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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 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 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.

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.