MHT CET · Maths · Complex Number
The real part of the principle value of \(2^{-i}\) is
- A \(\sin (\log 2)\)
- B \(\cos \left(\frac{1}{\log 2}\right)\)
- C \(\cos \left[\log \left(\frac{1}{2}\right)\right]\)
- D \(\cos (\log 2)\)
Answer & Solution
Correct Answer
(C) \(\cos \left[\log \left(\frac{1}{2}\right)\right]\)
Step-by-step Solution
Detailed explanation
Let \(z=2^{-i}\)
Taking \(\log\) on both sides, we get
\( \log z=\log \left(2^{-i}\right) \)
\( \Rightarrow \log z=-i \log 2 \)
\( \Rightarrow \log z=i \log \left(\frac{1}{2}\right) \)
\( \Rightarrow z=e^{i \log (1 / 2)} \quad\left(\because e^{i \theta}=\cos \theta+i \sin \theta\right) \)
\( \Rightarrow z=\cos \left(\log \frac{1}{2}\right)+i \sin \left(\log \frac{1}{2}\right) \)
\( \therefore \text { The real part of } z=2^{-i} \text { is } \cos \left(\log \frac{1}{2}\right) \)
Taking \(\log\) on both sides, we get
\( \log z=\log \left(2^{-i}\right) \)
\( \Rightarrow \log z=-i \log 2 \)
\( \Rightarrow \log z=i \log \left(\frac{1}{2}\right) \)
\( \Rightarrow z=e^{i \log (1 / 2)} \quad\left(\because e^{i \theta}=\cos \theta+i \sin \theta\right) \)
\( \Rightarrow z=\cos \left(\log \frac{1}{2}\right)+i \sin \left(\log \frac{1}{2}\right) \)
\( \therefore \text { The real part of } z=2^{-i} \text { is } \cos \left(\log \frac{1}{2}\right) \)
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