# complex rational expressions

Introduction :

The complex number is a form of a + ib where ‘a’ and ‘b’ are real numbers and then i is called the imaginary unit, having the property that i2 = - 1. The complex analysis is the part of math that investigates functions of the complex numbers. The complex analysis is the usually known as the theory of functions of the complex variable. It is the useful in many part of math, that including number theory and applied mathematics.
Example Problems for Solving Complex Analysis Problems:

Solving complex analysis problems – Example: 1

Evaluate `-2 int_r (z)/((z+2)(z-1)) dz` where r is the circle |z| = 3 traversed twice in the clockwise direction.

Solution:

The solution is times the sum of the residues of the integrand.

`-2 int_r (z)/((z+2)(z-1)) dz = -2 * 2 pi i(2/3 + 1/3) = -4 pi i`

Solving complex analysis problems – Example: 2

Evaluate `sum_(k=0)^100 e^(kz)`

Solution:

`sum_(k=0)^100 e^(kz)`

`sum_(k=0)^100 e^(kz) = 1 + e^z + e^(2z) +.... + e^(100z) = Sum`

If `z = 2n pi i` then `Sum = 1 + 1 + ... + 1 = 101`

If  `z != 2n pi i` then multiplying the sum by  `1 - e^z` reveals:

`Sum(1-e^z) = 1-e^z+e^z-e^2z+e^2z+....+e^(100z)-e^(101z) = 1 - e^(101z)`

Therefore `Sum = (1-e^(101z))/(1-e^z)`

Also, this can be obtained from the familiar formula

`sum_(k=0)^100 x^k = 1 + x + x^2 +....+x^(100) = (1-x^(100))/(1-x)`

where `x = e^x`

Solving complex analysis problems – Example: 3

Solve the Laurent series about `z_0 = 1` for the function `e^(2z)/(z-1)^3 `

Solution:

The only singularity is at z = 1 so the series will look like

`sum_(n=-oo)^oo A_n(z-1)^n , 0 < |z-1| < oo`

`A_n = (1)/(2 pi i) oint)_C (f(z))/(z-1)^(n+1) dz = (1)/(2 pi i) oint_C (e^(2z))/(z-1)^(n+4) dz, n = 0,+-1, +-2` ....

where C is any simple closed curve around the point z=1.

Cauchy's theorem gives `A_n = 0, n = -4,-5` ....

For `n>=3`

`A_n = (1)/(n+3)! (d^(n+3))/(dz^(n+3)) e^(2z) |_(z=1)`

`= (e^2 2^(n+3))/(n+3)! , n>=3`

The Laurent series is

`f(z) = e^(2z)/(z-1)^3 = 8e^2 sum_(n=-3)^oo (2^n)/(n+3)! (z-1)^n, 0<|z-1|<oo.`
Practice Problems for Solving Complex Analysis Problems:

1. Evaluate `int_C sinz dz, C ` starts at the origin, traverses the bottom half of a unit circle centered at z0 = 1 / 2 and then the line from z = 1 to z = i p.