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rvc
 one year ago
Complex Numbers.
Please help.
rvc
 one year ago
Complex Numbers. Please help.

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rvc
 one year ago
Best ResponseYou've already chosen the best response.3\[\rm Prove~that~: \tan\large ({ilog \large \frac{\large aib }{\large a+ib}})=\large\frac{2ab}{a^2b^2}\]

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3ok, 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?

rvc
 one year ago
Best ResponseYou've already chosen the best response.3nope i did not start with that i assumed re^ix=a+ib re^{ix}=aib

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3wonderful start! using that substitution, what will be tan x ?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3the left hand side is getting simplified nicely, \(\tan (i \log e^{(2ix)}) \) got upto this point?

rvc
 one year ago
Best ResponseYou've already chosen the best response.3\[\tan(ilog(e^{2ix}))>\tan(i(2ix))>\tan(2x)\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.3\[\tan(2x)=\frac{2tanx}{1tanx}\]

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3yep, to get tan 2x, we will need tan x from the substitution, re^(ix) = a+ib

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3go on, i think you'll solve it all by yourself!

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3you just need to get tan x = b/a

rvc
 one year ago
Best ResponseYou've already chosen the best response.3i can get that because i have a+ib

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Remember that \(e^{ix} = cosx +i sin x \)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3re^(ix) = a+ib = r (cos x + i sin x) remember the conversion from cartesian to polar? r = \(\sqrt{a^2+b^2}\) \(x = \arctan(b/a)\)
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