Complex Numbers And Quadratic Equations question 709
Question: Let $ z={\log_2}(1+i), $ then $ (z+\bar{z})+i(z-\bar{z})= $
Options:
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} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ z=log_2( 1+i )=log_2( \sqrt{2}{e^{i\pi /4}} ) $ $ =\frac{1}{2}+i\frac{\pi }{4}{\log_2}e $
$ \therefore z+\bar{z}=1andz-\bar{z}=i\frac{\pi }{2}{\log_2}e $ Hence, $ ( z+\bar{z} )+i( z-\bar{z} ) $ $ =1-\frac{\pi }{2}{\log_2}e=1-\frac{\pi }{2\ln 2}=\frac{\ln 4-\pi }{\ln 4} $
BETA
