pi} {2}$ and $\tan x$ is increasing on $\left (-\frac {\pi} {2}, \frac {\pi} {2} \right ).$ But the problem is that the last inequality what I have obtained doesn't match properly with that of the desired inequality. Did I make any mistake here? Could anybody please check it once?
Thanks for your time.
 [/INST] Your derivation of
$$
 \frac {|s|} {\sigma} \leq \sqrt {1 + \tan^2 \left (\frac {\pi} {2} - \delta \right )} 
$$
is correct, and equality holds if $|\arg(s)| = \pi/2 - \delta$.
If $|\arg(s)| = \pi/2 - \delta$ and $0 <\delta < \pi/4$ then
$$
\frac {|s|} {\sigma} = \sqrt {1 + \tan^2 \left (\frac {\pi} {2} - \delta \right )} > \sqrt {1 + \tan \left (\frac {\pi} {2} - \delta \right )} 
$$
so that the first inequality does not hold in general.
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