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