rvc
  • rvc
Complex Numbers. Please help.
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

rvc
  • rvc
\[\rm Prove~that~: \tan\large ({ilog \large \frac{\large a-ib }{\large a+ib}})=\large\frac{2ab}{a^2-b^2}\]
hartnn
  • hartnn
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?
rvc
  • rvc
nope i did not start with that i assumed re^ix=a+ib re^{-ix}=a-ib

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

hartnn
  • hartnn
wonderful start! using that substitution, what will be tan x ?
hartnn
  • hartnn
the left hand side is getting simplified nicely, \(\tan (i \log e^{(2ix)}) \) got upto this point?
rvc
  • rvc
\[\tan(ilog(e^{-2ix}))-->\tan(i(-2ix))-->\tan(2x)\]
rvc
  • rvc
\[\tan(2x)=\frac{2tanx}{1-tanx}\]
hartnn
  • hartnn
yep, to get tan 2x, we will need tan x from the substitution, re^(ix) = a+ib
rvc
  • rvc
tan^2x
hartnn
  • hartnn
go on, i think you'll solve it all by yourself!
hartnn
  • hartnn
you just need to get tan x = b/a
rvc
  • rvc
i can get that because i have a+ib
anonymous
  • anonymous
Remember that \(e^{ix} = cosx +i sin x \)
rvc
  • rvc
got it :)
hartnn
  • hartnn
re^(ix) = a+ib = r (cos x + i sin x) remember the conversion from cartesian to polar? r = \(\sqrt{a^2+b^2}\) \(x = \arctan(b/a)\)
rvc
  • rvc
thanks @hartnn
hartnn
  • hartnn
welcome ^_^

Looking for something else?

Not the answer you are looking for? Search for more explanations.