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Q : 4       If     \small x-iy=\sqrt{\frac{a-ib}{c-id}}  ,   prove that     \small (x^2+y^2)^2=\frac{a^2+b^2}{c^2+d^2}.

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 the given problem is 

\small x-iy=\sqrt{\frac{a-ib}{c-id}}
Now, multiply the numerator and denominator by

 \sqrt{c+id}
x-iy = \sqrt{\frac{a-ib}{c-id}\times \frac{c+id}{c+id}}
               = \sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2-i^2d^2}}= \sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}}                                
Now, square both the sides 
(x-iy)^2=\left ( \sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}} \right )^2
                    =\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}
x^2-y^2-2ixy=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}
On comparing the real and imaginary part, we obtain

x^2-y^2 = \frac{ac+bd}{c^2+d^2} \ \ and \ \ -2xy = \frac{ad-bc}{c^2+d^2} \ \ \ -(i)

Now,
(x^2+y^2)^2= (x^2-y^2)^2+4x^2y^2
                      = \left ( \frac{ac+bd}{c^2+d^2} \right )^2+\left ( \frac{ad-bc}{c^2+d^2} \right )^2 \ \ \ \ (using \ (i))
                     =\frac{a^2c^2+b^2d^2+2acbd+a^2d^2+b^2c^2-2adbc}{(c^2+d^2)^2}
                     =\frac{a^2c^2+b^2d^2+a^2d^2+b^2c^2}{(c^2+d^2)^2}
                     =\frac{a^2(c^2+d^2)+b^2(c^2+d^2)}{(c^2+d^2)^2}
                    =\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}
                   =\frac{(a^2+b^2)}{(c^2+d^2)}

Hence proved
 

Posted by

Gautam harsolia

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