Question

If (a2+1)2/(2a-i)=p+iq prove that p2+q2=(a2+1)2/4a2+1

Answer 1

Here is your ANSWER

hope it helps you

Answer 2

Answer:

In the question,

We have been provided the equation,

$\frac{(a+i)^{2}}{2a-i}=p+iq$

Now,

We need to show that,

$p^{2}+q^{2}=\frac{(a^{2}+1)^{2}}{4a^{2}+1}$

Therefore, for solving that.

On taking the conjugate of the complex number, we get,

$\frac{(a-i)^{2}}{2a+i}=p-iq$

So,

On multiplying both the equations, we get,

$\left[\frac{(a+i)^{2}}{2a-i}\right]\times \left[\frac{(a-i)^{2}}{2a+i}\right]=(p+iq)\times (p-iq)\\\frac{[(a+i)(a-i)]^{2}}{4a^{2}-i^{2}}=p^{2}-(iq)^{2}\\\frac{(a^{2}-i^{2})^{2}}{4a^{2}+1}=p^{2}+q^{2}$

Now, on simplifying this equation solved, further, we get,

$\frac{(a^{2}-i^{2})^{2}}{4a^{2}+1}=p^{2}+q^{2}\\\frac{(a^{2}+1)^{2}}{4a^{2}+1}=p^{2}+q^{2}$

Therefore, the final solved equation is given by,

$\frac{(a^{2}+1)^{2}}{4a^{2}+1}=p^{2}+q^{2}$

This is what we needed to prove.

Hence, Proved.