# The limit of $z\cdot\overline{z}$ as $z\to i$

How can I compute this limit: $$\lim_{z\rightarrow i} (z\cdot \overline{z})$$ (where $\overline{z}$ is the conjugate of $z$)?

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Please make your titles informative, and please make the body of your post self-contained. In particular, since you are asking a question, the body of the message should actually contain a question. – Arturo Magidin Feb 16 '11 at 3:52
Sory, I'll try that next time. – Tomas Sironi Feb 16 '11 at 3:54

As $\mathbb{C}$ comes nicely equipped with a metric, limits behave as nice as ever. Therefore, the limit of the product is the product of the limits. And the limit of the conjugate is the conjugate of the limit (this can be proved using a standard epsilon-delta argument). Hence, in your example: $$\lim_{z \to i} (z \cdot \overline{z})=\lim_{z \to i}(z) \lim_{z \to i}(\overline{z})= i \cdot(-i) = 1.$$
Hint: Is $z'$ the conjugate of $z$ and the lower dot a multiplication? If so, what is the problem? You might try expressing $z=a+bi$, with $a,b$ real and trying a real limit.
I am used to $\overline{z}$ but as people are used to different things it is worth defining. – Ross Millikan Feb 16 '11 at 3:59
I'm also used to that, but again, I didn't remember what's the command therefore I used $'$. – Tomas Sironi Feb 16 '11 at 4:01
\overline, just got it! – Tomas Sironi Feb 16 '11 at 4:02