If z -cos r - n 2- i sin r - n 2- where r -1-2-3- - n- then n lim- z 1 z 2 z 3- z n is equal toA- cos-i sin-B- cos-2-i sin-2C- ei a-2D- -ei a

Z_r=cosrα/n^2+ i sinrα/n^2 z_1=cosα/n^2+i sinα/n^2;z_2=cos2α/n^2+i sin2α/n^2;..... ⇒ z_n=cosnα/n^2+i nα/n^2⇒limn →∞ (z_1z_2z_3.....z_n) = limn →∞ [cos {α/n^2 ( ...

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Byju's Answer

Standard XII

Mathematics

Distance Formula Using Complex Numbers

if z ,=cos r ...

Question

$i f z, = c o s \frac{r α}{n^{2}} + i s i n \frac{r α}{n^{2}}, w h e r e r = 1, 2, 3, . . . . n, t h e n l i m n \to \infty z_{1} z_{2} z_{3} . . . . . z_{n} i s e q u a l t o$

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Solution

The correct option is C
$z_{r} = c o s \frac{r α}{n^{2}} + i s i n \frac{r α}{n^{2}} z_{1} = c o s \frac{α}{n^{2}} + i s i n \frac{α}{n^{2}}; z_{2} = c o s \frac{2 α}{n^{2}} + i s i n \frac{2 α}{n^{2}}; . . . . . \Rightarrow z_{n} = c o s \frac{n α}{n^{2}} + i \frac{n α}{n^{2}} \Rightarrow l i m n \to \infty (z_{1} z_{2} z_{3} . . . . . z_{n})$
$= l i m n \to \infty [c o s {\frac{α}{n^{2}} (1 + 2 + 3 + . . . . + n)} + i s i n {\frac{α}{n^{2}} (1 + 2 + 3 + . . . . + n)}] = l i m n \to \infty [c o s \frac{α b (n + 1)}{2 n^{2}} + i s i n \frac{α n (n + 1)}{2 n^{2}}] = c o s \frac{α}{2} + i s i n \frac{α}{2} = e^{\frac{i a}{2}}$ .

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if z,=cosrαn2+i sinrαn2,where r=1,2,3,....n,then limn→∞ z1z2z3.....zn is equal to

The correct option is C zr=cosrαn2+ i sinrαn2z1=cosαn2+i sinαn2;z2=cos2αn2+i sin2αn2;.....⇒zn=cosnαn2+i nαn2⇒limn→∞ (z1z2z3.....zn) =limn→∞[cos{αn2(1+2+3+....+n)}+i sin{αn2(1+2+3+....+n)}]=limn→∞[cos αb(n+1)2n2+i sin αn(n+1)2n2]=cosα2+i sinα2=eia2.

$i f z, = c o s \frac{r α}{n^{2}} + i s i n \frac{r α}{n^{2}}, w h e r e r = 1, 2, 3, . . . . n, t h e n l i m n \to \infty z_{1} z_{2} z_{3} . . . . . z_{n} i s e q u a l t o$