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Question

Question: If \(z_{r} = \cos\frac{r\alpha}{n^{2}} + i\sin\frac{r\alpha}{n^{2}},\) where \(r = 1,2,3,.....,n,\) ...

If zr=cos⁡rαn2+isin⁡rαn2,z_{r} = \cos\frac{r\alpha}{n^{2}} + i\sin\frac{r\alpha}{n^{2}}, where r=1,2,3,.....,n,r = 1,2,3,.....,n, then

lim⁡n→∞z1z2z3.....zn\lim_{n \rightarrow \infty}z_{1}z_{2}z_{3}.....z_{n}is equal to

A

cos⁡α+isin⁡α\cos\alpha + i\sin\alpha

B

cos⁡(α/2)−isin⁡(α/2)\cos(\alpha/2) - i\sin(\alpha/2)

C

eiα/2e^{i\alpha/2}

D

eiα3\sqrt[3]{e^{i\alpha}}

Answer

eiα/2e^{i\alpha/2}

Explanation

Solution

Sol. zr=cos⁡rαn2+isin⁡rαn2z_{r} = \cos\frac{r\alpha}{n^{2}} + i\sin\frac{r\alpha}{n^{2}}⇒ z1=cos⁡αn2+isin⁡αn2z_{1} = \cos\frac{\alpha}{n^{2}} + i\sin\frac{\alpha}{n^{2}};

z2=cos⁡2αn2+isin⁡2αn2z_{2} = \cos\frac{2\alpha}{n^{2}} + i\sin\frac{2\alpha}{n^{2}};………………..

⇒ zn=cos⁡nαn2+isin⁡nαn2z_{n} = \cos\frac{n\alpha}{n^{2}} + i\sin\frac{n\alpha}{n^{2}} ⇒ lim⁡n→∞(z1,z2,z3,.....zn)=lim⁡n→∞[cos⁡{αn2(1+2+3+.....+n)}+isin⁡{αn2(1+2+3+.....+n)}]\lim_{n \rightarrow \infty}(z_{1},z_{2},z_{3},.....z_{n}) = \lim_{n \rightarrow \infty}\left\lbrack \cos\left\{ \frac{\alpha}{n^{2}}(1 + 2 + 3 + ..... + n) \right\} + i\sin\left\{ \frac{\alpha}{n^{2}}(1 + 2 + 3 + ..... + n) \right\} \right\rbrack =lim⁡n→∞[cos⁡αn(n+1)2n2+isin⁡αn(n+1)2n2]= \lim_{n \rightarrow \infty}\left\lbrack \cos\frac{\alpha n(n + 1)}{2n^{2}} + i\sin\frac{\alpha n(n + 1)}{2n^{2}} \right\rbrack =cos⁡α2+isin⁡α2=eiα/2= \cos\frac{\alpha}{2} + i\sin\frac{\alpha}{2} = e^{i\alpha/2}.