If z r -cos - 3 r -isin - 3 r -r-1-2-3- then z 1 - z 2 - z 3 - -

The correct option is A i Since #160; z _ r =cos( π #160;/ 3 ^ r #160;) +isin( π #160;/ 3 ^ r #160;) #160; ,r=1,2,3,... we have z _ ...

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Standard XII

Mathematics

Algebra of Complex Numbers

If z r =cos...

Question

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

i

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Solution

The correct option is A $i$
Since $z_{r} = cos (\frac{π}{3^{r}}) + i sin (\frac{π}{3^{r}}), r = 1, 2, 3, . . .$

we have $z_{1} . z_{2} . z_{3} . . . \infty$

$= (cos \frac{π}{3} + i sin \frac{π}{3}) (cos (\frac{π}{3^{2}}) + i sin (\frac{π}{3^{2}})) (cos (\frac{π}{3^{3}}) + i sin (\frac{π}{3^{3}})) . . . \infty$

$= cos (\frac{π}{3} + \frac{π}{3^{2}} + \frac{π}{3^{3}} + . . .) + i sin (\frac{π}{3} + \frac{π}{3^{2}} + \frac{π}{3^{3}} + . . .)$

$= cos ⎛ ⎝ \frac{\frac{π}{3}}{1 - \frac{1}{3}} ⎞ ⎠ + i sin ⎛ ⎝ \frac{\frac{π}{3}}{1 - \frac{1}{3}} ⎞ ⎠$

$= cos (\frac{π}{2}) + i sin (\frac{π}{2}) = 0 + i .1 = i$

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Similar questions

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

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

Q. If

A_{r} = cos \frac{π}{3^{r}} + i sin \frac{π}{3^{r}},

then the value of

A_{1} \cdot A_{2} \cdot A_{3} \cdot \dots \cdot A_{\infty}

is (where

i = \sqrt{- 1}

)

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

Q. If

A_{r} = cos \frac{π}{3^{r}} + i sin \frac{π}{3^{r}},

then the value of

A_{1} \cdot A_{2} \cdot A_{3} \cdot \dots \cdot A_{\infty}

is (where

i = \sqrt{- 1}

)

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If zr=cos(π3r)+isin(π3r),r=1,2,3,..., then z1.z2.z3...∞=

The correct option is A iSince zr=cos(π3r)+isin(π3r),r=1,2,3,...we have z1.z2.z3...∞=(cosπ3+isinπ3)(cos(π32)+isin(π32))(cos(π33)+isin(π33))...∞=cos(π3+π32+π33+...)+isin(π3+π32+π33+...)=cos⎛⎝π31−13⎞⎠+isin⎛⎝π31−13⎞⎠=cos(π2)+isin(π2)=0+i.1=i

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