Understanding the Residue of a Complex Function at (z 2i)

Let's delve into the analysis of the residue of the complex function (mathcal{f}(z) frac{z(z - a)}{(z - a)^2(z^2 - a^2)}) at the point (z 2i). It is important to first understand that determining the residue involves examining the behavior of the function around the given point.

Step-by-Step Solution

The function we are considering is:

(mathcal{f}(z) frac{z(z - a)}{(z - a)^2(z^2 - a^2)})

We need to find the residue of this function at (z 2i). To do this, we will follow a series of steps involving limits and simplification.

Step 1: Simplification of the Function

First, let's simplify the function:

(mathcal{f}(z) frac{z(z - a)}{(z - a)^2(z^2 - a^2)} frac{z(z - a)}{(z - a)^2(z - a)(z a)} frac{z}{(z - a)(z a)})

Step 2: Determining the Residue

To find the residue at (z 2i), we must evaluate:

(lim_{z to 2i} (z - 2i)mathcal{f}(z))

Substituting the simplified function:

(lim_{z to 2i} (z - 2i) cdot frac{z}{(z - 2i)(z 2i)} lim_{z to 2i} frac{z}{z 2i})

Cancelling the (z - 2i) term, we get:

(lim_{z to 2i} frac{z}{z 2i} frac{2i}{2i 2i} frac{2i}{4i} frac{1}{2})

This tells us that the residue of (mathcal{f}(z)) at (z 2i) is (frac{1}{2}).

Step 3: Conclusion

Therefore, the residue of the given function at (z 2i) is (frac{1}{2}).

Additional Insights

It is worth noting that the function (mathcal{f}(z)) does not have a pole at (z 2i). This is a critical distinction in the analysis of complex functions. A pole is a point where a function becomes infinite, whereas the residue is a way of quantifying the strength and behavior of the function around such points.

Further Exploration

For a deeper understanding, consider studying the following related concepts:

1. Computation of Limits

Practice computing limits similar to (lim_{z to 2i} (z - 2i)mathcal{f}(z)).

2. Understanding Poles and Residues

Explore the differences between poles, removable singularities, and essential singularities in complex analysis.

3. Application of L'H?pital's Rule

Apply L'H?pital's rule where necessary to evaluate indeterminate forms.

Conclusion

In conclusion, the residue of the function (mathcal{f}(z) frac{z(z - a)}{(z - a)^2(z^2 - a^2)}) at (z 2i) is (frac{1}{2}). Understanding the behavior of complex functions around singular points is crucial in advanced mathematics and physics.