Introduction to Residue Integration in Complex Analysis

Integration in complex analysis often involves evaluating integrals along complex contours. One such method, known as the residue theorem, is particularly useful in solving integrals with singularities inside the contour. This article will guide you through the process of finding all possible values of the pair (a, b) for the given integral by analyzing the behavior of its poles within a specified contour.

Step-by-Step Guide: Analyzing the Integral

We are tasked with evaluating the integral:

∫CZ-4-i51Z-A-BZ-3-4i dZ

where the contour ( C ) is a circle centered at ( 4 i ) with a radius of 5.

Step 1: Identifying Singularities

The integrand has singularities (poles) at ( z a bi ) and ( z 3 4i ). To evaluate the integral, we need to check whether these poles lie inside or outside the contour ( C ).

Step 2: Determining the Position of the Poles

We need to determine the position of the poles:

Pole at ( 3 4i ): Pole at ( a bi ):

Pole at ( 3 4i ):

Calculate the distance from the center ( 4 i ) to the point ( 3 4i ):

(3   4i) - (4   i)  -1   3i
sqrt{(-1)^2   (3)^2}  sqrt{10}

The radius of the contour is 5, so ( sqrt{10} approx 3.16

Pole at ( a bi ):

Calculate the distance from the center ( 4 i ) to the point ( a bi ):

Let the distance be ( sqrt{(a - 4)^2 (b - 1)^2} ). For the pole to be inside the contour, the following must hold:

( sqrt{(a - 4)^2 (b - 1)^2} leq 5 )

Squaring both sides:

( (a - 4)^2 (b - 1)^2 leq 25 )

Step 3: Calculating the Integral

Using the residue theorem, the integral can be calculated as follows:

The residue at ( z 3 4i ) is:

( Residue left( frac{1}{z - a - bZ - 3 - 4i} bigg|_{z 3 4i} right) frac{1}{3 4i - a - bi} frac{1}{3 - a 4 - bi} )

The residue at ( z a bi ) is:

( Residue left( frac{1}{z - a - bZ - 3 - 4i} bigg|_{z a bi} right) frac{1}{a bi - 3 - 4i} frac{1}{a - 3 b - 4i} )

By the residue theorem:

( int_C f(z) , dz 2pi i left( text{Residue at } 3 4i text{ and Residue at } a bi right) )

Setting the integral equal to ( frac{2pi i}{1 i} ):

( 2pi i left( frac{1}{3 - a 4 - bi} frac{1}{a - 3 b - 4i} right) frac{2pi i}{1 i} )

Dividing by ( 2pi i ):

( frac{1}{3 - a 4 - bi} frac{1}{a - 3 b - 4i} frac{1}{1 i} )

The right-hand side can be simplified:

( frac{1 i}{(1 i)(1 - i)} frac{1}{2} frac{1}{2}i )

( frac{1}{3 - a 4 - bi} frac{1}{a - 3 b - 4i} frac{1}{2} - frac{1}{2}i )

Step 4: Solving the Equation

The right-hand side can be simplified:

( frac{1}{2} frac{1}{2}i )

Let:

( x 3 - a )

( y 4 - b )

Then we rewrite the equation:

( frac{1}{x yi} frac{1}{-x - yi} frac{1}{2} - frac{1}{2}i )

This leads to a system of equations that can be solved for ( x ) and ( y ).

Conclusion

The final values for ( a ) and ( b ) will depend on solving the resulting equations while ensuring that the condition ( (a - 4)^2 (b - 1)^2 leq 25 ) is satisfied.

In summary, to find the possible pairs ( (a, b) ):

Solve the system of equations derived from the residues. Ensure ( (a, b) ) lies within the circle defined by ( (a - 4)^2 (b - 1)^2 leq 25 ).

This approach will yield the possible pairs ( (a, b) ).