Proving that $w^n-z=0$, for non-zero complex $z$ and positive integer $n$, has exactly $n$ distinct complex roots

Question

I recently have been reading Basic Mathematics by Serge Lang and encountered this question

Let $z$ be a complex number not equal to $0$. Let $n$ be a positive integer. Show that there are $n$ distinct complex numbers $w$ such that $w^n = z$. Write these complex numbers in polar form. The proof given that a polynomial of degree $n$ has at most $n$ roots applies to the complex case, and thus we see that there are no other complex numbers $w$ such that $w^n = z$ other than those you have presumably written down.

I tried relating it with polynomial in form $x^n -z = 0$, and proceeding with that same pattern and tried to come up with solutions, but it gets quite complicated pretty quick and I wonder if it will worth to consider odd/even powers and their effect on imaginary part.

I would also love to see if there is any way to prove the $n$ roots thing for complex number case.

I will appreciate the help. Thanks!

Answer 1

The fundamental theorem of algebra guarantees the $n$ solutions.

Take a primitive $n$-th root of unity: $\zeta=e^{2\pi i/n}$. Then the $n$ solutions are $\zeta^kz^{1/n},\,k=0,2,\dots n-1$.

If $z=re^{i\theta}$, we get $r^{1/n}e^{i(\theta+2\pi k)/n}$.