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

$\displaystyle |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 | $

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

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

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

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

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