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

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

3 0

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.

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