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1. rvc

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

2. 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?

3. rvc

nope i did not start with that i assumed re^ix=a+ib re^{-ix}=a-ib

4. hartnn

wonderful start! using that substitution, what will be tan x ?

5. hartnn

the left hand side is getting simplified nicely, $$\tan (i \log e^{(2ix)})$$ got upto this point?

6. rvc

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

7. rvc

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

8. hartnn

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

9. rvc

tan^2x

10. hartnn

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

11. hartnn

you just need to get tan x = b/a

12. rvc

i can get that because i have a+ib

13. anonymous

Remember that $$e^{ix} = cosx +i sin x$$

14. rvc

got it :)

15. hartnn

re^(ix) = a+ib = r (cos x + i sin x) remember the conversion from cartesian to polar? r = $$\sqrt{a^2+b^2}$$ $$x = \arctan(b/a)$$

16. rvc

thanks @hartnn

17. hartnn

welcome ^_^