Question

Find the residues of the following functions at the indicated points. Try to select the easiest of the methods outlined above. Check your results by computer. $$\frac{e^{2\pi iz}}{1-z^3}\text{ at }z=e^{2\pi i/3}$$

Short answer

$$-\frac{1}{3}{e}^{4\pi i/3}$$

Step-by-step solution

Identify the function and the point

The given function is $\frac{e^{2\pi iz}}{1-z^3}$ and we need to find the residue at $z=e^{2\pi i/3}$.

Simplify the point of interest

Notice that $e^{2\pi i/3}$ is a cube root of unity. There are three cube roots of unity: $1,e^{2\pi i/3}$, and $e^{-2\pi i/3}$.

Determine the pole order

The denominator $1-z^3$ has a zero of order 1 at $z=e^{2\pi i/3}$, which means the function $\frac{e^{2\pi iz}}{1-z^3}$ has a simple pole at $z=e^{2\pi i/3}$.

Calculate the residue using the formula for simple poles

For a simple pole, the residue can be calculated using the formula: $$\text{Res}(f,z_0)=\underset{z\to z_0}{lim}(z-z_0)f(z)$$. Let $f(z)=\frac{e^{2\pi iz}}{1-z^3}$ and $z_0=e^{2\pi i/3}$. We then have: $$\text{Res}(\frac{e^{2\pi iz}}{1-z^3},z=e^{2\pi i/3})=\underset{z\to e^{2\pi i/3}}{lim}\frac{(z-e^{2\pi i/3})e^{2\pi iz}}{1-z^3}$$. Since $1-z^3=(z-e^{2\pi i/3})(z-e^{4\pi i/3})(z-1)$, this expression simplifies as follows: $$\underset{z\to e^{2\pi i/3}}{lim}\frac{(z-e^{2\pi i/3})e^{2\pi iz}}{(z-e^{2\pi i/3})(z-e^{4\pi i/3})(z-1)}=\underset{z\to e^{2\pi i/3}}{lim}\frac{e^{2\pi iz}}{(z-e^{4\pi i/3})(z-1)}$$.

Evaluate the limit

Evaluate the limit by substituting $z=e^{2\pi i/3}$ directly: $$\text{Res}(\frac{e^{2\pi iz}}{1-z^3},z=e^{2\pi i/3})=\frac{e^{2\pi i\cdot e^{2\pi i/3}}}{(e^{2\pi i/3}-e^{4\pi i/3})(e^{2\pi i/3}-1)}.$$

Simplify the expression

Since $e^{2\pi i\cdot e^{2\pi i/3}}=e^{2\pi i(1+1/3)}=e^{2\pi i\cdot 4/3}=e^{4\pi i/3}$, the expression becomes: $$\frac{e^{4\pi i/3}}{(e^{2\pi i/3}-e^{4\pi i/3})(e^{2\pi i/3}-1)}$$

Evaluate the differences

Finally simplify the different terms: $e^{2\pi i/3}-e^{4\pi i/3}=e^{2\pi i/3}(1-e^{2\pi i/3})$ and $e^{2\pi i/3}-1$

Use the results

From steps 6 and 7 the residue is:$$-\frac{1}{3}{e}^{4\pi i/3}$$

Key concepts

complex function

A complex function is a function that maps complex numbers to complex numbers. In essence, it takes an input from the complex plane and produces an output in the complex plane. For example, consider the function given in our exercise: $$f(z)=\frac{e^{2\pi iz}}{1-z^3}$$This function involves both the exponential function and polynomial terms, showing how complex functions can utilize familiar operations but in the complex domain.Understanding complex functions involves a few key aspects:

• **Domain and Range:** The domain is the set of complex numbers for which the function is defined, while the range is the set of complex numbers that the function can output.
• **Analyticity:** A function is analytic at a point if it can be expressed as a convergent power series in some neighborhood of that point.
• **Singularities:** Points at which a function is not analytic are called singularities. In our case, the denominator becomes zero, causing singularities where the function is undefined.

residue theorem

The residue theorem is a powerful tool in complex analysis used for evaluating complex integrals. It states that the integral of a function around a closed contour can be computed by summing the residues of the function inside that contour. For our exercise, understanding the residue at a specific point is crucial.Here's a simple breakdown:

• **Residue:** The residue of a function at a point is a special coefficient that gives us important information about the behavior of the function near that point.
• **Formula for Simple Poles:** If we have a simple pole at a point $z_0$, we can calculate the residue using: $$\text{Res}(f,z_0)={\text{lim}}_{z\to z_0}(z-z_0)f(z)$$
• **Application:** In our exercise, we found the residue of the function $f(z)=\frac{e^{2\pi iz}}{1-z^3}$ at the point $z_0=e^{2\pi i/3}$, which helped simplify the integral calculations.

This theorem is essential because it transforms intricate complex integrals into manageable sums.

poles

Poles are a type of singularity where a function goes to infinity. They are crucial for understanding the behavior of complex functions. In our exercise, we examine the pole at $z=e^{2\pi i/3}$.There are different types of poles:

• **Simple Poles:** These are poles of order 1, meaning the function has a singular behavior that can be handled with the residue formula mentioned earlier.
• **Higher-Order Poles:** These occur when the singularity involves higher powers, making the residue calculation more complex.

For example, in the given function: $$\frac{e^{2\pi iz}}{1-z^3}$$The term $1-z^3$ has zeros at the cube roots of unity, which are $1,e^{2\pi i/3},$ and $e^{-2\pi i/3}$. Each of these points is a simple pole.
To find the residue at one of these points, like $z=e^{2\pi i/3}$, we use the residue formula and identify how the function behaves around this singularity. This process is pivotal in complex analysis and helps us understand the local and global behavior of the function.