# Taylor series expansion of $\frac{1}{1+j-z^2}$

I know to manipulate $$\frac{1}{1+j-z^2}$$ to get a result in the form of $$\frac{1}{1-z}$$. However, I have only managed to get to $$\frac{1}{(-z+1)(-z-1)+j}$$. I'm not sure how to manipulate this further

• Try the substitution $u = z^2/(1+j)$ – Damien Jan 11 at 11:59
• thank you, the substitution worked – nastyapples Jan 11 at 14:25
• You welcome, nice having helped you – Damien Jan 11 at 14:41