If $f_r(\alpha) = \left(\cos\left(\frac{\alpha}{r^2}\right) + i \sin\left(\frac{\alpha}{r^2}\right)\right) \left(\cos\left(\frac{2\alpha}{r^2}\right) + i \sin\left(\frac{2\alpha}{r^2}\right)\right) \ldots \left(\cos\left(\frac{\alpha}{r}\right) + i \sin\left(\frac{\alpha}{r}\right)\right)$ then…

Question

If $f_r(\alpha) = \left(\cos\left(\frac{\alpha}{r^2}ight) + i \sin\left(\frac{\alpha}{r^2}ight)ight) \left(\cos\left(\frac{2\alpha}{r^2}ight) + i \sin\left(\frac{2\alpha}{r^2}ight)ight) \ldots \left(\cos\left(\frac{\alpha}{r}ight) + i \sin\left(\frac{\alpha}{r}ight)ight)$ then $\lim_{n 	o \infty} f_n(\pi) $equals (where $i = \sqrt{-1}$)

Aicharya · Accepted Answer

Using De Moivre's theorem f r α = e i α r 2 . e 2 i α r 2 … . e i α r = e i α r 2 1 + 2 + … + r = e i α r 2 r r + 1 2 = e i α 2 1 + 1 r ∴ lim n ⟶ ∞ ⁡ f n π = l i m n ⟶ ∞ e i π 2 1 + 1 n = e i π 2 = cos ⁡ π 2 + i sin ⁡ π 2 = i

If $f_r(\alpha) = \left(\cos\left(\frac{\alpha}{r^2}\right) + i \sin\left(\frac{\alpha}{r^2}\right)\right)…

Solution