Question 762274
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\frac{\sqrt{3}}{2}\ +\ i\frac{1}{2}\right)^{91}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ x^2\ +\ y^2\ =\ \left(-\frac{sqrt{3}}{2}\right)^2\ +\ \left(\frac{1}{2}\right)\ =\ \frac{3}{4}\ +\ \frac{1}{4}\ =\ 1]


Since *[tex \LARGE r\ =\ 1], and we know that *[tex \LARGE x\ =\ r\cos\varphi] and *[tex \LARGE y\ =\ r\sin\varphi], it is evident that *[tex \LARGE \cos\varphi\ =\ -\frac{\sqrt{3}}{2}] and *[tex \LARGE \sin\varphi\ =\ \frac{1}{2}]


From the unit circle we deduce *[tex \LARGE \varphi\ =\ \frac{5\pi}{6}]


So we now have


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ \left(-\frac{\sqrt{3}}{2}\ +\ i\frac{1}{2}\right)^{91}]


Then De Moivre says:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\left(\cos(\alpha)\ +\ i\sin(\alpha)\right)^n\ =\ r^n\left(\cos(n\alpha)\ +\ i\sin(n\alpha)\right)]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z^{91}\ =\ r^{91}\left(\cos\left(91\left(\frac{5\pi}{6}\right)\right)\ +\ i\sin\left(91\left(\frac{5\pi}{6}\right)\right)\right)]


Since *[tex \LARGE r\ =\ 1], that falls off.  Then *[tex \LARGE \frac{91\cdot 5\pi}{6}\ =\ \frac{455\pi}{6}\ =\ \frac{11\pi}{6}], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z^{91}\ =\ \cos\left(\frac{11\pi}{6}\right)\ +\ i\sin\left(\frac{11\pi}{6}\right)\ =\ \frac{\sqrt{3}}{2}\ -\ i\frac{1}{2}]




John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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