AP EAMCET · Maths · Complex Number
If \(\omega\) is a complex cube root of unity, then \(\omega^{\left(\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots \infty\right)}+\omega^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\ldots \infty\right)}\) is equal to
- A \(1\)
- B \(-1\)
- C \(\omega\)
- D \(i\)
Answer & Solution
Correct Answer
(B) \(-1\)
Step-by-step Solution
Detailed explanation
We have, \(\omega^{\left(\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots . \infty\right)}+\omega^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\ldots . \infty\right)}=\) ? Here, \(\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots \infty\) Follows infinite G.P. series so its sum,…
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