The value of - k -1100-sin2 - k -101- i cos2 - k -101- isA- iB- - iC- 1D 1- i

S= nbsp; ∑_k=1^100[sin(2π k101) icos(2π k101)] = i∑_k=1^100[cos(2π k101) +isin(2π k101)] = i ∑_k=1^100e^2π ki/101 nbsp; = i∑_k=1^100z^k where z=e^2π i/101 ...

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Standard XII

Mathematics

Euler's Representation

The value of ...

Question

The value of $100 \sum k = 1 [sin (\frac{2 π k}{101}) - i cos (\frac{2 π k}{101})]$ is

i

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Solution

The correct option is A $i$
$S = 100 \sum k = 1 [sin (\frac{2 π k}{101}) - i cos (\frac{2 π k}{101})]$

$= - i 100 \sum k = 1 [cos (\frac{2 π k}{101}) + i sin (\frac{2 π k}{101})]$

$= - i 100 \sum k = 1 e^{\frac{2 π k i}{101}}$
$= - i 100 \sum k = 1 z^{k}$ where $z = e^{\frac{2 π i}{101}}$
$= - i (z + z^{2} + z^{3} + \dots + z^{100})$
$= - i \frac{z (z^{100} - 1)}{z - 1}$
$= - i (\frac{z^{101} - z}{z - 1})$

$Now, z^{101} = e^{(\frac{2 π i}{101} \times 101)} = e^{2 π i} = 1$
$∴ S = - i \frac{(1 - z)}{z - 1} = i$

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The value of 100∑k=1[sin(2πk101)−icos(2πk101)] is

The correct option is A i S=100∑k=1[sin(2πk101)−icos(2πk101)] =−i100∑k=1[cos(2πk101)+isin(2πk101)] =−i100∑k=1e2πki101 =−i100∑k=1zk where z=e2πi101 =−i(z+z2+z3+⋯+z100) =−iz(z100−1)z−1 =−i(z101−zz−1) Now, z101=e(2πi101×101)=e2πi=1 ∴S=−i(1−z)z−1=i

The value of $100 \sum k = 1 [sin (\frac{2 π k}{101}) - i cos (\frac{2 π k}{101})]$ is