TS EAMCET · Maths · Complex Number
\(\left(\frac{1+i}{1-i}\right)^{228}=\)
- A \(-4\left(\frac{1-i}{1+i}\right)^{226}\)
- B \(4\left(\frac{1-i}{1+i}\right)^{226}\)
- C \(\left(\frac{1-i}{1+i}\right)^{228}\)
- D \(-\left(\frac{1-i}{1+i}\right)^{228}\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{1-i}{1+i}\right)^{228}\)
Step-by-step Solution
Detailed explanation
\(\left(\frac{1+i}{1-i}\right)^{228} = \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{228} = \left(\frac{1+2i+i^2}{1-i^2}\right)^{228} = \left(\frac{1+2i-1}{1-(-1)}\right)^{228} = \left(\frac{2i}{2}\right)^{228} = i^{228}\) \(i^{228} = (i^4)^{57} = 1^{57} = 1\)…
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\(
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\)
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\begin{gathered}
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\end{gathered}
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