Question

If $\frac{(a+ib)}{(c+id)}=x+iy$

Show that,
i) $\frac{a-ib}{c-id}=x-iy$
ii)$\frac{a^2+b^2}{c^2+d^2}=x^2+y^2$

Answer 1

Consider the given that,

$\frac{(a+ib)}{(c+id)}=x+iy........\left(1\right)$

Part (1)

prove that, $\frac{(a-ib)}{(c-id)}=x-iy$

Proof:- By equation (1),

Take conjugate and we get,

$\overline{\frac{(a+ib)}{(c+id)}}=\overline{x+iy}$

$\Rightarrow \frac{\overline{(a+ib)}}{\overline{(c+id)}}=\overline{x+iy}$

$\Rightarrow \frac{a-ib}{c-id}=x-iyHenceproved.......\left(2\right)$

Part (2),

Multiplying (1) and (2) to and we get,

$\frac{(a+ib)}{(c+id)}\times \frac{(a-ib)}{(c-id)}=(x+iy)(x-iy)$

$\frac{a^2-i^2{b}^2}{c^2-i^2{d}^2}=x^2-i^2{y}^2$

$\frac{a^2+b^2}{c^2+d^2}=x^2+y^{2}Henceproved.$