1. ## inequality

Show that $|(Re^{i \theta})^{2} + a^{2} | \ge R^{2} -a^{2} \ \ (R, a >0, \ 0 \le \theta \le \pi)$

$|R^{2}e^{2i \theta} + a^{2} | = |R^{2}(cos 2 \theta + i \sin 2 \theta) + a^{2} |= |(R^{2} \cos 2 \theta + a^{2}) + i R^{2} \sin 2 \theta |$

$= \sqrt{(R^{2} \cos 2 \theta +a^{2})^{2} + R^{4} \sin^{2} 2 \theta}$

$= \sqrt{R^4 \cos^{2} 2 \theta + 2R^{2}a^{2}\cos 2 \theta + a^{4} + R^{4} \sin^{2} \theta}$

$= \sqrt{R^{4} + 2R^{2}a^{2} \cos 2 \theta + a^{4}}$

$\ge \sqrt{(R^{2}-a^{2})^{2}}$ ?

2. Originally Posted by Random Variable
Show that $|(Re^{i \theta})^{2} + a^{2} | \ge R^{2} -a^{2} \ \ (R, a >0, \ 0 \le \theta \le \pi)$

$|R^{2}e^{2i \theta} + a^{2} | = |R^{2}(cos 2 \theta + i \sin 2 \theta) + a^{2} |= |(R^{2} \cos 2 \theta + a^{2}) + i R^{2} \sin 2 \theta |$

$= \sqrt{(R^{2} \cos 2 \theta +a^{2})^{2} + R^{4} \sin^{2} 2 \theta}$

$= \sqrt{R^4 \cos^{2} 2 \theta + 2R^{2}a^{2}\cos 2 \theta + a^{4} + R^{4} \sin^{2} \theta}$

$= \sqrt{R^{4} + 2R^{2}a^{2} \cos 2 \theta + a^{4}}$

$\ge \sqrt{(R^{2}-a^{2})^{2}}$ ?
Apply a variation of the triangle inequality to get

\begin{aligned}\left|\left(Re^{i\theta}\right)^2+a ^2\right|\geq\left|\left(Re^{i\theta}\right)^2\rig ht|-\left|a^2\right| & =\left|Re^{i\theta}\right|^2-a^2\\ & =R^2\left|e^{i\theta}\right|^2-a^2\\ & =R^2\left(\cos^2\theta+\sin^2\theta\right)-a^2\\ &= R^2-a^2\end{aligned}

3. Originally Posted by Chris L T521
Apply a variation of the triangle inequality to get

\begin{aligned}\left|\left(Re^{i\theta}\right)^2+a ^2\right|\geq\left|\left(Re^{i\theta}\right)^2\rig ht|-\left|a^2\right| & =\left|Re^{i\theta}\right|^2-a^2\\ & =R^2\left|e^{i\theta}\right|^2-a^2\\ & =R^2\left(\cos^2\theta+\sin^2\theta\right)-a^2\\ &= R^2-a^2\end{aligned}
Thanks. I don't why I never heard of the reverse triangle inequality (the proof of which is quite simple).

4. Originally Posted by Random Variable
Thanks. I don't why I never heard of the reverse triangle inequality (the proof of which is quite simple).
I'm not sure if this will change things...

I used the part highlighted in red:

${\color{red}\left|\left|a\right|-\left|b\right|\right|\leq\left|a+b\right|}\leq\lef t|a\right|+\left|b\right|$.

So it should technically be $\left|\left(Re^{i\theta}\right)^2+a^2\right|\geq\l eft|R^2-a^2\right|$.

If $R\geq a$, then it would be ok to leave it as $\left|\left(Re^{i\theta}\right)^2+a^2\right|\geq R^2-a^2$...

5. Originally Posted by Chris L T521
I'm not sure if this will change things...

I used the part highlighted in red:

${\color{red}\left|\left|a\right|-\left|b\right|\right|\leq\left|a+b\right|}\leq\lef t|a\right|+\left|b\right|$.

So it should technically be $\left|\left(Re^{i\theta}\right)^2+a^2\right|\geq\l eft|R^2-a^2\right|$.

If $R\geq a$, then it would be ok to leave it as $\left|\left(Re^{i\theta}\right)^2+a^2\right|\geq R^2-a^2$...
I was just about to ask if it matters if $a$ is greater than or less than $R$. For this particular problem it's less than $R$.