If zk-cos k-10 -isin k-10 - then z1z2z3z4 -

The correct option is B 1 z_k=e^ikπ/10 (euler form) ∴ z_1× z_2× nbsp;z_3× z_4= nbsp;e^i ( π/10+2π/10+3π/10+4 π/10 ) nbsp;=e^i ( 10 π/10 ) ...

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Variable Separable Method

If zk=cos k...

Question

If $z_{k} = c o s (\frac{k π}{10}) + i s i n (\frac{k π}{10})$ , then $z_{1} z_{2} z_{3} z_{4} =$

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Solution

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k

a_{k} = cos (\frac{k π}{7}) + i sin (\frac{k π}{7})

\frac{12 \sum k = 1 | a_{k + 1} - a_{k} |}{3 \sum k = 1 | a_{4 k - 1} - a_{4 k - 2} |}

k

a_{k} = cos (\frac{k π}{7}) + i sin (\frac{k π}{7})

\frac{12 \sum k = 1 | a_{k + 1} - a_{k} |}{3 \sum k = 1 | a_{4 k - 1} - a_{4 k - 2} |}

k

α_{k} = cos (\frac{k π}{7}) + i sin (\frac{k π}{7})

i = \sqrt{- 1}

\frac{12 \sum k = 1 | α_{k + 1} - α_{k} |}{3 \sum k = 1 | α_{4 k - 1} - α_{4 k - 2} |} =

z_{1}, z_{2}, z_{3} . . . . . n_{n}

α_{k} = cos \frac{k π}{7} + i sin \frac{k π}{7},

i = \sqrt{- 1}

\frac{\sum_{k = 1}^{12} | α_{k + 1} - α_{k} |}{\sum_{k = 1}^{3} | α_{4 k - 1} - α_{4 k - 2} |}

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