Question

Using De-Moivre's theorem, prove that i^2= -1.​

Answer 1

Write i in trigonometric form. Since |i| = 1 and arg(i) = π/2, we have

i = exp(i π/2) = cos(π/2) + i sin(π/2)

By DeMoivre's theorem,

i² = exp(i π/2)² = exp(i π) = cos(π) + i sin(π)

and it follows that i² = -1 since cos(π) = -1 and sin(π) = 0.