Let \( z = re^{i\theta} \) where \( r = |z| \) and \( \theta = \arg(z) \). Then, the conjugate \( \bar{z} \) is given by \( \bar{z} = re^{-i\theta} \). Thus, we have:
\[
f(z) = \frac{1}{\bar{z}} = \frac{1}{re^{-i\theta}} = \frac{e^{i\theta}}{r}
\]
This shows that
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