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|(\bar{z})^n| = |z|^n,
|(\bar{z})^n| = |z|^n,
\arg(z^n) = n \arg(z)</math>
\arg(z^n) = n \arg(z)</math>
<math>f(x) =
  \begin{cases}
    1 & -1 \le x < 0 \\
    \frac{1}{2} & x = 0 \\
    1 - x^2 & \text{otherwise}
  \end{cases}</math>

Revision as of 23:03, 17 August 2022