If arg(a/b)= pi/2, then find the value of((a+b)/(a-b)) where a,b are complex numbers.

let a=me^ix
     b=ne^iy
then arg(a/b)= x-y = pi/2
a/b= m/ne^ipi/2= i      
now a+b/a-b =  a/b +1                  
                        _______                                   
                          a/b -1
= i+1/i-1
= (i+1)^2/2
= 2i/2= i (ANS)

Arg(a/b)=Pi/2

Tan^-1   (a/b)=   Pi/2

a/b = tanP/2 ,therefore a/b=infinity.

Or b=0

As per your question a+b/a-b= 1

As a and b are complex numbers the answer is “i”

sorry a little mistake,
 you can solve these sums by hit and trial,
let a=i^2 b=i
such that arg(a/b)=pi/2
now a+b/a-b=i^2+1/i^2-1=i+1/i-1=2i / -2= -i (ANS)