Question

If \[\frac{( a^2 + 1 )^2}{2a - i} = x + iy, \text { then } x^2 + y^2\] is equal to

पर्याय

• \[\frac{( a^2 + 1 )^4}{4 a^2 + 1}\]

• \[\frac{(a + 1 )^2}{4 a^2 + 1}\]

• \[\frac{( a^2 - 1 )^2}{(4 a^2 - 1 )^2}\]

• none of these

Answer 1

\[\frac{( a^2 + 1 )^4}{4 a^2 + 1}\]

\[x + iy = \frac{\left( a^2 + 1 \right)^2}{2a - i}\]

Taking modulus on both the sides, we get:

`sqrt(x^2 +y^2) = ((a^2+1)^2)/(sqrt(4a^2+1))`

\[\text { Squaring both sides, we get,} \]

\[ x^2 + y^2 = \frac{\left( a^2 + 1 \right)^4}{4 a^2 + 1}\]