Question 230237
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We know that *[tex \LARGE \cos\left(\frac{\pi}{2}\right)\ =\ 0] and *[tex \LARGE \sin\left(\frac{\pi}{2}\right)\ =\ 1], therefore we can say that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i\ =\ \cos\left(\frac{\pi}{2}\right)\ +\ i\sin\left(\frac{\pi}{2}\right)\]


But we know from Euler's Formula that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{i\Theta}\ =\ \cos\Theta\ +\ i\sin\Theta]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{i\left(\frac{\pi}{2}\right)}\ =\ i]


Taking the square root of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{i\left(\frac{\pi}{4}\right)}\ =\ i^{\frac{1}{2}}]


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{i\left(\frac{\pi}{4}\right)}\ =\ \pm\sqrt{i}\ =\ \cos\left(\frac{\pi}{4}\right)\ +\ i\sin\left(\frac{\pi}{4}\right)\]


Finally:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pm\sqrt{i}\ =\ \frac{1}{\pm\sqrt{2}}\ +\ \frac{i}{\pm\sqrt{2}}\ =\ \pm\frac{\sqrt{2}}{2}\left(1\ +\ i\right)]


after rationalizing the denominators.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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