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.