rvc
  • rvc
Please help :)
Mathematics
schrodinger
  • schrodinger
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rvc
  • rvc
\[\rm Show~that~: \tan^{-1}i(\frac{ x-a }{ x+a })=\frac{ i }{ 2 }\log \frac{ x }{ a }\]
ganeshie8
  • ganeshie8
is \(x\) a complex number ?
rvc
  • rvc
i think so no it is not mentioned in the question

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hartnn
  • hartnn
its not, i recently solved same/similar question...trying to recollect what i had done :P
rvc
  • rvc
do we have formula for \[2\tan^{-1} x\]
hartnn
  • hartnn
one long way to do it is: write \(\tan^{-1} iA = i\tanh^{-1}A \) and use the formula for \(\tanh^{-1}A\) \(\tanh^{-1}p = (1/2) \ln (\dfrac{1+p}{1-p})\)
rvc
  • rvc
hartnn
  • hartnn
A(or p) = (x-a)/(x+a) (1+p)/(1-p) indeed gives you x/a
rvc
  • rvc
what about i?
hartnn
  • hartnn
\(\tan^{-1} iA = i\tanh^{-1}A\)
anonymous
  • anonymous
Have you studied of Hyperbolic Functions yet?
rvc
  • rvc
yep
rvc
  • rvc
got you @hartnn
anonymous
  • anonymous
Then, yes, you can do like what hartnn wrote above...
rvc
  • rvc
yay!!!!!!! solved :) @hartnn thanks
hartnn
  • hartnn
welcome ^_^
IrishBoy123
  • IrishBoy123
bit late, but alternative approach: let \(\large \theta = tan^{-1} i (\frac{x-a}{x+a})\) |dw:1439128404910:dw| \(\large e^{i\theta} = cis \ \theta \implies \theta = i \ ln \frac {1}{cis \ \theta}\) from \(\large tan \theta = \frac{i \ (x-a)}{x+a}\) \(\large cos \theta = \frac{x+a}{\sqrt{4ax}}\) \(\large sin \theta = \frac{i \ (x-a)}{\sqrt{4ax}}\) \(\large cis \ \theta = \frac{2a}{ \sqrt{4ax} } = \sqrt{\frac{a}{x}} \) \(\large \theta = i \ ln \sqrt{\frac{x}{a}} = \frac{i}{2} \ ln \frac{x}{a}\)
rvc
  • rvc
thanks @IrishBoy123 :)
anonymous
  • anonymous
Very nice approach @IrishBoy123 ..

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