Consider the points andÂ . Let be a point of the plane such that . Set to be the angle that is defined by and . ( the one that is less than .) Prove that the function is harmonic.

**Solution **[by Demetres Skouteris]

Skip to content
# Tag: Complex Analysis

## Harmonic function

## A convergent series

## An integral with trigonometric and rational function

## A logarithmic integral

## A contour integral

A site of university mathematics

Consider the points andÂ . Let be a point of the plane such that . Set to be the angle that is defined by and . ( the one that is less than .) Prove that the function is harmonic.

**Solution **[by Demetres Skouteris]

The complex function has a holomorphic branch in the half plane and its imaginary part is the desired angle. Hence, the function of the angle is harmonic.

Let be holomorphic on the open unit disk and suppose that the integral converges. If the Taylor expansion of is of the form then prove that the sum

converges.

**Solution**

We evaluate the integral using the standard orthogonality results for . Thus:

and thus the series converges.

**Note: **The set of functions satisfying this is a Hilbert space of functions, but it is *not* the same as the Hardy space .

Evaluate the integral

**Solution**

Firstly , we begin by noticing the following

Let us now consider the function as well as the contour

The function has three simple poles of which only is included within the contour. The residue at turns out to be . Thus

The contribution of the large circle as is whereas the contribution of the small circle as is . This can be seen by parametrising the small circle ( ). Hence:

Using we get that

Prove that

**Solution**

We begin by making the substitution thus:

Now let us consider the complex function where the principal arguement of lies within the interval as well as the contour below

It is clear that has two poles of order at and . The residue at is equal to whereas the residue at is equal to . Thus

Sending and the contribution of both the large and the small circle is . Hence:

Thus the conclusion follows.

Let such that . Prove that:

**Solution**

We have successively: