De Moivre’s Theorem application


If f_{r}(\alpha)=(\cos{\frac{\alpha}{r^{2}}}+i\sin{\frac{\alpha}{r^{2}}}) \times (\cos{\frac{2\alpha}{r^{2}}}+i\sin{\frac{2\alpha}{r^{2}}}) \ldots (\cos{\frac{\alpha}{r}}+i\sin{\frac{\alpha}{r}}), then

\lim_{n \rightarrow \infty}f_{n}{\pi} equals

(a) -1

(b) 1

(c) -i

(d) i


Using De Moivre’s theorem,

f_{r}{\alpha}=e^{i\frac{\alpha}{r^{2}}}e^{i\frac{2\alpha}{r^{2}}}\ldots e^{i\frac{\alpha}{r}}

which in turn equals e^{(i \frac{\alpha}{r^{2}})(1+2+\ldots+r)}=e^{(i\frac{\alpha}{r^{2}})(\frac{r(r+1)}{2})}=e^{i(\frac{\alpha}{2})(1+\frac{1}{r})}

Hence, \lim_{n \rightarrow \infty}f_{n}(\pi)=\lim_{n \rightarrow \infty}e^{(i)(\frac{\pi}{2})(1+\frac{1}{n})}=e^{i(\frac{\pi}{2})}=\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}=i.

More complex stuff to be continued in next blog (pun intended) 🙂

Nalin Pithwa

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