TS EAMCET · Maths · Complex Number
If \(\omega\) is a complex cube root of unity, then \(\sum_{x=1}^{10}\left((\omega x+2)\left(\omega^2 x+2\right)-3\right)\)
- A 285
- B 945
- C 1025
- D 705
Answer & Solution
Correct Answer
(A) 285
Step-by-step Solution
Detailed explanation
As, \(\omega\) is complex root of unity, then \(1+\omega+\omega^2=0, \omega^3=1\) \(\therefore \quad(\omega x+2)\left(\omega^2 x+2\right)-3\) \(=\omega^3 x^2+2 \omega x+2 \omega^2 x+4-3\) \(=x^2+2 x\left(\omega+\omega^2\right)+1 \quad\left(\because \omega^3=1\right)\)…
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