By Beck M., Marchesi G., Pixton G.

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**Example text**

Let σ be the counterclockwise semicircle formed by the segment S of the real axis from −R to R, followed by the circular arc T of radius R in the upper half plane from R to −R, where R > 1. We shall integrate the function f (z) = 1 1/(z + i) g(z) 1 = = , where g(z) = z2 + 1 z−i z−i z+i Since g(z) is analytic inside and on σ and i is inside σ, we can apply Cauchy’s formula: 1 2πi and so S σ dz 1 = +1 2πi z2 dz + z2 + 1 T σ g(z) 1 1 dz = g(i) = = , z−i i+i 2i dz = z2 + 1 σ dz 1 = 2πi · = π. z2 + 1 2i (∗∗) Now this formula holds for all R > 1, so we can take the limit as R → ∞.

J. 1 Definition and Basic Properties We will now spend a chapter on certain functions defined on subsets of the complex plane which are real valued. The main motivation for studying them is that the partial differential equation they satisfy is very common in the physical sciences. 1. Let G ⊆ C be a region. A function u : G → R is harmonic in G if it has continuous second partials in G and satisfies the Laplace1 equation uxx + uyy = 0 in G. There are (at least) two reasons why harmonic functions are part of the study of complex analysis, and they can be found in the next two theorems.

1 Definition and Basic Properties We will now spend a chapter on certain functions defined on subsets of the complex plane which are real valued. The main motivation for studying them is that the partial differential equation they satisfy is very common in the physical sciences. 1. Let G ⊆ C be a region. A function u : G → R is harmonic in G if it has continuous second partials in G and satisfies the Laplace1 equation uxx + uyy = 0 in G. There are (at least) two reasons why harmonic functions are part of the study of complex analysis, and they can be found in the next two theorems.

### A first course in complex analysis by Beck M., Marchesi G., Pixton G.

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