1.9. CONVERGENCE IN C 23 subcollection U1,...,Un that covers {x}×[c, d]. The open set Nx = U1∪···∪ Un contains a set of the form Ix ×[c, d] by the preceding discussion, where Ix is an open interval containing x. The collection {Ix}x∈[a,b] covers [a, b], and hence by the Heine-Borel theorem for R, there exists a finite subcollection Ix 1 , . . . , Ix m that covers [a, b]. We take our finite subcollection of the original open cover {Uj}j∈J to be {U | for some xi the set U is one of the elements in the union that defines Nx i }. Exercise 1.9.18. Show that a subset of C is closed iff it contains all its accumulation points. Exercise 1.9.19. Define the notion of sequentially compact for a subset of C, and show that a subset of C is sequentially compact if and only if it is closed and bounded. Definition 1.9.20. If z = x + iy ∈ C, z = 0, and r = |z|, then the polar form of z is z = r(cos θ + i sin θ) where θ is the unique solution to the equations x = r cos θ, y = r sin θ in the interval [0, 2π). The angle θ is called the principal branch of the argument of z and is denoted Arg(z). For z as above, we often write z = reiθ where eiθ is defined to be cos θ + i sin θ. In fact, this is the value of the complex exponential function f(z) = ez when z = iθ. Exercise 1.9.21. Suppose that n ∈ N. Prove that if z = e 2πik n , for k ∈ Z and 0 ≤ k ≤ n − 1, then zn = 1. Such a z is called an n-th root of unity. Note that these are all distinct. The n-th roots of unity form a cyclic group of order n under multiplica- tion. An n-th root of unity is primitive if it is a generator of this group. Exercise 1.9.22. Show that the primitive n-th roots of unity are of the form e2πik/n where k and n are relatively prime. Proposition 1.9.23. If n 1, the sum of the n distinct n-th roots of unity is 0. Proof. For any z ∈ C, 1 − zn = (1 − z)(1 + z + z2 + · · · + zn−1). Now let z be a primitive n-th root of unity. Exercise 1.9.24. Suppose z is a nonzero complex number, and write z = reiθ. Show that z has exactly n distinct complex n-th roots given by r1/nei(2πk+θ)/n for 0 ≤ k ≤ n − 1.

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