`Z_1!=Z_2` are two points in an Argand plane. If `a|Z_1|=b|Z_2|,` then prove that `(a Z_1-b Z_2)/(a Z_1+b Z_2)` is purely imaginary.

Explanation

Let `Z_(1) = r_(1) e^(itheta),Z_(2) =r_(2)e^(i(theta + alpha))` Given that `ar_(1) = br_(2)`
`therefore Z = (aZ_(1) - bZ_(2))/(aZ_(1) + bZ_(2))=(e^(itheta)-e^(i(theta + alpha)))/(e^(itheta) + e^(itheta+alpha))`
`= (1-e^(ialpha))/(1+e^(ialpha))" "("Dividing Nr. and Dr. by" e^(itheta))`
`(e^(-ialpha//2)-e^(ialpha//2))/(e^(-ialpha//2) + e^(ialpha//2))" "("Dividing Nr. and Dr.by " e^(ialpha//2))`
`= (-2i sin.(alpha)/(2))/(2cos.(alpha)/(2))`
`=-i tan .(alpha)/(2)`
Hence, Z is purely imaginary.