Question

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

Answer 1

the value is log(i) is correct answer

Answer 2

Rewriting the equation using the logarithmic laws we have:

Log([1+i]/[1-i])

We need to solve:

(1+i)/(1-i).....this is a complex number thus is solved specially by rationalization.

Let Z=(1+i) /(1-i)

Rationalizing the same we have:

{(1+i)/(1-i)} × {(1+i)/(1+i)}

This gives:

(1+i)^2/{(1)^2-(i)^2}


Using (a+b) ^2=a^2+b^2+2ab

We have:

{(1)^2+(i)^2+2i}/(1)^2-(i)^2

Setting i^2=-1

We get {1-1+2i}/{1+1}

Which gives:

2i/2=i

Z=0+i

Thus the solution is

Log (i)