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I would like to make sure I'm doing everything right and not missing anything, since I know that some familiar functions do crazy things in the complex setting.

Since $|z|^2 = \overline{z}z$ I simplified my function to $\frac{1}{z}$.

Then if I take the limit as $z$ tends to infinity I get zero. So the limit of $f(z)$ is also zero. Does this seem legit? Also, would anything be different if the numerator of $f(z)$ was simply $z$?

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1 Answer

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Your evaluation of the limit is correct. It may help you to think of the modulus: as long as $|f(z)|$ tends to zero as $|z|$ tends to infinity, $f(z)$ tends to zero as well.

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