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0 Applying this with k = 0 and using fn → f pointwise, we obtain 1 g0 (t, z)dt for z ∈ D(z0 , δ). 23 that the right-hand side, and hence f , is analytic on D(z0 , δ), and moreover, 1 f (k) (z) = 1 (k) gk (t, z)dt = lim fn(k) (z) for z ∈ D(z0 , δ), k g0 (t, z)dt = k! 0 n→∞ 0 1, (k) where g0 is the k-th derivative of g0 with respect to z. Indeed, g0 (t, z) is measurable on [0, 1] × D(z0 , δ) and for every fixed t, the function z → g0 (t, z) is analytic on D(z0 , δ). 4) and since gn,0 (t, z) → g0 (t, z), we have |g0 (t, z)| 2C for t ∈ [0, 1], z ∈ D(z0 , δ).

Let f be a complex function that is analytic on an open set containing γ and the interior of γ minus {z1 , . . , zq }. Then 1 2πi q f (z)dz = γ res(zi , f ). i=1 47 Proof. We proceed by induction on q. First let q = 1. Choose r > 0 such that γz1 ,r lies in the interior of γ. 4, 1 2πi f (z)dz = γ 1 2πi f (z)dz = res(z1 , f ). γz1 ,r Now let q > 1 and assume the Residue Theorem is true for fewer than q points. We cut γ into two pieces, the piece γ1 from a point w0 to w1 and the piece γ2 from w1 to w0 so that γ = γ1 + γ2 .

Let q, a be integers with q 2 and gcd(a, q) = 1. Denote by π(x; q, a) the number of primes p with p x and p ≡ a (mod q). Then π(x; q, a) ∼ 1 x · as x → ∞. 3. Let q be an integer 2. Then for all integers a coprime with q we have π(x; q, a) 1 lim = . x→∞ π(x) ϕ(q) This shows that in some sense, the primes are evenly distributed over the prime residue classes modulo q. 3 An elementary result for prime numbers We finish this introduction with an elementary proof of a weaker version of the Prime Number Theorem.