Question

The general value of $\log (1+i)+\log (1-i)$ is

• A. $\log 2+i2k\pi ,k\in Z$
• B. $\log \sqrt{2}+i2k\pi ,k\in Z$
• C. $2\log 2+i2k\pi ,k\in Z$
• D. $\log 3+i2k\pi ,k\in Z$

Answer 1

The correct option is A $\log 2+i2k\pi ,k\in Z$

$\log (1+i)+\log (1-i)$
$=[\log \sqrt{2}+i(2\pi n+\frac{\pi }{4})]+[\log \sqrt{2}-i(2\pi m+\frac{\pi }{4})]$
$(\because 1+i=\sqrt{2}[\cos (\frac{\pi }{4}+2n\pi )+i\sin (\frac{\pi }{4}+2n\pi )]n\in Z)$
$(\because 1-i=\sqrt{2}[\cos (\frac{\pi }{4}+2m\pi )-i\sin (\frac{\pi }{4}+2m\pi )]m\in Z)$
$=2\log \sqrt{2}+i(2n\pi -2m\pi )$
$=\log 2+i2k\pi ,(\text{where}k=(n-m),k\in Z)$