1. Introduction

1.1. Cardano Formula

Are there only real and positive numbers in the world? This is a question that has plagued mathematicians for a long time. Some examples of this problem were given by early mathematicians in the face of imaginary numbers. Over the centuries, as complex systems such as complex numbers became more widely accepted, so did the study of their contents. Early on, Tartaglia proposed a unique method of solving cubic equations, a breakthrough in the algebraic age, as it began to use abstract reasoning in place of so-called numerical examples. First the Cubic function is express like this:

\( a{x^{3}}+b{x^{2}}+cx+d=0 \)

we get a new function

\( A{x^{3}}+Cx+D \)

We can get the solution:

\( A=x+\frac{b}{3a} \)

\( C=\frac{3ac−{b^{2}}}{3{a^{2}}} \)

\( D=\frac{2{b^{3}}−9abc+27{a^{2}}d}{27{a^{2}}} \)

When we have the function

\( {x^{3}}=15x+4 \)

\( x=\sqrt[3]{2+11i}+\sqrt[3]{2−11i} \)

So that if we accepted this solution, we must accept the existence of i which means imaginary number.

1.2. Euler formula

Leonhard Euler coined the system of complex numbers - one of the most important and fundamental terms about complex numbers. He said i= \( \sqrt[]{−1} \) , which means the square root of a negative number. Euler said that the symbol I is the primary determinant that means the equation cannot really be solved.

As is shown in figure 1, one type to the equation of complex analysis:

\( z=x+iy \)

The x is the real part and the latter is the imaginary part, using the Euler formula, we can get the other function type like this:

\( x=rcosθ \)

\( y=rsinθ \)

The Euler formula

\( {e^{ix}}=cosx+isinx \)

So that

\( z=r{e^{iθ}} \)

2. Cauchy-Goursat therorem

Accoring to the Cauchy-Goursat therorem

\( \frac{მu}{მx}=\frac{მv}{მy} \)

\( \frac{მu}{მy}=−\frac{მv}{მx} \)

3. Cauchy integral formula

\( \int _{c}f(z)dz=0 \)

\( f(z) \) can spilt into u+iv, and dz can spilt to dx+idy

\( \int _{c}u+iv(dx+idy) \)

First we use Green's theorem and then cauchy goursat theorem.

Let \( D=\lbrace z∈ℂ:|z|≤r\rbrace \) be the closed disk of radius r in ℂ.

Let 𝑓:𝑈→ℂ be holomorphic on some open set containing D.

Then for each a in the interior of D:

\( f(a)=\frac{1}{2πi}\oint _{მD}\frac{f(z)}{(z−a)}dz \)

where ∂𝐷 is the boundary of 𝐷, and is traversed anticlockwise.

4. Residue

In figure 2, we have a loop C1 and the clockwise C3 which is negetative.

According to the Cauchy integral formula

\( f(z)=\frac{1}{2πi}\cdot \int _{c1−c3}\frac{f(w)}{w−z}dw \)

\( \int _{c1}\frac{f(w)}{w−z+zo−zo}dw \)

\( =\int _{c1}\frac{f(w)}{(w−z)(1−\frac{z−z0}{w−zo})}dw \)

\( =\int _{c1}\frac{f(w)}{(w−z)}\sum _{0}^{∞}{(\frac{z−zo}{w−zo})^{n}}dw \)

\( =\sum _{0}^{∞}{(z−zo)^{n}}\cdot \int _{c1}\frac{f(w)}{{(w−z)^{n+1}}}dw \)

\( \int _{c3}\frac{f(w)}{z−zo}\cdot \frac{1}{1−\frac{w−zo}{z−zo}}dw \)

\( =\int _{c3}\frac{f(w)}{(z−zo)}\sum _{0}^{∞}\frac{w−zo}{z−zo}dw \)

\( =\sum _{−∞}^{1}{(z−zo)^{n}}\int _{c3}\frac{f(w)}{{(w−zo)^{n}}}dw \)

5. Residue Theorem

\( u1,u2,...un⊆U \) (open)

\( {a_{2}}∈u2 \)

\( {a_{2}}∉uj i≠j \)

\( u2∩uj=ϕ \)

Then by the ezistence of Layrent Series

\( f(z)=\sum _{−∞}^{∞}{a_{j}}{(z−{z_{u}})^{n}} \)

\( \int _{{მu_{i}}}f(z)dz=\int _{{მu_{i}}}\sum _{−∞}^{∞}{a_{2}}{(z−zo)^{n}}dz \)

\( =\sum _{−∞}^{−i}\int {a_{2}}{(z−zo)^{n}}dz+\sum _{0}^{∞}\int {a_{2}}{(z−zo)^{n}}dz+\int \frac{{a_{−1}}}{z−zo}dz \)

\( \int \frac{{a_{−1}}}{z−zo}dz=2πi{a_{−1}} \)

\( =2πi[Res] \)

6. Conclusion

The residue theorem can be used in many ways, that is to say, by obtaining residual formulas for certain functions of several variables. Represent the sum of the values of the polynomial function of a rational convex polyhedron by the Euler-McLaughlin summation formula and integrating [1]. The Residue Theorem can also be used for quadratic refinement by counting the curves on the hypersurface and the complete intersection in the projected space [2]. Also in “Counting Lattice Points by Means of the Residue Theorem” [3], the general residue theorem is used to help derive a new expression such as using the residual theorem to derive an Earhart polynomial. The inner Earhart polynomial and the closure of the tetrahedron are then linked by proving the Earhart-McDonald reciprocity law for the tetrahedron. In “Notes on n-point Witten diagrams in AdS2” [4], the one-dimensional boundary of AdS2 in the Witten diagram allows for a simple setup, and we can also obtain the cause of the error with the residual theorem. In the article “Parametric Euler T-sums of odd harmonic numbers” [5], when we want to establish several formulas for the T-sum of Euler odd harmonic numbers, we need to go through the contour integration method and the remainder theorem. Lipstein et al. [6] discovered that using the remainder theorem to obtain a Witten diagram allows to check multiple points of the formula, which provides more details to solve the tree-level single-ring cosmic scattering equation.

The residue theorem also has many implications such as the fact that at zero temperature and a finite chemical potential calculated by integrating over time will yield results consistent with those obtained at small but non-zero temperatures. And applying the remainder theorem to time integration yields a different answer [7]. Simplified analysis in continuous and discrete-time systems may be an alternative to Cauchy's remainder theorem. Simplified methods give analytical results that are more explicit and less restrictive [8]. The method of parentheses (MoB), proposed for evaluating Feynman integrals, gives results in terms of combinations of series(s), and it can also be used to prove the residue theorem [9]. Through A simple evaluation of a theta value and the Kronecker limit formula, we can verify complex functions not only by computing the integral by the residue theorem but also by using elliptic functions or analytical continuations [10].

For example, of the application by Residue Theorem. When account for the anomalous integrals:

\( \int _{−∞}^{∞}\frac{dx}{1+{x^{6}}} \)

then we know

\( {z_{k}}=exp(\frac{π+2kπ}{6}i),k=0,1,2,3,4,5 \)

\( Res(f;{z_{k}})=−\frac{{z_{k}}}{6} \)

\( \int _{−∞}^{∞}\frac{dx}{1+{x^{6}}}=−2πi·\frac{1}{6}·\sum _{k=0}^{2}{z_{k}}=\frac{2π}{3} \)