Question
Let $$z = {\log _2}\left( {1 + i} \right),$$ then $$\left( {z + \bar z} \right) + i\left( {z - \bar z} \right) = $$
A.
$$\frac{{\ln 4 + \pi }}{{\ln 4}}$$
B.
$$\frac{{\pi - \ln 4}}{{\ln 2}}$$
C.
$$\frac{{\ln 4 - \pi }}{{\ln 4}}$$
D.
$$\frac{{\pi + \ln 4}}{{\ln 2}}$$
Answer :
$$\frac{{\ln 4 - \pi }}{{\ln 4}}$$
Solution :
$$\eqalign{
& z = {\log _2}\left( {1 + i} \right) \cr
& = {\log _2}\left( {\sqrt 2 {e^{\frac{{i\pi }}{4}}}} \right) \cr
& = \frac{1}{2} + i\frac{\pi }{4}{\log _2}e \cr
& \therefore z + \bar z = 1{\text{ and }}z - \bar z = i\frac{\pi }{2}{\log _2}e \cr
& {\text{Hence, }}\left( {z + \bar z} \right) + i\left( {z - \bar z} \right) \cr
& = 1 - \frac{\pi }{2}{\log _2}e \cr
& = 1 - \frac{\pi }{{2\ln 2}} \cr
& = \frac{{\ln 4 - \pi }}{{\ln 4}} \cr} $$