Question

Find the residues of the given function at all poles. Take \(z=r e^{i \theta}\) \(0 \leq \theta<2 \pi\). $$\frac{\ln z}{1+z^{2}}$$

Short answer

Residue at z = i: -1/2i and Residue at z = -i: -3i \(2}.

Step-by-step solution

Identify the Poles

First, find the poles of the function \(\frac{\text{ln} \, z}{1 + z^2}\). The poles are the values of \z\ for which the denominator, \(1 + z^2\), equals zero. \(1 + z^2 = 0\) leads to \(z^2 = -1\), so \(z = i\) or \(z = -i\).

Express Around Poles

Express the function around each pole. For \(z = i\), write \(z = i + \frac{1}{w}\), then the function becomes \(\frac{\text{ln} \, (i + \frac{1}{w})}{1 + (i + \frac{1}{w})^2} \). Similarly for \(z = -i\), \(z = -i + \frac{1}{w}\).

Residue Calculation at \(z = i\)

For the pole at \(z = i\), decompose \(1 + z^2\) around \(z = i\): \(1 + (i + w)^2 \). The residue is the limit as \(z \to i\) of \( (z - i) \frac{\text{ln} \, z}{1 + z^2}\). Calculate this as: \( \frac{i \frac{\text{ln} \, i}{2i}}{-2i} (z \to i). \)

Simplified Calculation for \(z = i\)

Simplify the function above: \( \frac{\text{ln} \, i}{-4i} \). Using \( \text{ln} \, i = \frac{\text{ln} \, e^{i \frac{\text{\text\}}{\text{\text\}}}}\), we get \( \text{ln} \, i = i \frac{\text{\text\}}{\text{\text\}}. \) Finally, the residue at \(z = i\) is \( - \frac{\text{\text\}}{\text{\text\}} \).

Residue Calculation at \(z = -i\)

Similarly, for \(z = -i\), follow the same process. The residue is the limit as \(z \to -i\) of \( (z + i) \frac{\text{ln} \, z}{1 + z^2}\). Calculate this as: \( \frac{-i \frac{\text{ln} \, (-i)}{2i}}{-2i}\).

Simplified Calculation for \(z = -i\)

Simplify the function above: \( \frac{\text{ln} \, (-i)}{-4i} \), Recall \( \text{ln} \, (-i) = \text{ln} \, i + \text{ln} \, (-1) = i \frac{\text{\text\}}{\text{\text\}} + i \text{\text\}. \) Thus, the residue at \(z = -i\) is \( -\frac{3 i \text{\text\}}{\text{\text\}} \).

Conclusion Residue

Summarize the residues found: \(z = i\) and \(z = -i\) have residues: \(1\)

Key concepts

Residues

In complex analysis, a residue is a complex number that describes the behavior of a function near a singularity, often a point where the function goes to infinity. Calculating residues is crucial in evaluating complex integrals using the residue theorem. For example, in the function given, \( \frac{\ln z}{1+z^2} \), identifying residues helps us understand its behavior at poles, specifically at \( z = i \) and \( z = -i \). When computing residues, we typically use limits. The residue at a singularity \( z = a \) is given by: \ \text {Res} (f, a) = \text {lim}_{z \to a} \, (z - a) f(z) \ For our given function, residues are found by simplifying the expression around each pole.

Poles

Poles are points where a complex function takes an infinite value. In the given function \( \frac{\ln z}{1+z^2} \), the poles are where the denominator equals zero. Solving for \( 1 + z^2 = 0 \), we determine that \( z^2 = -1 \) which gives us \( z = i \) and \( z = -i \). These are first order poles because \( 1 + z^2 \) has a simple zero (it approaches zero linearly) at these points. Around each pole, we can rewrite the function in a form that makes it easier to calculate the residue.

Logarithmic Functions

Logarithmic functions are fundamental in complex analysis and are essential in the given function \( \frac{\ln z}{1+z^2} \. \ \ln z \) represents the natural logarithm of a complex number \ z \, which has a branch cut, typically along the negative real axis. For example, \( \ln i = i \frac{\pi}{2} \). The complex logarithm is defined as \ \ln z = \ln |z| + i \, \text{arg}(z) \. Using properties of the logarithmic function, such as \ \ln i = i \frac{ \pi}{2} \ and \ \ln(-i) = i \frac{ \pi}{2} + i \pi = -3i \frac{\pi}{2} \ , we calculate residues by simplifying the function around the poles.

Calculus of Residues

The calculus of residues is a powerful technique in complex analysis for evaluating contour integrals. It centers around the residue theorem, which states that for a meromorphic function \( f \) within a closed contour \( C \), the integral of \( f \) along \( C \) is \[ \, \int_{C} f(z) dz = 2 \pi i \sum_{\text{residues}} \]. This simplifies computations significantly by converting a complex contour integral into a summation of residues. In practice, once poles (like \ z = i \ and \ z = -i \ in our given function) are found, the residues at these points incorporate the behavior of \ f \ near the singularities. Utilizing this technique, integrals that would otherwise be difficult to compute become manageable and straightforward.