Prove that `tan(ilog_e((a-ib)/(a+ib)))=(2ab)/(a^2-b^2)` (where `a, b in R^+`)

Prove that tan(i log_(e)((a-ib)/(a+ib)))=(2ab)/(a^(2)-b^(2)) (where a,b in R^(+))

tan(ilog((a+ib)/(a-ib)))=

tan{ilog((a-ib)/(a+ib))}=

tan (i log ((a-ib)/(a+ib))) = (i) ab (ii) (2ab)/(a^(2) -b^(2)) (iii) (a^( 2) -b^(2))/(ab) (iv) (2ab)/(a^(2)+b^(2))

Write ((a+ib)/(a-ib))^(2)-((a-ib)/(a+ib))^(2) in the fom of x+iy

If a+ ib= 0, prove that a= b= 0 where a, b in R .

If a and b are real and i=sqrt(-1) then sin[i ln((a+ib)/(a-ib))] is equal to 1) (2ab)/(a^(2)-b^(2)) 2) (-2ab)/(a^(2)-b^(2)) 3) (2ab)/(a^(2)+b^(2)) 4) (-2ab)/(a^(2)+b^(2))

tan [ i log ((a - ib)/(a + ib )) ] is equal to : a) ab b) (2 ab)/( a ^(2) - b ^(2)) c) (a ^(2) - b ^(2))/( 2 ab) d) (2 ab)/( a ^(2) + b ^(2))

The value of Arg [i ln ((a-ib)/(a+ib))] , where a and b are real numbers, is