Question

Simplify:\[\iota \log \left( \frac{x-i}{x+i} \right)\]

Answer 1

@Callisto

Answer 2

@ganeshie8

Answer 3

.

Answer 4

\[\Huge *\]

Answer 5

ok so we need to find the logarithm of a complex number of the form a+ib

we can write \[a+ib=r(\cos \theta+i \sin \theta) =re^\left( i \theta \right)\] where\[r=\sqrt{a^2+b^2}\]\[\theta=\tan^{-1} \frac{ b }{ a }\]

so we have
\[a+ib=re^\left( i \theta \right)\]take log on both sides
\[\log(a+ib)=\log(re^\left( i \theta \right))=\log(r)+\log(e^\left( i \theta \right))\]i took the base of logarithm as e :)
so we get\[\log_{e}(a+ib)=\log_{e}(r)+i \theta=\log_{e}(\sqrt{a^2+b^2})+i \tan^{-1} \left( \frac{ b }{ a } \right)\]


now we have
i log((x-i)/(x+i))
=i[log(x-i)-log(x+i)]

now we apply the results which we got above^
\[i[\log_{e}\sqrt{x^2+(-1)^2}]+i \tan^{-1} \left( \frac{ -1 }{ x} \right)-i[\log_{e}\sqrt{x^2+1^2]}-i \tan^{-1} \left( \frac{ 1 }{ x } \right)\]

=\[i[\tan^{-1} \left( \frac{ -1 }{ x } \right)-\tan^{-1} \left( \frac{ 1 }{ x} \right)]\]

:)

Answer 6

wait after the second last step i forgot to write the i which was outside the bracket so multiply that too

\[i \left[ i \left[ \tan^{-1} \left( \frac{ -1 }{ x} \right)-\tan^{-1} \left( \frac{ 1 }{ x} \right) \right] \right]\]

Answer 7

bravo!


\[= 2 \cot^{-1}x\]

Answer 8

(:

Answer 9

Nice one @imqwerty

Answer 10

haha thanks :)