"How can (1+sqrt(-3))^(1/5) + (1 - sqrt(-3))^*1/5) be real?"

Let z := 1 + sqrt(-3) = 1 + sqrt(3) * i with i being the complex unit. Notice that

(1+sqrt(-3))^(1/5) + (1 - sqrt(-3))^*1/5) = z^(1/5) + (conjugate(z))^(1/5)

with

conjugate(a + b * i) = a - b * i

being the complex conjugate (a,b real numbers).

It can be shown (will do below) that

(conjugate(z))^(1/5) = conjugate(z^1/5)

Noticing that for any real number a, b

(a + b * i) + conjugate(a + b * i)

= (a + b * i) + (a - b * i)

= 2 * a

is a real number, we conclude

(1+sqrt(-3))^(1/5) + (1 - sqrt(-3))^*1/5)

= z^(1/5) + (conjugate(z))^(1/5)

= z^(1/5) + conjugate(z^(1/5))

= w + conjugate(w)

is a real number if we set w := z^(1/5), q.e.d.

Now, what's left to show is that for any complex number z, we have

(conjugate(z))^(1/5) = conjugate(z^(1/5)).

You can write z as a * exp(i * phi)

with a = |z| (the norm of z) and phi = arg(z), both being real numbers. Observe then that

z^(1/5) = a^(1/5) * exp(i * phi/5)

and

conjugate(z^(1/5)) = conjugate(a^(1/5) * exp(i * phi/5))

= a^(1/5) * conjugate(exp(i * phi/5))

= a^(1/5) * exp(-i * phi/5)

= (a * exp(-i * phi)^(1/5)

= (conjugate(z))^1/5

which was to shown.

Yeah, math major. Hope that helps.