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jiteshmeghwal9
 one year ago
Simplify:\[\iota \log \left( \frac{xi}{x+i} \right)\]
jiteshmeghwal9
 one year ago
Simplify:\[\iota \log \left( \frac{xi}{x+i} \right)\]

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imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4ok 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((xi)/(x+i)) =i[log(xi)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
 one year ago
Best ResponseYou've already chosen the best response.4wait 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
 one year ago
Best ResponseYou've already chosen the best response.1bravo! \[= 2 \cot^{1}x\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Nice one @imqwerty
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