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It was known in the 17th century that the roots of polynomial equations, i.e., when f(x) = 0 in: f(x) = a _{n}x^{n} + a_{n-1}x^{n-1} + +a_{1}x + a_{0} (such as f(x) = x^{2} + 2 = 0 at x = ),often involve square roots of negative numbers. It was realized subsequently that the complex number, which is defined as the combination of a real number and a square root of negative number, is more useful than the real number alone. In the Cartesian form the complex number z is expressed as z = x + iy, where the negative sign in the square root has been absorbed into the symbol i = . In terms of the polar coordinates r and , z = r e ^{i}, where r^{2} = x^{2} + y^{2}, and tan = y/x (Figure 08). The complex conjugate of z is defined as z* = x - iy or z* = r e^{-i}, such that z z* = r^{2}. The term "imaginary" for the part associated with "i" was coined by | |

## Figure 08 Complex Number [view large image] |
René Descartes and was meant to be derogatory. The study of functions of a complex variable is known as complex analysis. Unlike real functions, which are commonly represented as two-dimensional graphs, complex functions have four-dimensional graphs. |

Contour integral is related to a special class of complex function called analytic function denoted here by f(z) = u + iv, with the complex variable z = x + iy. It has the property that the derivative df(z)/dz exists uniquely at every point. In other words, the way z 0 is indepen-dent of the path, e.g., we may choose a path with x=0 or y=0. It follows that the real and imaginary parts of an analytic function are solutions of the two-dimensional Laplace equation: | |

## Figure 09 Contour Integral [view large image] |
, . |

---------- (18) |

---------- (20) |

---------- (21) |

A technique to evaluate ordinary integration is to use the contour integral with poles whenever Eq.(17) is satisfied. For example, considering the following integral:

I = _{} dx / (x^{2} + 1) ---------- (22). This integral is part of a closed contour integral, which has a pole at z = i (see Figure 10): I_{c} = _{} dx / (x^{2} + 1) + _{}_{cR} dz / (z^{2} + 1) = I = 2i [1/(z + i)]_{z=i} = 2i (1/2i) = ------------ (23), | |

## Figure 10 A Contour Loop |
where the second term vanishes if we let the radius of the semi-circle R . |

There are two poles at x = +1 and x = -1, right on the real axis. In order to apply the loop in Figure 10 to evaluate this integral we can artificially move one of the poles a wee bit up and the other one down by an amount i as shown in Figure 10, and then let 0 at the end. Thus,

Actually, the answer can be obtained as easily by substituting y

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