If Zr-sin2 - r-11-i cos2 - r-11 then -r-010 Z r -A- -1B- 0C- iD- - i

The correct option is B 0Z_r = sin2 π r11 i cos2 π r11 = i(cos2 π r11+i sin2 π r11) = ie^ikr, (k=2π11) nbsp; Since, e^ikr is the 11th root of unity. There ...

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

Byju's Answer

Standard XII

Mathematics

Complex Numbers

If Zr=sin2 π ...

Question

If $Z_{r} = sin \frac{2 π r}{11} - i cos \frac{2 π r}{11}$ then $10 \sum r = 0 Z_{r} =$

- 1

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

0

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

i

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

- i

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

Open in App

Solution

The correct option is B $0$
$Z_{r} = sin \frac{2 π r}{11} - i cos \frac{2 π r}{11}$
$= - i (cos \frac{2 π r}{11} + i sin \frac{2 π r}{11})$
$= - i e^{i k r}, (k = \frac{2 π}{11})$
Since, $e^{i k r}$ is the $11 th$ root of unity.
Therefore, sum of roots is zero.

$\Rightarrow 10 \sum r = 0 Z_{r} = - i [10 \sum r = 0 e^{i k r}]$
$= - i \times 0 = 0$

Suggest Corrections

Similar questions

Q. If

Z_{r} = sin \frac{2 π r}{11} - i cos \frac{2 π r}{11}

then

\sum_{r = 0}^{10} Z_{r} =

Q. Let

C_{n} = \frac{1}{n} \int \frac{1}{n + 1} \frac{{tan}^{- 1} (n x)}{{sin}^{- 1} (n x)} d x .

Then

lim n \to \infty n^{2} . C_{n}

equals

Q. The value of

\int_{0}^{1} \tan^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x,

is

(a) 1
(b) 0
(c) −1
(d) π/4

Q. l, m, n are the

p^{t h}, q^{t h}

and

r^{t h}

term of a G.P., all positive, then

∣ ∣ ∣ ∣ \begin{matrix} l o g l & p & 1 l o g m & q & 1 l o g n & r & 1 \end{matrix} ∣ ∣ ∣ ∣

equals

Find the value of $x$ such that $\frac{- 11}{x}$ is not a rational number.

Join BYJU'S Learning Program

If Zr=sin2πr11−icos2πr11 then 10∑r=0Zr=

The correct option is B 0Zr=sin2πr11−icos2πr11 =−i(cos2πr11+isin2πr11) =−ieikr, (k=2π11) Since, eikr is the 11th root of unity. Therefore, sum of roots is zero. ⇒10∑r=0Zr=−i[10∑r=0eikr] =−i×0=0

If $Z_{r} = sin \frac{2 π r}{11} - i cos \frac{2 π r}{11}$ then $10 \sum r = 0 Z_{r} =$