tan ( i log (a-ib/a+ib) ) is equal to, in terms of a& b where i is iota.. ?

Say a = r cos C , b= r sin C where C is some angle
tan C = a/b
a -ib/a+ib = cos C - i sin C / cos C + i sin C = ( cos c - i sin c )^ 2 = cos 2C - i sin 2C = e^(-2Ci)
Therefore, tan ( i log (e^(-2Ci) ) ) = tan (i X -2Ci ) = tan 2C = 2ab/a^2 - b^2