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rvc

  • one year ago

Complex Numbers. Please help.

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  1. rvc
    • one year ago
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    \[\rm Prove~that~: \tan\large ({ilog \large \frac{\large a-ib }{\large a+ib}})=\large\frac{2ab}{a^2-b^2}\]

  2. hartnn
    • one year ago
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    ok, what have you tried? I would start by using tan ix = i tanh x and then use the definition of tanh x have you tried this approach?

  3. rvc
    • one year ago
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    nope i did not start with that i assumed re^ix=a+ib re^{-ix}=a-ib

  4. hartnn
    • one year ago
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    wonderful start! using that substitution, what will be tan x ?

  5. hartnn
    • one year ago
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    the left hand side is getting simplified nicely, \(\tan (i \log e^{(2ix)}) \) got upto this point?

  6. rvc
    • one year ago
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    \[\tan(ilog(e^{-2ix}))-->\tan(i(-2ix))-->\tan(2x)\]

  7. rvc
    • one year ago
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    \[\tan(2x)=\frac{2tanx}{1-tanx}\]

  8. hartnn
    • one year ago
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    yep, to get tan 2x, we will need tan x from the substitution, re^(ix) = a+ib

  9. rvc
    • one year ago
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    tan^2x

  10. hartnn
    • one year ago
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    go on, i think you'll solve it all by yourself!

  11. hartnn
    • one year ago
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    you just need to get tan x = b/a

  12. rvc
    • one year ago
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    i can get that because i have a+ib

  13. anonymous
    • one year ago
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    Remember that \(e^{ix} = cosx +i sin x \)

  14. rvc
    • one year ago
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    got it :)

  15. hartnn
    • one year ago
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    re^(ix) = a+ib = r (cos x + i sin x) remember the conversion from cartesian to polar? r = \(\sqrt{a^2+b^2}\) \(x = \arctan(b/a)\)

  16. rvc
    • one year ago
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    thanks @hartnn

  17. hartnn
    • one year ago
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    welcome ^_^

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