find the value.

am getting 2¹²

we have to find [(b + ib)^4]^4
or, [b^4 + (1/b)^4 + 6b^2(i/b)^2 + 4b^3(i/b) + 4b(i/b)^3]^4.
or, [6 + (-6) + 4(ib^2 - i/b^2)]^4
or,256(i^4) [b^2 - 1/b^2]^4
also b^4 + 1/b^4 = 6
or, (b^2 - 1/b^2)^2 = 4
or, (b^2 - 1/b^2) = + 2 or -2.

we can use either value to get the answer.
i am looking for a better solution if possible...
correct me if something is wrong.

or we can go the usual way,
(b+\frac{i}{b})^{16} = ((b+\frac{i}{b})^2)^8 = (b^2-\frac{1}{b^2}+2i)^8 = (b^4+\frac{1}{b^4}-4-2-\frac{4i}{b^2}+4ib^2)^4 = 2^8 * (b^2 - \frac{1}{b^2})^4 = 2^8 * (b^4+\frac{1}{b^4}-2)^2 = 2^{12}

very nise solution from rishabh