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A) Suppose first that Ω is open and pathwise connected, and that it can be written as Ω = Ω1 ∪ Ω2 where Ω1 and Ω2 are disjoint non-empty open sets. Choose two points w1 ∈ Ω1 and w2 ∈ Ω2 and let γ denote a curve in Ω joining w1 to w2 . Consider a parametrization z : [0, 1] → Ω of this curve with z(0) = w1 and z(1) = w2 , and let t∗ = sup {t : z(s) ∈ Ω1 for all 0 ≤ s < t}. 0≤t≤1 Arrive at a contradiction by considering the point z(t∗ ). (b) Conversely, suppose that Ω is open and connected. Fix a point w ∈ Ω and let Ω1 ⊂ Ω denote the set of all points that can be joined to w by a curve contained in Ω.

Consider the function defined by f (x + iy) = |x||y|, whenever x, y ∈ R. 28 Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS Show that f satisfies the Cauchy-Riemann equations at the origin, yet f is not holomorphic at 0. 13. Suppose that f is holomorphic in an open set Ω. Prove that in any one of the following cases: (a) Re(f ) is constant; (b) Im(f ) is constant; (c) |f | is constant; one can conclude that f is constant. N 14. Suppose {an }N n=1 and {bn }n=1 are two finite sequences of complex numbers.

If γ is piecewise smooth, then the integral of f over γ is simply the sum of the integrals of f over the smooth parts of γ, so if z(t) is a piecewise-smooth parametrization as before, then n−1 ak+1 f (z) dz = γ f (z(t))z (t) dt. k=0 ak By definition, the length of the smooth curve γ is b |z (t)| dt. length(γ) = a Arguing as we just did, it is clear that this definition is also independent of the parametrization. Also, if γ is only piecewise-smooth, then its length is the sum of the lengths of its smooth parts.

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