Q.Prove that $(1+i)^4 \times (1-i)^4 = 16$.

Complex numbers combine under addition, subtraction, multiplication, division and conjugation using rules that extend ordinary real-number algebra, treating $i$ as a symbol satisfying $i^2=-1$. Addition and subtraction act componentwise: $(a+ib)\pm(c+id)=(a\pm c)+i(b\pm d)$ — real parts combine with real parts, imaginary with imaginary. Scalar multiplication by a real $k$ scales both parts: $k(a+ib)=ka+i(kb)$. Multiplication expands like two binomials and then uses $i^2=-1$ to collapse the $i^2$ term: $(a+ib)(c+id)=ac+adi+bci+bdi^2=(ac-bd)+i(ad+bc)$. All four operations retain the familiar commutative, associative and identity properties from real-number arithmetic. The conjugate of $z=a+ib$ is $\bar z=a-ib$ — obtained by flipping only the sign of the imaginary part — and satisfies $\overline{\overline z}=z$, $z=\bar z$ exactly when $z$ is real, $z=-\bar z$ exactly when $z$ is purely imaginary, and the useful product identity $z …