A) 0
B) 2
C) 4
D) \[-2\]
Correct Answer: B
Solution :
\[z=i\log (2-\sqrt{3})\] \[\Rightarrow \] \[{{e}^{-iz}}={{e}^{{{i}^{2}}\log (2-\sqrt{3})}}\] \[={{e}^{-\log (2-\sqrt{3})}}\] \[={{e}^{\log {{(2-\sqrt{3})}^{-1}}}}\] \[\Rightarrow \] \[{{e}^{iz}}={{(2-\sqrt{3})}^{-1}}=2+\sqrt{3}\] Similarly, \[{{e}^{-iz}}=2-\sqrt{3}\] \[\because \] \[\cos z=\frac{{{e}^{iz}}+{{e}^{-iz}}}{2}\] \[=\frac{2+\sqrt{3}+2-\sqrt{3}}{2}=2\]
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