Understanding Simple Poles and Residues in Function 1/(z^4 + 1)

[philip041]

I have a function 1/(z^4 + 1)

I know the poles are pi/4, 3pi/4, 5pi/4, 7pi/4

It says they are simple poles, I thought I understood why, but now I am totally confused. How does my lecturer just know that they are simple?

A simple pole is a pole of order 1, but I thought this meant you look at the series and get the order from the highest indice of z on the denominator?

Help!

Cheers

 

[Count Iblis]

The numerator has simple zeroes, so the function is pof the form:

1/[(z-a)(z-b)(z-c)(z-d)]

where a, b, c, and d are different. If you expand this around, say, z = a, you can proceed as follows. You can write the function as:

1/(z-a) * 1/[(z-b)(z-c)(z-d)]

The factor 1/[(z-b)(z-c)(z-d)] is not singular at z = a, so it has a regular Taylor expansion. So, the Laurent expansion of the function around z = a is given by the Taylor expansion of that factor times 1/(z-a).

 

[philip041]

What is a simple zero?

 

[HallsofIvy]

I have a function 1/(z^4 + 1)

I know the poles are pi/4, 3pi/4, 5pi/4, 7pi/4

Then you know wrong. The poles are (\sqrt{2}/2)(1+ i), -(\sqrt{2}/2)(1- i), (-\sqrt{2}/2)(1+ i), and (\sqrt{2}/2)(1- i), which can be written e^{i\pi/4}, e^{3i\pi/4}, e^{5i\pi/4}, and e^{7i\pi/4}.

It says they are simple poles, I thought I understood why, but now I am totally confused. How does my lecturer just know that they are simple?

A simple pole is a pole of order 1, but I thought this meant you look at the series and get the order from the highest indice of z on the denominator?

Help!

Cheers

\frac{1}{z^4+ 1}= \frac{1}{z- (\sqrt{2}/2)(1+ i)}\frac{1}{z+ (\sqrt{2}/2)(1-i)}\frac{1}{z+(\sqrt{2}/2)(1+ i)}\frac{1}{z-(\sqrt{2}/2)(1-i)}
Each of those numbers is a "simple pole" or "pole of order 1" because each has a power of 1 in the denominator.

 

[philip041]

Cheers, this helps!