If (a + ib)/(c + id) is purely real complex number then prove that: ad=bc​

Question

If


`(a + ib)/(c + id)`
is purely real complex number then prove that: ad=bc​

Answer 1

Rewrite the given number as

`(a+ib)/(c+id)=((a+ib)(c-id))/((c+id)(c-id))=(ac+bd+i(bc-ad))/(c^2+d^2)`

If it's purely real, then the complex part should be 0, so that

`(bc-ad)/(c^2+d^2)=0\implies bc-ad=0\implies\boxed{ad-bc}`

as required.