Question

If b}{c + id} " alt=" \frac{a + ib}{c + id} " align="absmiddle" class="latex-formula">
is purely real complex number then prove that: ad=bc​

Answer 1

Rewrite the given number as

$\dfrac{a+ib}{c+id}=\dfrac{(a+ib)(c-id)}{(c+id)(c-id)}=\dfrac{ac+bd+i(bc-ad)}{c^2+d^2}$

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

$\dfrac{bc-ad}{c^2+d^2}=0\implies bc-ad=0\implies\boxed{ad-bc}$

as required.