Residue Calculation Of (z-2)/(cos(πz)-1) At Z=0

Hey guys! Let's dive into the fascinating world of complex analysis and tackle a classic problem: calculating the residue of a complex function. Today, we're going to figure out the residue of the function h(z) = (z-2) / (cos(πz) - 1) at the point z = 0. This is a super important concept in complex analysis, especially when you're dealing with integrals and series. Residue calculus, in particular, provides powerful tools for evaluating contour integrals, which show up all over the place in physics and engineering. So, buckle up, and let's get started!

Identifying the Singularity

First things first, we need to understand what's happening at z = 0. Singularities are points where a function goes a bit wild, usually by becoming undefined. To kick things off, we need to figure out what kind of singularity we're dealing with at z = 0. A singularity occurs when the denominator of our function is zero, or when some other funky behavior arises. For the function h(z) = (z - 2) / (cos(πz) - 1), the denominator is cos(πz) - 1. Setting this equal to zero gives us:

cos(πz) - 1 = 0 cos(πz) = 1

This equation is satisfied when πz = 2πn, where n is an integer. Thus, z = 2n. This means that our function has singularities at z = 0, ±2, ±4, and so on. Now, let's hone in on z = 0. To determine the order of the pole at z = 0, we need to investigate how many times we can factor out (z - 0) from the denominator. We know that cos(x) has a Taylor series expansion around x = 0 given by:

cos(x) = 1 - (x^2 / 2!) + (x^4 / 4!) - (x^6 / 6!) + ...

So, cos(πz) will be:

cos(πz) = 1 - ((πz)^2 / 2!) + ((πz)^4 / 4!) - ...

Now, plug this into our denominator:

cos(πz) - 1 = (1 - ((πz)^2 / 2!) + ((πz)^4 / 4!) - ...) - 1 cos(πz) - 1 = - (π^2 z^2 / 2) + (π^4 z^4 / 24) - ...

We can factor out a z^2 term from this expression:

cos(πz) - 1 = z^2 [-(π^2 / 2) + (π^4 z^2 / 24) - ...]

Since we factored out z^2, this tells us that z = 0 is a pole of order 2. Understanding the order of the pole is critical because it determines the method we'll use to compute the residue. A pole of order 2 requires a slightly more complex approach than a simple pole (order 1).

Calculating the Residue: Method for a Pole of Order 2

Alright, now that we've established that z = 0 is a pole of order 2 for our function, we need to pull out the big guns to calculate the residue. For a pole of order n at a point z = a, the residue can be found using the following formula:

Res(h, a) = (1 / (n - 1)!) * lim (z→a) d^(n-1) / dz^(n-1) [(z - a)^n * h(z)]

In our case, n = 2 and a = 0, so the formula simplifies to:

Res(h, 0) = (1 / (2 - 1)!) * lim (z→0) d / dz [z^2 * h(z)] Res(h, 0) = lim (z→0) d / dz [z^2 * (z - 2) / (cos(πz) - 1)]

Let's break this down step by step. First, we need to compute z^2 * h(z):

z^2 * h(z) = z^2 * (z - 2) / (cos(πz) - 1)

We already know from our earlier analysis that cos(πz) - 1 = - (π^2 z^2 / 2) + (π^4 z^4 / 24) - .... So, we have:

z^2 * h(z) = z^2 * (z - 2) / (- (π^2 z^2 / 2) + (π^4 z^4 / 24) - ...)

We can simplify this by dividing both the numerator and the denominator by z^2:

z^2 * h(z) = (z - 2) / (- (π^2 / 2) + (π^4 z^2 / 24) - ...)

Now, we need to take the derivative of this expression with respect to z. Let's call this expression g(z):

g(z) = (z - 2) / (- (π^2 / 2) + (π^4 z^2 / 24) - ...)

