A) \[\frac{\ln \,4+\pi }{\ln \,\,4}\]
B) \[\frac{\pi -\ln \,4}{\ln \,\,2}\]
C) \[\frac{\ln \,4-\pi }{\ln \,\,4}\]
D) \[\frac{\pi +\ln \,\,4}{\ln \,\,2}\]
Correct Answer: C
Solution :
\[\operatorname{z}=lo{{g}_{2}}\left( 1+i \right)=lo{{g}_{2}}\left( \sqrt{2}{{e}^{i\pi /4}} \right)\] \[=\,\,\,\frac{1}{2}+i\frac{\pi }{4}{{\log }_{2}}e\] \[\therefore \,\,\,z+\bar{z}=1\,\,and\,\,z-\bar{z}=i\frac{\pi }{2}{{\log }_{2}}e\] Hence, \[\left( z+\bar{z} \right)+i\left( z\,-\bar{z} \right)\] \[=\,\,1-\frac{\pi }{2}{{\log }_{2}}e=1-\frac{\pi }{2\,\,\ln 2}=\frac{\ln \,\,4-\pi }{\ln \,\,4}\]
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