Show that ... arctan(1/v) = (pi/2) - arctan(v)

[Calixto]

Kudos to whoever can explain this!

The original question was... how can I show that ... arctan(1/v) = (pi/2) - arctan(v) ?

I understand how to do this the easy way... by forming a right triangle and so on and so forth...

But could someone please explain to me what this is about? Mhill posted this and said it would work, but I don't understand...if you use the log representation for artan (1/x) and artan (x) so { artan(x)= (2i)^{-1}(log(1+ix)-log(1-ix)) }

and the same replacing x--> 1/x you

get the accurate result.

 

[Dick]

If you understand the trig way to do this, then you don't need to do it using complex variables. If you do want to do it that way you'll want to be careful about where the branch cuts are for your definition of log. If you don't know what a branch cut is then don't do it. It's needlessly complicated.