If Zr-cos-2r-i sin-2r for r-1- 2- 3- - then Z1Z2Z3 - -

The correct option is D 1We have Z_r=cos(π2^r)+isin( π2^r)=e^i(π2^r) ∴ Z_1Z_2Z_3..........∞ =e^i(π2)(1+12+12^2+.......∞) = e^(i·π2·1/1 1/2) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Integration of Piecewise Continuous Functions

If Zr=cosπ/...

Question

If $Z_{r} = c o s (\frac{π}{2^{r}}) + i s i n (\frac{π}{2^{r}})$ for $r = 1, 2, 3, . . . .$ then $Z_{1} Z_{2} Z_{3} . . . . . . . \infty =$

-2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D -1
We have $Z_{r} = cos (\frac{π}{2^{r}}) + i sin (\frac{π}{2^{r}}) = e^{i (\frac{π}{2^{r}})}$
$∴ Z_{1} Z_{2} Z_{3} . . . . . . . . . . \infty = e^{i (\frac{π}{2}) ⎛ ⎝ 1 + \frac{1}{2} + \frac{1}{2^{2}} + . . . . . . . \infty ⎞ ⎠}$
$= e^{⎛ ⎜ ⎝ i \cdot \frac{π}{2} \cdot \frac{1}{1 - \frac{1}{2}} ⎞ ⎟ ⎠}$
$= e^{i π} = (cos π + i sin π) = - 1$

Suggest Corrections

Similar questions

State whether the following statement is true or false.

Z_{r} = c o s \frac{π}{3^{r}} + i s i n \frac{π}{3^{r}}, r = 1, 2, 3, . . . . . .

then

z_{1} z_{2} z_{3} . . . . . . . . \infty = i

If $z_{r}$ = cos $(\frac{π}{2^{r}})$ + i sin $(\frac{π}{2^{r}})$ where i = $\sqrt{- 1}$ . Find the value of $z_{1}$ . $z_{2}$ . $z_{3}$ . $z_{4} \dots \dots \infty$

Q. If

z_{r} = cos (\frac{π}{3^{r}}) + i sin (\frac{π}{3^{r}}), r = 1, 2, 3, . . .

, then

z_{1} . z_{2} . z_{3} . . . \infty =

x_{r} = cos \frac{π}{2^{r}} + i sin \frac{π}{2^{r}}

then

x_{1} . x_{2} . x_{3} . . . . . . \infty = - 1

Q. If

X_{r} = cos \frac{π}{2^{r}} + i sin \frac{π}{2^{r}}

, prove that

X_{1} X_{2} X_{3} . . . . \infty

= -1

Join BYJU'S Learning Program

If Zr=cos(π2r)+isin(π2r) for r=1,2,3,.... then Z1Z2Z3.......∞=

The correct option is D -1We have Zr=cos(π2r)+isin(π2r)=ei(π2r)∴Z1Z2Z3..........∞=ei(π2)⎛⎝1+12+122+.......∞⎞⎠=e⎛⎜⎝i⋅π2⋅11−12⎞⎟⎠=eiπ=(cosπ+isinπ)=−1

If $Z_{r} = c o s (\frac{π}{2^{r}}) + i s i n (\frac{π}{2^{r}})$ for $r = 1, 2, 3, . . . .$ then $Z_{1} Z_{2} Z_{3} . . . . . . . \infty =$