Question

Complex Numbers.
Please help.

Answer 1

\[\rm Prove~that~: \tan\large ({ilog \large \frac{\large a-ib }{\large a+ib}})=\large\frac{2ab}{a^2-b^2}\]

Answer 2

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?

Answer 3

nope
i did not start with that

i assumed re^ix=a+ib
re^{-ix}=a-ib

Answer 4

wonderful start!

using that substitution, what will be tan x ?

Answer 5

the left hand side is getting simplified nicely,

\(\tan (i \log e^{(2ix)}) \)

got upto this point?

Answer 6

\[\tan(ilog(e^{-2ix}))-->\tan(i(-2ix))-->\tan(2x)\]

Answer 7

\[\tan(2x)=\frac{2tanx}{1-tanx}\]

Answer 8

yep, to get tan 2x,
we will need tan x from the substitution,
re^(ix) = a+ib

Answer 9

tan^2x

Answer 10

go on, i think you'll solve it all by yourself!

Answer 11

you just need to get
tan x = b/a

Answer 12

i can get that because i have a+ib

Answer 13

Remember that \(e^{ix} = cosx +i sin x \)

Answer 14

got it :)

Answer 15

re^(ix) = a+ib = r (cos x + i sin x)

remember the conversion from cartesian to polar?

r = \(\sqrt{a^2+b^2}\)
\(x = \arctan(b/a)\)

Answer 16

thanks @hartnn

Answer 17

welcome ^_^