Find the vertices of regular polygon of n sides if its centre is located at $z = 0$ and one of its vertices is known.

z_{k + 1} = z_{1} (c o s π k / n + i s i n π k / n), k = 1, 2, 3, . . . ., (n - 1)

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

z_{k + 1} = z_{1} (c o s 2 π k / n + i s i n 2 π k / n), k = 1, 2, 3, . . . ., (n - 1)

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

z_{k + 1} = z_{1} (- c o s π k / n + i s i n π k / n), k = 1, 2, 3, . . . ., (n - 1)

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

z_{k + 1} = z_{1} (c o s 2 π k / n - i s i n 2 π k / n), k = 1, 2, 3, . . . ., (n - 1)

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

Open in App

Solution

The correct option is B $z_{k + 1} = z_{1} (c o s 2 π k / n + i s i n 2 π k / n), k = 1, 2, 3, . . . ., (n - 1)$
The points will be $n t h$ roots of unity.
Hence the points will be given by $z^{n} = 1$
$z^{n} = e^{i 2 k π}$
Or
$z = e^{i \frac{2 k π}{n}}$
$= c o s (\frac{2 k π}{n}) + i s i n (\frac{2 k π}{n})$ where K=1,2,3...
Now these are point for a polygon of $n$ sides with unit length. In case the length is not one unit length,
$z_{k + 1} = z_{1} (c o s (\frac{2 k π}{n}) + i s i n (\frac{2 k π}{n}))$

Suggest Corrections

Similar questions

S_{n} = n \sum k = 1 \frac{n}{n^{2} + k n + k^{2}}

T_{n} = n - 1 \sum k = 0 \frac{n}{n^{2} + k n + k^{2}}

n = 1, 2, 3, . .

S_{n} = n \sum k = 1 \frac{n}{n^{2} + k n + k^{2}}

T_{n} = n - 1 \sum k = 0 \frac{n}{n^{2} + k n + k^{2}}

n = 1, 2, 3, . . . . T h e n,

z^{2} + z + 1 = 0

n

n \sum k = 1 (z^{k} + z^{- k})^{2} = n + 3 [\frac{n}{3}]

[x]

\leq x

w \neq 1

w^{k} + w^{- k} = {\begin{matrix} 0 - 1 & if k is not a multiple of 3 2 & if k is a multiple of 3 \end{matrix}

a_{k} = \frac{1}{k (k + 1)}

k = 1, 2, 3, \dots n t h e n (n \sum k = 1 a_{k})^{2}

a_{k} = \frac{1}{k (k + 1)}, f o r k = 1, 2, 3..... n, t h e n {(n \sum k = 1 a_{k})}_{k}^{2} =

Join BYJU'S Learning Program