If $\frac{(a+i)^2}{2 a-1}=p+i q$, then $p^2+q^2$ is equal to

$\frac{\left(a^2+1\right)^2}{2 a^2+1}$

$\frac{\left(a^2+1\right)^2}{2 a^2-1}$

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

\frac{(a+i)^2}{2 a-1}=p+i q

, then

p^2+q^2

is equal to

View answer and explanation

Correct option: C

\frac{\left(a^2+1\right)^2}{4 a^2-1}

Option C,

\frac{\left(a^2+1\right)^2}{4 a^2-1}

, is the correct answer because it results from multiplying the complex conjugate of the given expression.

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