jiteshmeghwal9
  • jiteshmeghwal9
Simplify:\[\iota \log \left( \frac{x-i}{x+i} \right)\]
Mathematics
jamiebookeater
  • jamiebookeater
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jiteshmeghwal9
  • jiteshmeghwal9
jiteshmeghwal9
  • jiteshmeghwal9
IrishBoy123
  • IrishBoy123
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amilapsn
  • amilapsn
\[\Huge *\]
imqwerty
  • imqwerty
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^ \[i[\log_{e}\sqrt{x^2+(-1)^2}]+i \tan^{-1} \left( \frac{ -1 }{ x} \right)-i[\log_{e}\sqrt{x^2+1^2]}-i \tan^{-1} \left( \frac{ 1 }{ x } \right)\] =\[i[\tan^{-1} \left( \frac{ -1 }{ x } \right)-\tan^{-1} \left( \frac{ 1 }{ x} \right)]\] :)
imqwerty
  • imqwerty
wait after the second last step i forgot to write the i which was outside the bracket so multiply that too \[i \left[ i \left[ \tan^{-1} \left( \frac{ -1 }{ x} \right)-\tan^{-1} \left( \frac{ 1 }{ x} \right) \right] \right]\]
IrishBoy123
  • IrishBoy123
bravo! \[= 2 \cot^{-1}x\]
imqwerty
  • imqwerty
(:
Astrophysics
  • Astrophysics
Nice one @imqwerty
imqwerty
  • imqwerty
haha thanks :)

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