- #1
sinkersub
- 3
- 0
Problem A now solved!
Problem B:
I am working with two equations:
The first gives me the coefficients for the Laurent Series expansion of a complex function, which is:
f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n
This first equation for the coefficients is:
a_n = \frac{1}{2πi} \oint \frac{f(z)}{(z-z_0)^{n+1}} \, dz
The second equation is Cauchy's Integral formula for the nth derivative of a complex function:
f^{(n)}(z_0) = \frac{n!}{2πi} \oint \frac{f(z)}{(z-z_0)^{n+1}} \, dz
My problem is as follows:
We can clearly re-arrange these two equations to get the following expression:
a_n = \frac{f^{(n)}(z_0)}{n!}
Now, in this instance, f(z) = \frac{z^3 + 2z^2 + 4}{(z-1)^3}.
If we try to calculate a_n for this function, it keeps getting sent to zero, due to the function being evaluated at z_0!
What am I missing? Where have I gone wrong in this derivation/problem?
Ultimately in this problem I'm trying to find the Laurent Series expansion of the function.
Any help/tips would be much appreciated!
sinkersub
Problem B:
I am working with two equations:
The first gives me the coefficients for the Laurent Series expansion of a complex function, which is:
f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n
This first equation for the coefficients is:
a_n = \frac{1}{2πi} \oint \frac{f(z)}{(z-z_0)^{n+1}} \, dz
The second equation is Cauchy's Integral formula for the nth derivative of a complex function:
f^{(n)}(z_0) = \frac{n!}{2πi} \oint \frac{f(z)}{(z-z_0)^{n+1}} \, dz
My problem is as follows:
We can clearly re-arrange these two equations to get the following expression:
a_n = \frac{f^{(n)}(z_0)}{n!}
Now, in this instance, f(z) = \frac{z^3 + 2z^2 + 4}{(z-1)^3}.
If we try to calculate a_n for this function, it keeps getting sent to zero, due to the function being evaluated at z_0!
What am I missing? Where have I gone wrong in this derivation/problem?
Ultimately in this problem I'm trying to find the Laurent Series expansion of the function.
Any help/tips would be much appreciated!
sinkersub
Last edited:
