Informally, the idea is that if you integrate $f(z)$ around a circle of large radius $R$, then $|f(z)|$ will be approximately $1/R^2$, and the length of the contour will be $2\pi R$. So $\left|\oint f(z)\,dz\right|$ will be approximately $2\pi/R$, which you can make arbitrarily small by taking $R$ large enough. Now use the residue theorem to conclude that the sum of the residues is 0.Poirot said:Let $f(z)=\frac{z^{100}}{z^{102}+1}$. Prove that the sum of the residues of f is 0. (hint: consider the integral of f around a circular contour centrered at zero)
The purpose of proving the residue sum of f(z) is to determine if the function has any singularities (points where the function is not defined) within a given contour or closed curve. This information can then be used to evaluate the contour integral of the function.
The residue sum of f(z) is calculated by finding the poles of the function (points where the denominator is equal to 0) within the given contour. The residue at each pole is then calculated using the formula Res(f(z), z0) = lim(z→z0) [(z-z0)f(z)]. The residue sum is then the sum of all the individual residues.
If the residue sum of f(z) is 0, it means that there are no poles within the given contour. This can be interpreted as the function being analytical (continuous and differentiable) within the contour, which makes the evaluation of the contour integral much simpler.
Yes, the residue sum of f(z) can be a non-zero value if there are poles within the given contour. In this case, the function is not analytical within the contour and the evaluation of the contour integral becomes more complex.
The residue sum of f(z) is commonly used in the field of complex analysis, which has applications in various fields of science and engineering such as physics, electrical engineering, and fluid mechanics. It is also used in the study of differential equations and in the development of numerical methods for solving mathematical problems.