De Moivre’s theorem states, “If z = r(cos u + i sin u), then zn = rn(cos nu + i sin nu).”
• Verify de Moivre’s theorem for n = 2.
a. Provide a correct proof that includes written justification for each step.
To verify de Moivre's theorem for n = 2, we need to show that if z = r(cos u + i sin u), then zn = rn(cos 2u + i sin 2u).
Let's start by expressing zn as (r(cos u + i sin u))^2:
zn = (r(cos u + i sin u))^2
Now, we can expand this expression using the properties of complex numbers:
zn = r^2(cos u + i sin u)(cos u + i sin u)
To simplify further, we can use the product-to-sum identities:
zn = r^2[(cos u)(cos u) + (cos u)(i sin u) + (i sin u)(cos u) + (i sin u)(i sin u)]
Now, let's simplify each term:
zn = r^2[(cos^2 u - sin^2 u) + i(2 cos u sin u)]
Using the trigonometric identity cos^2 u - sin^2 u = cos 2u, we can then write:
zn = r^2[cos 2u + i(2 cos u sin u)]
Finally, let's factor out r^2:
zn = r^2(cos 2u + i(2 cos u sin u))
This is the same as rn(cos 2u + i sin 2u), which proves de Moivre's theorem for n = 2.
Therefore, the equation zn = rn(cos 2u + i sin 2u) holds true for n = 2.