Question

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

$\left( \cos\frac{\alpha}{r} + i\sin\frac{\alpha}{r} \right)$ hen $\lim_{n \rightarrow \infty}f_{n}$ (p) equals

• A. –1
• B. 1
• C. – I
• D. i

Answer 1

Answer: (D)

i

Answer 2

Using De Moivre's theorem

f_r (a) = $e^{2i\alpha/r^{2}}$….$e^{i\alpha/r}$

= $e^{(i\alpha/r^{2})(1 + 2 + ...... + r)}$

= $e^{(i\alpha/r^{2})\lbrack r(r + 1)/2\rbrack}$ = $e^{(i\alpha/2)(1 + 1/r)}$

\ $\lim_{n \rightarrow \infty}f_{n}(\pi)$ = $e^{i\pi/2(1 + 1/n)}$

= e^ip/2 = cos$\left( \frac{\pi}{2} \right)$ + i sin$\left( \frac{\pi}{2} \right)$ = i