Question

Find the general value of log(1+i)+log(1-i)

Answer 1

We know that, log(ab)=loga+logb

Now, we need to find the value of log(1+i)+log(1-i)

By following the rule of logarithm,

$\rightarrow\tt log(1 + i) + log(1 - i)$

$\rightarrow\tt log(1 + i)(1 - i)$

$\rightarrow\tt log[(1 - i + i - {i}^{2} )]$

$\rightarrow\tt log[(1 - {i}^{2} )]$

$\rightarrow\tt log(1 + 1)$

$\rightarrow\tt log2$

$\rightarrow\tt 0.30103$

Therefore reqd. general value of log(1+i)+log(1-i) is log2 or 0.30103