Question

if $\frac{a+ib}{c+id}=p+iq$ , prove that
i) $\frac{a-ib}{c-id}=p-iq$
ii) $p^2+q^2=\frac{a^2+b^2}{c^2+d^2}$

Answer 1

Solution-

$\frac{a+ib}{c+id}=p+iq$

$\frac{(a+ib)(c-id)}{c^2+d^2}=p+iq$

$\frac{(ac+bd)+i(bc-ad)}{c^2+d^2}=p+iq$

$P=\frac{ac+bd}{c^2+d^2}q=\frac{bc-ad}{c^2+d^2}$

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

$=p=iq$

also $p^2+q^2=\frac{(ac+bd{)}^2+(bc-ad{)}^2}{(c^2+d^2{)}^2}$

$=\frac{a^2{c}^2+b^2+d^2+2abcd+b^2{c}^2+a^2{d}^2-2abcd}{(c^2+d^2{)}^2}$

$=\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)}=\frac{a^2+b^2}{c^2+d^2}$