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

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

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