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Complex Numbers question

2006 · Shift 0 · Q68
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Complex Numbers question

2006 · Shift 0 · Q68

JEE MainMathematicsComplex NumbersMCQ+4 / −1
The value of ∑k=110(sin⁡2kπ11+i  cos⁡2kπ11)\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\,\,\cos {{2k\pi } \over {11}}} \right)}k=1∑10​(sin112kπ​+icos112kπ​) is :
  1. A
    i
  2. B
    1
  3. C
    - 1
  4. D
    - i
View written solutionFree

Correct answer: D

  1. Let
S=∑k=110(sin⁡2kπ11+icos⁡2kπ11).S=\sum_{k=1}^{10}\left(\sin\frac{2k\pi}{11}+i\cos\frac{2k\pi}{11}\right).S=k=1∑10​(sin112kπ​+icos112kπ​).
  1. Rewrite each term using exponentials.

Recall that

cos⁡θ+isin⁡θ=eiθ.\cos\theta+i\sin\theta=e^{i\theta}.cosθ+isinθ=eiθ.

We need

sin⁡θ+icos⁡θ.\sin\theta+i\cos\theta.sinθ+icosθ.

Observe that

sin⁡θ+icos⁡θ=i(cos⁡θ−isin⁡θ)=ie−iθ.\sin\theta+i\cos\theta=i(\cos\theta-i\sin\theta)=i e^{-i\theta}.sinθ+icosθ=i(cosθ−isinθ)=ie−iθ.

So,

sin⁡2kπ11+icos⁡2kπ11=ie−i2kπ11.\sin\frac{2k\pi}{11}+i\cos\frac{2k\pi}{11}=i e^{-i\frac{2k\pi}{11}}.sin112kπ​+icos112kπ​=ie−i112kπ​.

Hence

S=i∑k=110e−i2kπ11.S=i\sum_{k=1}^{10}e^{-i\frac{2k\pi}{11}}.S=ik=1∑10​e−i112kπ​.
  1. Now use the sum of 11th roots of unity.

Let

ω=e−i2π11.\omega=e^{-i\frac{2\pi}{11}}.ω=e−i112π​.

Then

S=i∑k=110ωk.S=i\sum_{k=1}^{10}\omega^k.S=ik=1∑10​ωk.

Since ω11=1\omega^{11}=1ω11=1 and ω≠1\omega\neq 1ω=1,

1+ω+ω2+⋯+ω10=0.1+\omega+\omega^2+\cdots+\omega^{10}=0.1+ω+ω2+⋯+ω10=0.

Therefore,

∑k=110ωk=−1.\sum_{k=1}^{10}\omega^k=-1.k=1∑10​ωk=−1.

So,

S=i(−1)=−i.S=i(-1)=-i.S=i(−1)=−i.
  1. Check options.
  • A: iii
  • B: 111
  • C: −1-1−1
  • D: −i-i−i

Thus the correct option is

−i.\boxed{-i}.−i​.
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