Show that tan(i log a-ib/a+ib)=2ab/a^2-b^2

We have to prove that: tan [i*log (a-ib/a+ib)] = 2ab/(a^2
- b^2)

Start with denoting a + ib in the form r*( cos x + i
sin x)

Equate the real and complex components to get r*cos
x = a and r*sin x = b

tan x = sin x / cos x =
b/a

In the same notation a – ib = r( cos x – i*sin
x)

The left hand side of what we have to prove
is:

tan [i*log
(a-ib/a+ib)]

=> tan [ i*log (r( cos x – i*sin x)/
r*( cos x + i sin x))]

we can write cos x – i*sin x as
e^(-ix) and cos x + i*sin x as e^(ix)

=> tan[ i*log
(e^(-ix)/e^(ix))]

=> tan[ i*log
(e^(-2ix))]

=> tan[i (-2ix) log
e]

log e = 1

=> tan [
-2*i^2*x]

=> tan(
-2*-1*x)

=> tan
2x

expanding tan 2x

=>
2 tan x / 1 – (tan x)^2

=> [2*(b/a)] / [ 1 –
(b/a)^2]

=> 2ab / a^2 +
b^2

which is the right hand
side.

Therefore we prove that  tan [i*log
(a-ib/a+ib)] = 2ab/(a^2 -
b^2)