- #1
nicolas.ard
- 4
- 0
Hello folks,
I have this function, un complex numbers
[itex]\frac{1}{(1+z^2)}[/itex]
I know that the Taylor serie of that function is
[itex]\frac{1}{(1+z^2)}[/itex] = [itex]\sum (-1)^k.z^(2.k) [/itex]
I have this function, un complex numbers
[itex]\frac{1}{(1+z^2)}[/itex]
I know that the Taylor serie of that function is
[itex]\frac{1}{(1+z^2)}[/itex] = [itex]\sum (-1)^k.z^(2.k) [/itex]