1. ## complex analysis

Thanks

Edgar

1. For each of the following functions, find the order of the pole that the
function has at the point a and find the residue.
(a) z−3/[(z−1)(z−2)] , a = 2
(b) exp[iz]/z^6+1 , a = exp(ipie/6)
(c) z^2/z^2+1 , a = i
(d) 1/(z−1)^3(z−2) , a = 1

2. The order shouldn't be too hard and I assume you have a formula (limit?) to calculate the residue? What have you tried?

3. Originally Posted by edgar davids

Thanks

Edgar

1. For each of the following functions, find the order of the pole that the
function has at the point a and find the residue.
(a) z−3/[(z−1)(z−2)] , a = 2
$
\frac{z-3}{(z-1)(z-2)}=\frac{1}{z-2}\ \left[\frac{z-3}{z-1}\right]
$

The second term on the right is holomorphic in a neighbourhood of $z=2$ and the pole is captured in the first term, so its order is 1.

The residue in this case is:

$\lim_{z \to 2} (z-2)f(z)=\lim_{z \to 2}\frac{z-3}{z-1}=-1$

RonL

4. Originally Posted by edgar davids

Thanks

Edgar

1. For each of the following functions, find the order of the pole that the
function has at the point a and find the residue.
(a) z−3/[(z−1)(z−2)] , a = 2
(b) exp[iz]/z^6+1 , a = exp(ipi/6)
(c) z^2/z^2+1 , a = i
(d) 1/(z−1)^3(z−2) , a = 1
(By the way, $\pi$ is "pi" not "pie.")

Technically what we want to do is expand each function in a Laurent series and note what largest (negative) power of (z - a) appears at. However we can simply eyeball this, as all we need to do is count how many powers of (z - a) appear in the denominator. (This is assuming the function doesn't contain a function g(z) that has poles of its own.)

The residue is the coefficient of the $(z - a)^{-1}$ term in the Laurent expansion. If the pole z = a is simple (order 1) then the residue is $a_{-1} = [(z - a)f(z)]|_{z = a}$.

a) $\frac{z - 3}{(z - 1)(z - 2)}$
This expression has a pole of order 1 at z = 2.
The residue then may be calculated as
$\left [ (z - 2) \cdot \frac{z - 3}{(z - 1)(z - 2)} \right ] _{z = 2} = \frac{2 - 3}{2 - 1} = -1$

b) $\frac{e^{iz}}{z^6+1}$
Note that $e^{iz}$ contains no poles.
So the only poles are from $z^6 + 1 = \left (z - e^{i \pi/6} \right ) \left (z - e^{3i \pi/6} \right )$ $\left (z - e^{5i \pi/6} \right ) \left (z - e^{7i \pi/6} \right ) \left (z - e^{9i \pi/6} \right ) \left (z - e^{11i \pi/6} \right )$

As $\left (z - e^{i \pi/6} \right )$ only appears once in the denominator, this is a pole of order 1.

So the residue will be:
$\left [ \left ( z - e^{i \pi/6} \right ) \frac{e^{iz}}{z^6+1} \right ] _{z = e^{i \pi/6}}$
I am having a problem calculating this for some reason. I'll get back to you on it once I figure out the problem.

c) $\frac{z^2}{z^2+1} = \frac{z^2}{(z+i)(z-i)}$
Again we see that z = i is a pole of order 1.

So the residue will be:
$\left [ (z - i) \frac{z^2}{(z+i)(z-i)} \right ] _{z = i} = \frac{-1}{2i}$.

-Dan

5. d) $\frac{1}{(z-1)^3(z-2)}$
By inspection this has a pole at z = 1 of order 3.

There might be a simpler way to figure this out (my Residues class was several years ago), but we can always write this out as:
$\frac{1}{(z-1)^3(z-2)} = \frac{A}{z - 2} + \frac{B}{z - 1} + \frac{C}{(z - 1)^2} + \frac{D}{(z - 1)^3}$

The residue will then be B. So:
$\frac{1}{(z-1)^3(z-2)} = \frac{1}{z - 2} + \frac{-1}{z - 1} + \frac{-1}{(z - 1)^2} + \frac{-1}{(z - 1)^3}$

So the residue is -1.

-Dan

6. Notice that the residue of c can be simplified to i/2: -1/(2i) = -i/(2i²) = i/2.

For b, normally you'd cancel the factor of the pole, but you can also apply l'Hôpital's rule.