To find g'(z), we'll use the quotient rule, which states that if g(z) = u(z) / v(z), then g'(z) = (u'(z)v(z) - u(z)v'(z)) / (v(z))^2. Here, u(z) = z - 2 and v(z) = - (π^2 / 2) + (π^4 z^2 / 24) - ....

First, let's find the derivatives of u(z) and v(z):

u'(z) = 1 v'(z) = (π^4 z / 12) - ...

Now, applying the quotient rule:

g'(z) = [1 * (- (π^2 / 2) + (π^4 z^2 / 24) - ... ) - (z - 2) * ((π^4 z / 12) - ...)] / (- (π^2 / 2) + (π^4 z^2 / 24) - ...)^2

Evaluating the Limit

Next, we need to evaluate the limit as z approaches 0:

Res(h, 0) = lim (z→0) g'(z)

As z approaches 0, the higher-order terms in z will vanish. So, we can focus on the leading terms:

Res(h, 0) = lim (z→0) [(- π^2 / 2) - (z - 2) * (π^4 z / 12)] / (- π^2 / 2 + (π^4 z^2 / 24))^2

As z → 0, the expression simplifies to:

Res(h, 0) = (- π^2 / 2) / (- π^2 / 2)^2 Res(h, 0) = (- π^2 / 2) / (π^4 / 4) Res(h, 0) = (- π^2 / 2) * (4 / π^4) Res(h, 0) = -2 / π^2

So, after all that math, we've found that the residue of h(z) at z = 0 is -2 / π^2. Awesome!

Alternative Approach: Using L'Hôpital's Rule

Now, just to make sure we've really nailed this, let's explore an alternative method for calculating the residue. This time, we'll use L'Hôpital's Rule, which can be super handy when dealing with limits of indeterminate forms like 0/0.

Recall that we need to find:

Res(h, 0) = lim (z→0) d / dz [z^2 * (z - 2) / (cos(πz) - 1)]

Let’s rewrite the expression inside the derivative as:

g(z) = z^2 * (z - 2) / (cos(πz) - 1)

We need to find lim (z→0) g'(z). Before we differentiate, let’s rewrite g(z) to make our lives a bit easier. We know that:

cos(πz) - 1 = -2sin^2(πz/2)

So,

g(z) = z^2 * (z - 2) / (-2sin^2(πz/2))

Now, let's rewrite this as:

g(z) = (z - 2) / (-2(sin(πz/2)/z)^2)

Now, let’s multiply the numerator and denominator by z^2:

g(z) = (z^2(z - 2)) / (-2sin^2(πz/2))

When z approaches 0, we have a 0/0 indeterminate form. So, we can apply L'Hôpital's Rule. But before we do that, let's simplify our function a bit more by recognizing a key limit:

lim (x→0) sin(x) / x = 1

So, as z → 0, we have sin(πz/2) / (πz/2) → 1. Therefore:

sin(πz/2) ≈ πz/2

Plugging this approximation into our expression for g(z):

g(z) ≈ (z^2(z - 2)) / (-2(πz/2)^2) g(z) ≈ (z^2(z - 2)) / (-2(π^2z2/4)) g(z) ≈ (z - 2) / (-π^2/2)

Now, let's differentiate g(z):

g'(z) = d/dz [(z - 2) / (-π^2/2)] g'(z) = 1 / (-π^2/2) g'(z) = -2/π^2

Taking the limit as z → 0, we get:

Res(h, 0) = lim (z→0) g'(z) = -2/π^2

Boom! We arrived at the same answer using a different method. This confirms our result and gives us even more confidence in our calculation.

Conclusion

So there you have it, folks! We successfully calculated the residue of the function h(z) = (z - 2) / (cos(πz) - 1) at z = 0 using both the formula for poles of order 2 and an alternative approach using L'Hôpital's Rule. The residue is -2 / π^2. Understanding residues is crucial in complex analysis, as they're the key to solving many complex integrals and series problems. I hope this breakdown helped you grasp the concept and the techniques involved. Keep practicing, and you'll become a residue-calculating pro in no time! Happy math-ing!