Given:

Taking the L.C.M, we get

[∵(a + b)(a – b) = (a² – b²)]

Putting i² = -1

= 0 + 0i

Hence, the given equation is purely real as there is no imaginary part.

(ii) Given:

Taking the L.C.M, we get

…(i)

[∵(a + b)(a – b) = (a² – b²)]

Now, we know that,

(a + b)² + (a – b)² = 2(a² + b²)

So, by applying the formula in eq. (i), we get

Putting i² = -1

Hence, the given equation is purely real as there is no imaginary part.