By Isac G.

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Suppose ∞ k=0 ak = A and convergent. Show that if 2 e−n z n , 2 ∞ b = B. Suppose further that each of the series is absolutely k=0 k k ck = a j bk− j j =0 then ∞ ck = AB. k=0 Outline: Use the fact that |ak | and |bk | converge to show that dk converges where k dk = |a j ||bk− j |. j =0 In particular, dn+1 + dn+2 + · · · → 0 as n → ∞. Note then that if An = a0 + a1 + · · · + an Bn = b0 + b1 + · · · + bn Cn = c0 + c1 + · · · + cn , An Bn = Cn + Rn , where |Rn | ≤ dn+1 + dn+2 + · · ·+ d2n , and the result follows by letting n → ∞.

7 Proposition −C f = f. C Proof −C b f =− f (z(b + a − t))˙z (b + a − t)dt. a Again, expanding the integral into real and imaginary parts and applying the changeof-variable theorem to each (real) integral, we ﬁnd −C a f = b f (z(t))˙z (t)dt = − f. C 48 4 Line Integrals and Entire Functions E XAMPLE 1 Suppose f (z) = x 2 + i y 2 (where x and y denote the real and imaginary parts of z, respectively), and consider C : z(t) = t + i t, 0 ≤ t ≤ 1. Then z˙ (t) = 1 + i and 1 f (z)dz = 0 C 1 (t 2 + i t 2 )(1 + i )dt = (1 + i )2 t 2 dt = 2i /3.

Setting f (i y) = A(y) + i B(y), it follows that f (z) = e x A(y) + i e x B(y). For f to be analytic, the Cauchy-Riemann equations must be satisﬁed; therefore A(y) = B (y) and A (y) = −B(y), so that A = −A. Thus we consider A(y) = α cos y + β sin y B(y) = −A (y) = −β cos y + α sin y. Since f (x) = e x , however, A(0) = α = 1 and B(0) = −β = 0, so that, ﬁnally, we are led to examine f (z) = e x cos y + i e x sin y. Indeed, it is easy to verify that f is an entire function with the desired properties (1) and (2).